1
ANALYSIS
Department of Quantitative Economics
School of Business and Economics
Maastricht University
Coordinator: Dr. Anna Zseleva
September 2021
CONTENTS
1 PRELIMINARIES
1 Sets
1
2 Subsets of the set of real numbers
4
3 Finite sums and finite products
9
4 Absolute value
11
5 Functions
17
5.1 Some definitions
17
5.2 Visualizing functions
18
6 Building new functions
19
7 Some families of functions
22
7.1 Polynomial functions
22
7.2 Trigonometric functions
23
7.3 Exponential and logarithmic functions
25
8 The inverse function
27
2 THE PRINCIPLE OF INDUCTION
1 Statements about natural numbers
34
2 Mathematical Induction
35
3 Some notation
40
3 (INFINITE) SEQUENCES
1 Sequences
46
2 Recursively defined sequences
51
3 The limit of a sequence
54
4 Properties of convergent sequences
62
5 Elementary arithmetic rules for limits of sequences
65
6 Other arithmetic rules for limits of sequences
68
4 BOUNDED SEQUENCES
1 Monotone sequences
73
2 The number e
77
3 Subsequences
78
4 The Theorem of Bolzano-Weierstrass
80
5 CONTINUITY OF A FUNCTION
1 Continuity of a function, a formal definition
85
2 Arithmetic rules for continuous functions
90
3 Two other tools to prove continuity
92
4 Continuity of a function, an alternative definition
96
Appendix
101
6 CONTINUOUS FUNCTIONS
1 Properties of continuous functions on a compact interval
103
2 Continuity of the inverse function
109
7 LIMITS OF FUNCTIONS
1 Limits of functions
113
2 One-sided limits
117
3 Discontinuities
118
4 Behavior of a function at infinity
119
8 DERIVATIVES
1 Differentiability
125
2 Higher-order derivatives
131
3 Arithmetic rules for differentiable functions
132
4 The derivative of the inverse function
137
5 Some economical concepts based on the derivative
141
6 Standard derivatives
144
9 SIGNIFICANCE OF THE DERIVATIVE
1 Extreme values
149
2 The Mean Value Theorem
151
3 Monotone functions
155
4 Convex and concave functions
159
5 Taylor’s Theorem
162
6 De l’Hôpital’s Rule
166
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Preliminaries
PRELIMINARIES
SETS
Sets play an important role in mathematics. The set IR of real numbers and the set IN
consisting of the natural numbers 1, 2, 3, 4, . . . are well-known examples of a set. But also
the letters in the alphabet and the first year students of econometrics in Maastricht at
September 3, 2021 form a set. So the members of a set – these members are called the
elements of the set – are not necessarily numbers. They may even be just ideas like in the
set of virtues.
A set is usually denoted by a capital letter, its elements by small letters.
DEFINITION
If S is a set and s an element of S, then we write this as s ∈ S (we also say:
s belongs to S or S contains the element s). If t is not an element of the set
S, then we write t ∈
/ S (in words: t does not belong to S).
EXAMPLE 1
/ IN.
Obviously, 5 ∈ IN, whereas 2 21 ∈
There are various ways to define a set. For instance, the set E of even natural numbers can
be defined by
– listing its elements (and to enclose the listed elements in curly brackets):
E = {2, 4, 6, 8, 10, . . .}
(in words: E is the set (consisting) of the numbers 2, 4, 6, 8, 10 and so on)
– using a property which selects the elements of the set:
E = {n ∈ IN| n is divisible by 2}
(in words: E is the set of all natural numbers which are divisible by 2)
– using a formula which produces the set:
E = {2n| n ∈ IN}
(in words: E is the set of all numbers of the form 2n where n is a natural number).
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2
DEFINITION
Preliminaries
Let S and T be two sets.
We say that S is a subset of T (or that T includes S) if every element of S
subsets
is an element of T ; we write S ⊂ T or T ⊃ S. Note that S ⊂ S.
We say that S and T are equal, written S = T , if S ⊂ T and T ⊂ S.
equal sets
EXAMPLE 2
EXERCISE 1
Obviously, {1, 2, 3} ⊂ IN and {x ∈ IN| x2 = 4} = {2}.
Let
A = {1, 2, 3, 4, 5},
B = {x| x = 2n for some n ∈ IN}
and
C = {x ∈ IN| x < 6}.
Which of the following statements are true?
(a) {4, 3, 2} ⊂ A
(b) 3 ∈ B
(c) A ⊂ C
(d) {2} ∈ A
(e) C ⊂ B
(f) {2, 4, 6, 8} ⊂ B
(g) C ⊂ A
(h) A = C.
There are three important ways of combining given sets to produce new ones. One can take
the union and the intersection of two sets and one can determine one set minus another
one. Intuitively, ’union’ may be thought of as putting together, ’intersection’ is like cutting
down, and ’minus’ corresponds to throwing out. The precise definitions are given below.
DEFINITION
union
Let S and T be two sets;
– the union of the sets S and T , denoted as S ∪ T , consists of all elements
which belong to at least one of the two sets S and T :
S ∪ T = {x| x ∈ S or x ∈ T },
[in mathematics ’or’ is always used in the sense ’one or the other, or both’,
which is equivalent with ’at least one of the two’]
intersection
– the intersection of the sets S and T , denoted as S ∩ T , consists of the
elements common to both sets:
S ∩ T = {x| x ∈ S and x ∈ T },
minus
– the set S minus the set T , denoted as S \ T , consists of the elements of S
not contained in T :
S \ T = {x| x ∈ S and x ∈
/ T }.
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Preliminaries
If S = {1, 3, 5, 7} and T = {3, 5, 9}, then S∪T = {1, 3, 5, 7, 9}, S∩T = {3, 5}
EXAMPLE 3
and S \ T = {1, 7}.
EXERCISE 2
Let A = {1, 5, 7, 10} and B = {2, 7, 11, 14}. Determine A ∪ B, A ∩ B, A \ B,
B \ A, A \ B ∪ B and A \ B ∪ A ∩ B .
A helpful way to visualize the foregoing set operations is by use of Venn diagrams. If we
portray the sets S and T as the sets of points inside two circles, then the sets S ∪ T , S ∩ T
and S \ T are the shaded areas represented below.
S
S
Venn diagrams
S
T
T
S ∪T
FIGURE 1
S∩T
T
S\T
Venn diagrams
While Venn diagrams (and other diagrams as well) are useful in understanding the relationship between sets, and may be helpful in getting ideas for developing a proof, they should
not be viewed as proofs themselves. A diagram necessarily represents only one case and it
may not be obvious whether there may be other cases as well. For instance, the set S ∩ T
can also be represented by two other diagrams: one where S is a subset of T and one where
the sets S and T have no elements in common. Can you find two other ways?
empty set
DEFINITION
The empty set is the set which contains no elements. It is denoted by φ.
We say that two sets S and T are disjoint if S ∩ T = φ, that is: the two sets
disjoint sets
have no elements in common.
EXAMPLE 4
The sets {1, 2, 3} and {52} are disjoint.
The set operations we introduced before satisfy all kinds of rules. In the following example
one such rule will be discussed. Another one is left as an exercise.
EXAMPLE 5
Let A, B and C be three sets. Then
A \ (B ∪ C) = (A \ B) ∩ (A \ C).
In order to prove this relation, we have to show that (a) A \ (B ∪ C) ⊂ (A \ B) ∩ (A \ C)
and that (b) (A \ B) ∩ (A \ C) ⊂ A \ (B ∪ C).
(a) We show the first inclusion. So let x ∈ A \ (B ∪ C). Then x ∈ A and x ∈
/ B ∪ C.
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Preliminaries
However if x ∈
/ B ∪ C, then x ∈
/ B and x ∈
/ C. Thus x ∈ A \ B and x ∈ A \ C. But this
implies that x ∈ (A \ B) ∩ (A \ C).
(b) We show the second inclusion. So let x ∈ (A \ B) ∩ (A \ C). Then x ∈ A \ B and
x ∈ A \ C. However if x ∈ A \ B, then x ∈ A and x ∈
/ B. Similarly, the inclusion
x ∈ A \ C implies that x ∈ A and x ∈
/ C. Since x ∈
/ B and x ∈
/ C, x ∈
/ B ∪ C. Hence,
x ∈ A \ (B ∪ C).
EXERCISE 3
Let A and B be two sets. Prove that A \ B ∪ A ∩ B) = A.
Up to this point we discussed combinations of two or three sets. However one can also
consider unions and intersections of any finite collection of sets.
DEFINITION
If S1 , S2 , . . . are sets and n ∈ IN, then
n
S
Si consists of all elements which belong to at least one of the sets
•
i=1
S1 , . . . , Sn :
n
[
i=1
•
n
T
Si = {x| x ∈ Sj for some j},
Si consists of the elements common to all the sets S1 , . . . , Sn :
i=1
n
\
i=1
EXERCISE 4
Si = {x| x ∈ Sj for all j}.
For two sets S and T the symmetric difference S △ T is defined as
S △ T = (S \ T ) ∪ (T \ S).
(a) Draw a Venn diagram for S △ T .
(b) What is S △ S?
(c) What is S △ φ?
2
SUBSETS OF THE SET OF REAL NUMBERS
It is hardly possible to overestimate the importance of the real number system both to pure
and applied mathematics. Anywhere measurements appear, we are certain to find the real
numbers in action. Furthermore all the truly important results in calculus involve the real
number system in an essential way.
natural numbers
The simplest real numbers are the ’counting’ numbers 1, 2, 3, 4, . . . which are also known as
the positive integers or the natural numbers. As observed before, IN will be used to denote
the set of all natural numbers.
integers
The set
{. . . , −3, −2, −1, 0, 1, 2, 3, . . .},
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Preliminaries
consisting of the natural numbers, their negatives and zero, will be denoted by Z; its elements
are called integers.
rational numbers
The next most important class of real numbers is the rationals. By definition, a rational
number is the ratio of two integers. Thus
1
2,
0.1 and 0.3333 . . . =
1
3
are rational numbers.
We shall use Q to denote the set of rational numbers:
nm
o
Q=
m, n ∈ Z and n 6= 0 .
n
Rational numbers first arose in the measurement of line segments. In fact, the Greek geometers thought that all numbers were rational. The discovery of the Pythagorean Theorem
shattered this belief and caused the Greek mathematicians to revise their thinking about
numbers. One particular consequence of the Pythagorean Theorem is the fact that an isosce√
les right triangle whose legs are of unit length has a hypothenuse of length 2. The Greek
found themselves facing a real problem when it was proved that no rational number could
exist whose square was two.
In thinking about the real numbers, it is often helpful to regard them as points on a straight
line. This geometric intuition allows us to interpret statements about numbers in terms of
this picture. Sometimes it may even suggest a method to prove some statement.
In order to identify points and numbers we suppose that we are given a straight line. First
origin
we select a point O on the line, to be called the origin. This point divides the line into two
half-lines, one of which we designate as positive, the other negative. It is conventional to
designate the right half-line as positive:
−
+
O
Next we choose a unit of length. Once these choices have been made, we can describe a
correspondence between points on the line and real numbers:
– we assign to an x > 0 that point on the positive half-line whose distance from O is x
– we assign to an x < 0 that point on the negative half-line whose distance from O is −x
– the number 0 is matched with the origin.
In this way we are able to match numbers with points:
the real line
−1
0
1
1 12
One can prove that this correspondence between points on the line and real numbers is
one-to-one, that is: there corresponds to each point on the line exactly one number, and
conversely, each number belongs to just one point.
Inequalities are easily interpreted geometrically: a < b if and only if the point corresponding
to a lies to the left of the point corresponding to b.
Although arbitrary subsets of the real numbers are seldom the focus of the study in analysis
it is important to imagine their variation. For instance it might be interesting to know for
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6
which numbers x strictly between 0 and 1 it holds that sin
then looking for the subset of real numbers defined by
{x ∈ IR| 0 < x < 1 and sin
π
π
x
x
Preliminaries
> 0. Mathematically we are
> 0}.
EXERCISE 5
Try to make a picture of this latter set. What happens if x is close to zero?
EXERCISE 6
Strictly speaking prices in our economy can only take specific values. You
cannot pay half a Euro cent for instance. Let the real number one indicate
one Euro. According to this, draw the set of real numbers that correspond to
a price.
In case arbitrary subsets are subject of our study, you should keep in mind that fantasy has
almost no limitation. This might become clear if one tries to image the following set
{x ∈ IR| 0 < x < 1 and the decimal description of x does not contain the number 5}.
So, neither
1
2
nor
1
7
is element of this set but
1
3
and
5
12
are.
In the study of calculus we are especially interested in those subsets of the real line which,
intuitively, have no gaps or breaks. The sets
{x ∈ IR| 0 ≤ x ≤ 1}
0
1
2
{x ∈ IR| 0 ≤ x < 1}
0
1
2
and
interval
are examples of such a set. These connected sets are called intervals. There are eight
different kinds of intervals.
If a and b are real numbers with a < b, then
(a, b) =
[a, b] =
bounded interval
(a, b]
=
[a, b)
=
{x ∈ IR| a < x < b}
{x ∈ IR| a ≤ x ≤ b}
is the open interval from a to b
is the closed interval from a to b
{x ∈ IR| a ≤ x < b}
is a half-open or half-closed interval from a to b.
{x ∈ IR| a < x ≤ b}
is a half-open or half-closed interval from a to b
These bounded intervals, thought of as points on a line, are simply line segments, with or
without endpoints. The length of all these bounded intervals equals b − a.
unbounded interval
The unbounded intervals consist of half-lines, with or without endpoints:
(a, ∞)
=
[a, ∞)
(−∞, a]
=
=
(−∞, a) =
{x ∈ IR| a < x}
{x ∈ IR| x < a}
{x ∈ IR| a ≤ x}
{x ∈ IR| x ≤ a}
the unbounded open interval to the right of a
the unbounded open interval to the left of a
the unbounded closed interval to the right of a
the unbounded closed interval to the left of a.
Note that the symbols ∞ and −∞, usually called ’infinity’ and ’minus infinity’, do not
denote some numbers. In particular, it is not true that ∞ is some number satisfying ∞ ≥ x
for all numbers x. Sometimes the set IR is considered as the interval (−∞, ∞).
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Preliminaries
After reading the following definition it will be clear why intervals were called bounded or
unbounded.
upper bound
DEFINITION
Let S be a nonempty subset of IR.
We say that a real number u is an upper bound of S if s ≤ u for every s ∈ S.
The set S is called bounded above if it has an upper bound.
We say that a real number ℓ is a lower bound of S if s ≥ ℓ for every s ∈ S.
lower bound
The set S is called bounded below if it has a lower bound.
The set S is called bounded if it is both bounded above and bounded below.
(un)bounded set
Otherwise it is called unbounded.
An unbounded subset of IR may or may not be bounded above or bounded below. For
instance the set of all real numbers has neither an upper bound nor a lower bound. The set
IN of natural numbers however – although it is unbounded – has a lower bound e.g. 0, −1,
or −eπ . Actually the number one is the largest lower bound.
EXAMPLE 6
The sets S = [0, 1] and T = [0, 1) are bounded above. The numbers 1, 2
and 13 are examples of upper bounds of both sets. Since the sets are also bounded below
(for example by 0), the sets S and T are bounded.
The sets U = {2n| n ∈ IN} and V = (0, ∞) are unbounded.
EXERCISE 7
There is a very useful way of describing the points in a closed interval [a, b].
(a) First consider the interval [0, b] for b > 0. Prove that if x is in [0, b], then
x = tb, for some t with 0 ≤ t ≤ 1. What is the mid-point of the interval
[0, b]?
(b) Now prove that if x is in [a, b], then x = (1 − t)a + tb, for some t with
0 ≤ t ≤ 1.
[Clue: the expression for x can also be written as a + t(b − a).]
(c) Prove, conversely, that (1 − t)a + tb is in [a, b] if 0 ≤ t ≤ 1.
(d) Describe the points in the open interval (a, b) in a similar way.
coordinate system
Of even greater interest to us than the method of drawing numbers is a method of drawing
(ordered) pairs of numbers. Here ’ordered’ means that the order of the numbers is relevant:
the pair (1, 2) is different from the pair (2, 1). Note however that the set {1, 2} is not different
from the set {2, 1}.
Cartesian
This procedure requires a (Cartesian) ’coordinate system’: two straight lines intersecting
at a right angle. One of these straight lines is called the horizontal axis, whereas the other
one is called the vertical axis. The intersection of the two lines is called the origin and,
after choosing a unit of length (one for each line), each of the two axes is labelled with real
numbers. In general, the right and upper half-lines are designated as positive. Now the pair
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Preliminaries
(a, b) can be identified with the point in the plane determined by the two axes as in the
following figure.
y
(a, b)
b
a
FIGURE 2
x,y-plane
x
The x, y-plane or Cartesian plane
Usually, the plane in the foregoing figure is called the x, y-plane or Cartesian plane, a is
called the first or x-coordinate and b is called the second or y-coordinate of the point (a, b).
Once we are able to draw pairs of numbers, we are able to define (and draw) the ’product’
of two sets of real numbers.
Cartesian product
DEFINITION
If I and J are sets, then the Cartesian product of I and J, written I × J, is
the set of all ordered pairs (x, y) such that x ∈ I and y ∈ J. In symbols:
I × J = {(x, y)| x ∈ I and y ∈ J}.
If I and J are intervals, then using the familiar Cartesian coordinate system with I on the
horizontal axis and J on the vertical axis, I × J is represented by a rectangle. For example,
if I = [1, 3] and J = [1, 2], then I × J is the rectangle shown in the following figure.
y
2
I ×J
J
1
1
I
3
x
FIGURE 3
The Cartesian product of the intervals I and J
EXERCISE 8
Draw, in the x, y-plane, the set of all points (x, y) satisfying the following
conditions.
(a) x > y
(b) y < x2
(c) x + y is an integer
(d) (x − 1)2 + (y − 2)2 < 1
(e) x4 < y < x2
(g) 4x2 + 9y 2 < 36
(f) x2 = y 2
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Preliminaries
FINITE SUMS AND FINITE PRODUCTS
In this section we will discuss some convenient and frequently used notations.
finite sum
The following definition involves a notation for a finite sum. Instead of writing
t1 + t2 + · · · + tn ,
for the sum of the numbers t1 , t2 , . . . , tn , we will employ the Greek capital letter
for ’sum’) and write
n
X
X
(sigma,
ti .
i=1
In other words:
Thus
Pn
i=1 ti
denotes the sum of the numbers obtained by letting i = 1, 2, . . . , n.
n
X
i=1
summation index
i = 1 + 2 + · · · + n.
Note that the letter i – which is called the summation index – can be replaced by any other
letter (except n, of course). So
n
X
ti =
i=1
n
X
tk =
n+4
X
tj−4 .
j=5
k=1
In practice, all sorts of modifications of this symbolism are used.
EXAMPLE 7
The expression
8
X
ti ,
i=1
i6=2
for example, is an obvious way of writing
t1 + t3 + t4 + t5 + t6 + t7 + t8 ,
or
t1 +
8
X
ti .
i=3
The following rules are helpful when manipulating sums. Instead of giving a formal proof,
we will give a more intuitive sketch of the proof.
THEOREM 1
Let n be a natural number. Then
n
X
s k + tk ) =
k=1
and
n
X
sk +
k=1
n
X
k=1
c tk = c
n
X
k=1
n
X
k=1
tk .
tk
10
PROOF
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Preliminaries
n
X
tk .
Rearranging the terms of the sum leads to
n
X
k=1
sk + tk ) = (s1 + t1 ) + (s2 + t2 ) + · · · + (sn + tn )
= (s1 + s2 + · · · + sn ) + (t1 + t2 + · · · + tn ) =
n
X
sk +
k=1
k=1
Finally,
n
X
k=1
EXERCISE 9
n
X
tk .
c tk = c t1 + c t2 + · · · + c tn = c (t1 + t2 + · · · + tn ) = c
k=1
Prove that
n
X
1
1
=1−
i(i − 1)
n
i=2
by using the identity
1
1
1
=
− .
i(i − 1)
i−1
i
Often one has to combine several summation signs. Such a situation will be considered in
the next example.
EXAMPLE 8
(DOUBLE SUMS)
Consider the following rectangular array of numbers:
a11
a12
a21
..
.
a22
..
.
am1
am2
···
a1n
···
..
.
a2n
..
.
· · · amn
Let us determine the sum of all the mn numbers in the array by first finding the sum of the
numbers in each of the m rows and then adding all these row sums.
n
n
n
X
X
X
amj , respectively and the sum of these row
a2j , . . . ,
a1j ,
The m row sums are
n
X
a1j +
j=1
which can be written as
j=1
j=1
j=1
sums is
n
X
j=1
a2j + · · · +
n
m X
X
i=1
n
X
amj ,
j=1
aij .
j=1
If instead we add the numbers in each of the n columns first and then take the sum of these
column sums, we get
m
X
i=1
ai1 +
m
X
i=1
ai2 + · · · +
m
X
ain =
i=1
m
n X
X
j=1
i=1
aij .
Obviously, the two results are equal. So
m X
n
X
aij =
i=1 j=1
n X
m
X
aij ,
j=1 i=1
where, according to usual practice, we have deleted the parentheses. So in a finite double
sum, the order of summation is immaterial.
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Preliminaries
EXAMPLE 9
For numbers a1 , a2 , . . . , am and b1 , b2 , . . . , bn ,
n
m n
m X
m X
n
X
X
X
X
=
bj
ai
ai b j
ai b j =
i=1
i=1 j=1
=
Theorem
Pn1
with c=
bj
j=1
finite product
Theorem 1
with c=ai
j=1
X
n
bj
X
m
i=1
j=1
ai
j=1
i=1
=
X
m
i=1
ai
X
n
j=1
bj .
Similar to the addition, we write the product of the numbers t1 , t2 , . . . , tn as
n
Y
ti .
i=1
4
ABSOLUTE VALUE
If x 6= 0 is a real number, then either x is positive or −x is positive. The absolute value of
x 6= 0 is defined as the positive number of the numbers x and −x.
absolute value
DEFINITION
EXAMPLE 10
Let x ∈ IR. The absolute value of x, written as |x|, is given by
−x if x < 0
|x| =
x
if x ≥ 0.
|6| = 6, since 6 > 0. Also | − 2| = −(−2) = 2, since −2 < 0.
If the real numbers are thought of as points on a line, then the absolute value of a number
can be seen as its distance from the origin:
z
y
|y|
}|
{z
|x|
}|
{
x
0
More generally, |x − y| represents the distance between x and y.
To give an example, in the following figure
z
y
|x − y|
}|
0
{
x
the distance between x and y is equal to the distance between the point x and the origin
plus the distance between the point y and the origin. This is equal to
|x| + |y| = x + (−y) = x − y = |x − y|.
The same result can be obtained if x and y are both positive or both negative.
In Chapter 5 we wish to describe (for a function f and a point c in its domain) the situation
that the images f (x) are getting close to some number as x gets close to c. Since x is ’close
to’ c in fact means that the distance between x and c is small, the absolute value will play
a prominent role in Chapter 5.
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Preliminaries
For d > 0, the equation |x| = d has two solutions: x = d and x = −d. These are, of course,
the two points on the real line that lie at distance d from the origin.
EXAMPLE 11
We consider the equation
|2x + 3| = 1.
According to the observation preceding this example,
|2x + 3| = 1 ⇐⇒ 2x + 3 = 1 or 2x + 3 = −1 ⇐⇒ x = −1 or x = −2.
Often equations (and inequalities) involving absolute values can be solved algebraically by
breaking them into cases according to the definition of absolute value. This will be
illustrated in the next example (and in a number of subsequent exercises).
EXAMPLE 12
If we want to solve the equation
|x − 1| = 1 − x,
it is not allowed to apply the method used in the foregoing example. Why? And what is
going to happen if you take the square of the left-hand side and the right-hand side of the
equality?
Instead we distinguish between two cases: x < 1 and x ≥ 1.
If x < 1, the expression |x − 1| is equal to 1 − x. So we obtain
|x − 1| = 1 − x ⇐⇒ 1 − x = 1 − x.
Hence, any x < 1 is a solution of the equation.
If x ≥ 1, the expression |x − 1| is equal to x − 1. So we obtain
|x − 1| = 1 − x ⇐⇒ x − 1 = 1 − x ⇐⇒ x = 1.
So x = 1 is a solution too.
The solution set of this equation is the interval (−∞, 1].
EXERCISE 10
Solve the equations
(b) |x2 − 1| = 2x.
(a) |x − 3| = 2|x|
EXERCISE 11
Let x be a real number. Prove that −|x| ≤ x ≤ |x|.
EXERCISE 12
The maximum of two numbers x and y is denoted by max{x, y}. Thus
max{−2, 2} = max{2, 2} = 2.
Prove that
max{x, y} =
x + y + |y − x|
.
2
1
13
Preliminaries
EXERCISE 13
Let x and z be real numbers such that x < z, and let y ∈ IR.
(a) Assume that x < y < z. Prove that |x − y| + |y − z| = |x − z|.
Give a geometrical interpretation of this result in terms of distances.
(b) Assume that y > z. Prove that |x − y| + |y − z| 6= |x − z|.
Give a geometrical interpretation of this result in terms of distances.
In the following theorem we summarize some properties of the absolute value.
THEOREM 2
For all x, y ∈ IR, z 6= 0 and each a > 0,
(a) | − x| = |x|
(b) |xy| = |x| |y|
(c)
1
1
=
z
|z|
(d) |x| < a is equivalent to −a < x < a (|x| ≤ a is equivalent to −a ≤ x ≤ a).
PROOF
(c)
The proofs of (a) and (b) are left as an exercise for the reader.
Note that property (b) implies that for z 6= 0,
1 = |1| = z ·
(d)
1
1
1
1
= |z| ·
=⇒
=
.
z
z
z
|z|
Let x be a real number and let a > 0. Because we have to prove an ’if and only if’
statement, the proof consists of two parts.
Suppose that |x| < a. Then −a < −|x| and together with exercise 11 this yields
−a < −|x| ≤ x ≤ |x|.
But then the assumption |x| < a yields −a < −|x| ≤ x ≤ |x| < a. Hence −a < x < a.
Conversely, suppose that −a < x < a. We will consider the same two cases as before.
If x ≥ 0, then |x| = x < a and if x < 0, then |x| = −x < −(−a) = a.
EXERCISE 14
Prove parts (a) and (b) of Theorem 2.
EXERCISE 15
(a) Prove that for all a ≥ 0 and b ≥ 0,
a = b ⇐⇒ a2 = b2 .
(b) Solve the equation in Exercise 10 (a) once again.
(c) Prove that for all real numbers x and y,
|x + y| = |x| + |y| ⇐⇒ xy ≥ 0.
EXERCISE 16
Prove that 2 |a||b| ≤ a2 + b2 for all real numbers a and b.
EXERCISE 17
Let x be a real number between −1 and 1.
(a) Prove that x2 ≤ |x|.
(b) Prove that |x3 | ≤ |x|.
1
14
EXERCISE 18
Preliminaries
Let (a, b) be an interval. Prove that |x − y| < b − a for all x, y ∈ (a, b).
Give a geometrical interpretation of this result in terms of distances.
The following (main) result will play an important role in the remainder of this book.
THEOREM 3
TRIANGLE INEQUALITY
For all x, y ∈ IR,
|x + y| ≤ |x| + |y|.
PROOF
Let x, y ∈ IR. According to Exercise 11,
−|x| ≤ x ≤ |x|
and
−|y| ≤ y ≤ |y|.
Adding these inequalities, we get
−|x| − |y| ≤ x + y ≤ |x| + |y| ⇐⇒ − |x| + |y| ≤ x + y ≤ |x| + |y|.
Hence, according to Theorem 2 (d), |x + y| ≤ |x| + |y|.
EXERCISE 19
Let x, y and z be real numbers.
(a) Prove that |x − y| ≤ |x| + |y|.
(b) Prove that |x + y + z| ≤ |x| + |y| + |z|.
EXERCISE 20 (THE TELESCOPE METHOD)
Let a, b and c be real numbers. Prove that |a − b| ≤ |a − c| + |c − b|.
EXERCISE 21
Find an upper bound of the set
{| 4x2 − 2x + 2 | | − 1 ≤ x < 2}.
Give a geometrical interpretation of this result for the curve y = 4x2 −2x+2.
The number you determined in the foregoing exercise is called an upper bound for the expression |4x2 − 2x + 2|. Finding such upper bounds plays an important role in Chapter 5.
Sometimes one is interested in an upper bound of a specific form. This situation will be
discussed in the following example and in a subsequent exercise.
EXAMPLE 13
We consider the expression |x2 (1 + 2x)| for − 21 < x <
2
find an upper bound for this expression of the form ax , where a > 0.
Note that |x2 (1 + 2x)| = |x2 | · |1 + 2x| = x2 |1 + 2x|.
1
2
and we want to
1
15
Preliminaries
So our problem will be solved if we can find an upper bound a > 0 for the expression |1+2x|.
Such an upper bound can be found by using the fact that x lies between − 21 and
1
2.
We
have
− 21 < x <
1
2
=⇒ −1 < 2x < 1 =⇒ 0 < 1 + 2x < 2 =⇒ |1 + 2x| < 2.
As a consequence |x2 (1 + 2x)| < 2x2 . So we may choose a = 2.
EXERCISE 22
Again consider the expression |x2 (1 + 2x)|. Now assume that −1 < x < 1.
Determine a number b such that
|x2 (1 + 2x)| ≤ b|x|.
THEOREM 4
REVERSE TRIANGLE INEQUALITY
For all x, y ∈ IR,
|x − y| ≥ |x| − |y| .
PROOF
Let x, y ∈ IR.
According to the Triangle Inequality, |x| = |x − y + y| ≤ |x − y| + |y|, which implies that
|x| − |y| ≤ |x − y|.
(1)
Similarly, |y| = |y − x + x| ≤ |y − x| + |x| = |x − y| + |x| leads to
−|x − y| ≤ |x| − |y|.
The inequalities (1) and (2) can be summarized as:
−|x − y| ≤ |x| − |y| ≤ |x − y|.
According to Theorem 2 (d) this is equivalent to
|x| − |y| ≤ |x − y| ⇐⇒ |x − y| ≥ |x| − |y| .
EXERCISE 23
Find a b > 0 such that
4x2
1
≤ b, for all 2 < x < 4.
− 2x − 2
Finally, we will discuss the solution of inequalities.
EXAMPLE 14
We will solve the inequality
|2x + 2| < 4.
According to Theorem 2 (d),
|2x + 2| < 4 ⇐⇒ −4 < 2x + 2 < 4 ⇐⇒ −6 < 2x < 2 ⇐⇒ −3 < x < 1.
Hence, the solution set of this inequality is the interval (−3, 1).
(2)
1
16
EXAMPLE 15
Preliminaries
We will solve the inequality
2x − 4
< 1.
x+1
According to Theorem 2 (d),
2x − 4
2x − 4
< 1.
< 1 ⇐⇒ −1 <
x+1
x+1
So in fact we must solve two inequalities and determine the intersection of the two solution
sets.
We start with the first inequality:
−1 <
2x − 4
2x − 4 x + 1
3x − 3
⇐⇒
+
> 0 ⇐⇒
> 0.
x+1
x+1
x+1
x+1
As you know from secondary school such an inequality has to be solved by constructing a
sign diagram for the numerator and denominator of the fraction at the left-hand side of the
inequality. These two diagrams are used to construct a sign diagram for the fraction itself.
We obtain
−
−
−
−
−
−
0
+
+
1
−
−
0
+
+
+
+
+
+
−
−
−
0
+
+
−1
+
+
×
−1
1
3x − 3
x+1
3x − 3
x+1
So the solution set for the first inequality is (−∞, −1, ) ∪ (1, ∞).
Next we solve the second inequality:
2x − 4 x + 1
x−5
2x − 4
< 1 ⇐⇒
−
< 0 ⇐⇒
< 0.
x+1
x+1
x+1
x+1
According to the sign diagrams
−
−
−
−
−
−
0
+
+
5
−
−
0
+
+
+
+
+
+
−
−
−
0
+
+
−1
+
+
×
−1
5
x−5
x+1
x−5
x+1
the solution set for this inequality is (−1, 5).
Hence, the solution set of the original inequality is the interval (1, 5).
Sometimes the following equivalence is helpful for solving inequalities.
1
17
Preliminaries
For all real numbers x and y,
|x| < |y| ⇐⇒ x2 < y 2 .
The fact that |y| > |x| ≥ 0 implies that |x| + |y| > 0. Similarly, the fact that y 2 > x2 ≥ 0
implies that |x| + |y| > 0. Hence,
|x| < |y| ⇐⇒ |x| − |y| < 0 ⇐⇒ (|x| − |y|)(|x| + |y|) < 0 ⇐⇒ |x|2 − |y|2 < 0 ⇐⇒ x2 < y 2 .
EXERCISE 24
Solve the following inequalities.
(a) 12 |5x − 3| < 4
(b) |x| + |x − 1| ≤ 0
(c) |x − 3| > 2|x|
(d) |x − 1| + |x − 2| > 1
(e) |x2 − 1| ≤ 2x − 2
5
5.1
(f) |x − 1| · |x + 2| ≥ 4.
FUNCTIONS
SOME DEFINITIONS
Given a subset A of IR, a function f is a rule which assigns to each element x in A a unique
real number, which we denote by f (x) (read: ’f of x’). This is indicated by
f : A → IR.
The number f (x) is called the value of f at x or the image of x under f . We also say that
f maps x into f (x).
The set A is called the domain of the function. Sometimes it is denoted by Df .
The set of all possible images is called the range of the function f . It is denoted by Rf .
EXAMPLE 16
Let A = IR and let f be the function that assigns to each real number its
square. So f (2) = 4, f (−1) = 1 and, more generally,
f (x) = x2 .
This can be visualized as follows:
IR
IR
f
4
f
1
2
−1
1
18
Preliminaries
Sometimes it is helpful to think of function as a machine with an input and an output:
device
output y
input x
f
In case of the ’square function’ we discussed in Example 16, you could choose your pocket
calculator as the device. If you type a number (that is not too large), after using the
x2 -button, your calculator produces its square.
5.2
graph
VISUALIZING FUNCTIONS
The usual way to visualize a function f is by its graph. It consists of those points (x, y) in
the Cartesian plane satisfying y = f (x) and x ∈ Df . It is some curve represented by the
equation y = f (x).
In set-theoretic terms, the graph of the function f is the set
{(x, y)| y = f (x) and x ∈ Df }.
zero
A number z ∈ Df satisfying f (z) = 0 is called a zero of the function f . Obviously, z is a
zero of the function f if and only if the graph of the function f intersects the horizontal axis
at the point (z, 0).
So the graph of the linear function ℓ defined by ℓ(x) = 2x + 1 is the straight line y = 2x + 1.
The graph of the quadratic function q defined by q(x) = x2 − 1 is the parabola y = x2 − 1.
y
y
q
ℓ
x
FIGURE 4
x
Some graphs
Note that − 21 is the only zero of the linear function ℓ, whereas −1 and 1 are the zeros of the
quadratic function q.
In the next example we discuss (graphs and zeros of) quadratic functions.
1
19
Preliminaries
We consider the quadratic function
EXAMPLE 17
f : x → ax2 + bx + c,
where a 6= 0. In order to determine the zeros of this function we rewrite f (x) by completing
the square:
b
c
f (x) = ax2 + bx + c = a x2 + x +
a
a
2 2
b
c
b
b
2
= a x +2 x+
−
+
2a
2a
2a
a
2
b2 − 4ac
b
.
−
=a x+
2a
4a2
(1)
Hence, the zeros of the function f can be found by solving the equation
2
b
b2 − 4ac
f (x) = 0 ⇐⇒ ax2 + bx + c = 0 ⇐⇒ x +
=
.
2a
4a2
Obviously, the numbers of zeros depends on the (sign of the) number
D = b2 − 4ac,
discriminant
which we call the discriminant of the equation ax2 + bx + c = 0.
b
.
2a
Finally, if D > 0, then the function has two zeros, which can be found by means of the
If D < 0, then the function has no zeros. If D = 0, then the function has one zero x = −
quadratic formula
quadratic formula
√
b2 − 4ac
2a
Formula (1) can also be used to describe the graph of the function f . If a > 0, then the
x=
−b ±
graph is a parabola like the one in Figure 4. If a < 0, it is a parabola which is turned upside
down compared to the one represented in Figure 4.
6
BUILDING NEW FUNCTIONS
Functions can be combined in a variety of ways to make new functions.
Like numbers, functions can be added, subtracted, multiplied, and divided (except where
the denominator is zero) to produce new functions.
DEFINITION
sum function
If f and g are functions, then on the intersection of the domains of both
f
functions the functions f + g, f g and are defined by
g
(f + g)(x) = f (x) + g(x)
product function
quotient function
and
(f g)(x) = f (x) g(x)
f f (x)
(x) =
provided that g(x) 6= 0.
g
g(x)
If c is a real number, then cf is the function defined by
scalar product
(cf )(x) = cf (x).
1
20
Given f (x) =
EXAMPLE 18
f g, and
Preliminaries
√
√
x and g(x) = 2 − x, we determine the functions f + g,
f
and specify their domains.
g
We have
√
√
x + 2 − x with domain [0, 2]
p
with domain [0, 2]
= x(2 − x)
r
x
=
with domain [0, 2).
2−x
(f + g)(x) = f (x) + g(x) =
(f g)(x)
f (x)
g
= f (x) g(x)
f (x)
=
g(x)
Note that the domain of first two functions is the intersection of the domains [0, ∞) and
(−∞, 2] of the functions f and g. In the case of the quotient of the two functions f and g,
we had to remove the point x = 2 where the function g is zero.
Finally, we examine the option of making new functions from old ones by ’composing’ them.
Consider the function h defined by
h(x) =
p
x2 + 79.
If you are are asked to determine the value of h at 39, then probably you will proceed in
two steps.
First you determine the number 392 + 79 and then you calculate its square root. So
p
√
h(39) = 392 + 79 = 1600 = 40.
Introducing the functions f : [0, ∞) → IR and g: IR → IR defined by
f (x) =
√
x
g(x) = x2 + 79
and
one can say that you first determine g(39) and then the function f is applied to g(39). This
can be visualized as follows:
IR
IR
IR
1600
f
g
40
39
In other words:
or, more generally, for any x
h(39) = f g(39)
h(x) = f g(x) .
In this situation h is called the composite function of f and g and this is denoted by h = f ◦ g
(read ’f circle g’)
1
composite function
21
Preliminaries
DEFINITION
If g: A → IR, f : B → IR and Rg ⊂ B, then the composite function f ◦ g is
defined by
(f ◦ g)(x) = f g(x) .
As we first calculate g(x) and then calculate f of this result, g is called the
inner function and f the outer function.
If we think of f and g as machines, then f ◦ g is obtained by arranging the machines f and
g in an ’assembly line’ so that the output of g becomes the input of f :
device
device
input x
output g(x)
g
f
√
x and g(x) = x + 2, we determine the functions f ◦ g,
Given f (x) =
EXAMPLE 19
output f g(x)
g ◦ f , f ◦ f and g ◦ g and specify their domains.
We have
√
(f ◦ g)(x) = f g(x) = f (x + 2) = x + 2
with domain [−2, ∞)
√
with domain [0, ∞)
(g ◦ f )(x) = g f (x) = f (x) + 2 = x + 2
√
√ p√
(f ◦ f )(x) = f f (x) = f x =
x = 4 x with domain [0, ∞)
(g ◦ g)(x) = g g(x) = g(x + 2) = x + 4
with domain IR
To see why the domain of f ◦ g is [−2, ∞), observe that g(x) = x + 2 belongs to the domain
of f only if x + 2 ≥ 0, that is if x ≥ −2.
1−x
, determine the function g ◦ g and specify its domain.
1+x
EXERCISE 25
Given g(x) =
EXERCISE 26
Determine the functions f ◦ g, g ◦ f , f ◦ f and g ◦ g and specify their domains
if
(a) f (x) =
(b) f (x) =
EXERCISE 27
2
x
and g(x) =
1
1−x
and
x
1−x
g(x) =
√
x − 1.
Find the missing entries in the following table.
f (x)
(a)
(b)
(c)
(d)
(e)
(f )
2
x
√
x
g(x)
(f ◦ g)(x)
x+1
x+4
x
√
3
x
|x|
2x + 3
x+1
x
x
x−1
1
x2
1
22
7
Preliminaries
SOME FAMILIES OF FUNCTIONS
Functions are often grouped into families according to the form of their defining formulas or
other common characteristics. In this section we will discuss some of the most basic families
of functions.
7.1
polynomial function
POLYNOMIAL FUNCTIONS
Let a0 , a1 , . . . , an ∈ IR, an 6= 0 and n ∈ IN. A function f defined by
f (x) = a0 + a1 x + a2 x2 + · · · + an xn
coefficient/degree
is called a polynomial function with coefficients a0 , a1 , . . . , an and degree n.
Functions like x 7→ 2x − 3, x 7→ x2 − 3x + 7 and x 7→ (x2 − 4)3 are polynomial functions of
degree 1, 2 and 6, respectively.
power function
A polynomial functions is obtained by combining the constant function x 7→ 1 and the power
functions x 7→ x, x 7→ x2 , . . . , x 7→ xn , . . .
The linear function ℓ defined by
ℓ(x) = mx + b,
is in fact a polynomial function of degree 1. Its graph is the straight line y = mx + b. The
number m is called the slope of this line; it is a measure for the steepness of the line. The
number b is the vertical intercept.
ℓ(x2 ) − ℓ(x1 )
.
x2 − x1
EXERCISE 28
Determine, for the linear function ℓ, the number
EXERCISE 29
(a) Find an equation of the line with slope 2 and containing the point (3, 1).
(b) Find an equation of the line containing the points (1, 1) and (2, 3).
(c) Find an equation of the line containing the points (x1 , y1 ) and (x2 , y2 ).
(d) Find an equation of the line containing the origin who is perpendicular
to the line you found in part (a).
rational function
A function f that can be expressed as a ratio of two polynomial functions is called a rational
function, i.e.
f (x) =
p(x)
,
q(x)
where p and q are polynomial functions. Observe that the domain of f consists of all value
of x such that q(x) 6= 0.
x2
1
1
are examples of rational functions.
and f3 : x →
So f1 : x → 3 , f2 : x →
x
4x − 4
(x − 1)2
The graphs of the first two functions are represented below.
1
23
Preliminaries
y
y
f1
f2
1
−1
1
x
1
An orthogonal hyperbola and a non-orthogonal hyperbola
FIGURE 5
Sketch the graph of the function f3 .
EXERCISE 30
7.2
x
TRIGONOMETRIC FUNCTIONS
Most students first encounter the quantities sin α and cos α as ratios of sides in a right-angled
triangle with α as one of its acute angles:
hypotenuse
opposite side
α
adjacent side
More specifically,
sin α =
the side opposite angle α
the hypotenuse
cos α =
the side adjacent to angle α
.
the hypotenuse
In calculus we need a more general definition of sin t and cos t, where t is an arbitrary real
number. Such a definition is formulated in terms of a circle rather than a triangle.
Consider the circle with center at the origin and radius one. This so-called unit circle can
be represented by the equation x2 + y 2 = 1.
y
P (cos t, sin t)
arc length t
t (radians)
(1, 0)
FIGURE 6
The unit circle
x
1
24
Preliminaries
For a real number t, let P be the point on the unit circle at distance | t | from the point
(1, 0), measured along the circle in the counterclockwise direction if t > 0 and in the clockwise
direction if t < 0.
We define sin t as the second coordinate of the point P and cos t as the first one. The angle
at the origin is an angle of t radians.
A third important trigonometric function is the function tan defined by
tan t =
sin t
,
cos t
provided that cos t 6= 0.
Many important properties of sin t and cos t follow from the fact that they are coordinates
of a point on the unit circle. For instance, for all t,
− 1 ≤ sin t ≤ 1
− 1 ≤ cos t ≤ 1
sin2 t + cos2 t = 1.
and
Since the unit circle has circumference 2π, adding 2π to t causes the point P to go one extra
revolution and end up in the same place: thus, for every t,
sin(t + 2π) = sin t
and
cos(t + 2π) = cos t.
That says that sine and cosine are periodic with period 2π. The graphs of these functions
restricted to the interval [0, 2π] are represented below.
y
y
sin
1
π
2π x
1
2π
−1
FIGURE 7
π
3
2π
2π x
cos
The graph of the sine and cosine
The following formulas enable us to express the sine and cosine of a sum of two angles in
terms of the sines and cosines of those angles.
THEOREM 5
ADDITION FORMULAS FOR SINE AND COSINE
For all x and y,
sin(x + y) = sin x cos y + cos x sin y
and
cos(x + y) = cos x cos y − sin x sin y.
1
25
Preliminaries
If we choose in the foregoing formulas y = x, we obtain the so-called double-angle-formulas:
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x
and
= 2 cos2 x − 1.
The following formulas enable explains how to add the two trigonometric functions.
THEOREM 6
ADDING TWO TRIGONOMETRIC FUNCTIONS
For all x and y,
x−y
x+y
cos
2
2
x+y
x−y
cos x + cos y = 2 cos
cos
.
2
2
sin x + sin y = 2 sin
and
7.3
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Note that integer and rational powers of a number b > 0 are defined by
b0 = 1, bn = b × b × · · · × b
|
{z
}
n factors
b
p/q
and b−n =
1
bn
√
√
p
q
q
b
= bp =
b−p/q =
1
bp/q
.
Observe that these definitions do not include irrational powers of b such as 5
√
2
and 3π . We
will not be concerned with this matter here and accept the following properties of powers:
for real numbers p and q (and positive b) it holds that
(1) bp bq = bp+q
bp
= bp−q
bq
q
(3) bp = bpq .
(2)
exponential function
Let b > 0. A function f defined by
f (t) = bt
is called an exponential function with base b.
If b > 1, we have exponential growth; if 0 < b < 1, we have exponential decay.
The most frequently used base for an exponential function is the number e = 2.7182 . . ..
1
26
3t
Preliminaries
(1.5)t
et
1
−2
−1
FIGURE 8
1
2
4 t
3
Exponential growth
As is obvious from Figure 8, each horizontal line y = b intersects the graph of the function
t → et once if b > 0 (in Section 4 of Chapter 8, we will discuss this property in more detail).
In other words: the equation ex = b with b > 0 has a unique solution. This solution, which
is denoted by ln b, is called natural logarithm of the number b.
More precise: the natural logarithm of x > 0, denoted as ln x is the power of e needed to
get x:
eln x = x.
The graph of the natural logarithm is given below.
y
ln
1
1
FIGURE 9
2
3
4
5
x
The natural logarithm
Note that ln 1 = 0, because e0 = 1. Other properties of the natural logarithm are
(1) ln(xy) = ln x + ln y
x
= ln x − ln y
(2) ln
y
t
(3) ln(b ) = t ln b.
The natural logarithm is useful when we have to solve for unknown exponents.
EXAMPLE 20
In order to solve the equation
2t = 6,
we take the natural logarithm of both sides of the equality sign. This leads to
2t = 6 ⇐⇒ ln(2t ) = ln 6 ⇐⇒ t ln 2 = ln 6 ⇐⇒ t =
ln 6
.
ln 2
1
27
Preliminaries
EXERCISE 31
Solve the following equations.
(a) 3t = 10
(b) 10 = 30(1.03)t
(d) 7t+2 = 5 e3−t
(e) 2t − 1 = eln t
2
(c) 2 · 3t = 5 · 7t
2
(f) 8t = 2 et .
The fact that the function t → ln t is strictly increasing can be used to solve inequalities
involving exponentials.
EXAMPLE 21
The inequality
2t > 6
can be solved as follows:
2t > 6 ⇐⇒ ln(2t ) > ln 6 ⇐⇒ t ln 2 > ln 6 ⇐⇒ t >
EXERCISE 32
ln 6
.
ln 2
Solve the following inequalities.
(a) et+2 < 3t
(b) 10 > 500 e−0.2t
(c) 2 e3t ≤ 4 e7t
Logarithms other than the natural one also exist. For b > 0 and b 6= 1, the logarithm to the
base b of x > 0, denoted as b log x, is the power of b needed to get x:
b
b
log x
= x.
The logarithm with base 10 is written as log x.
ln 6
In Example 20 we proved that
is the (unique) solution of the equation 2t = 6. The
ln 2
solution of this equation is also given by 2 log 6. Of course the two solutions are equal. One
can prove, more generally, that
b
log x =
ln x
.
ln b
This formula is useful if you have to evaluate, for instance, the logarithm 2 log 7 on your
calculator.
8
THE INVERSE FUNCTION
In Section 6 we have discussed various ways of forming new functions from old ones: addition,
multiplication and composition. However by using these alone we cannot produce a ’simple’
√
function like x → 3 x. In order to investigate all kinds of properties of this function (and
related ones) we have to introduce a quite sophisticated way of constructing new functions
from old ones.
In this section we introduce the concept of invertibility of a function and we investigate how
this property is related to strict monotonicity.
1
28
invertible function
DEFINITION
Preliminaries
Let f be a function on an interval I. The function f is called one-to-one or
invertible if for any two points x, x′ ∈ I
x 6= x′ =⇒ f (x) 6= f (x′ )
or, equivalently,
f (x) = f (x′ ) =⇒ x = x′ .
In other words: a one-to-one function is one which maps distinct points into
distinct points.
Geometrically, a function is one-to-one if each horizontal line intersects the graph of the
function at most once.
1
is not one-to-one, because f (1) =
1 + x2
However, the function g on (0, ∞), defined by
1
g(x) =
,
1 + x2
is one-to-one.
The function f : x →
EXAMPLE 22
Indeed, for x, x′ ∈ (0, ∞),
g(x) = g(x′ ) =⇒
1
2
= f (−1).
1
1
=
=⇒ x2 = (x′ )2 =⇒ x2 − (x′ )2 = 0
1 + x2
1 + (x′ )2
=⇒ (x − x′ )(x + x′ ) = 0 =⇒ x = x′ .
| {z }
>0
Hence, g(x) = g(x′ ) implies that x = x′ .
Next we will introduce for an invertible function on an interval a new function, called the
inverse function.
inverse function
DEFINITION
If f is an invertible function on an interval I, then the inverse of f is the
function f −1 on f (I), which is defined by
f −1 (y) = x,
where x is the unique element in I with f (x) = y. So f −1 f (x) = x for
all x ∈ I.
EXAMPLE 23
Let g be the function defined by
g(x) = 2x + 1.
Then g is invertible and for any y ∈ IR
y = g(x) ⇐⇒ y = 2x + 1 ⇐⇒ x = 21 (y − 1).
Hence, for any y ∈ IR,
g −1 (y) = 21 (y − 1).
1
29
Preliminaries
EXAMPLE 24
For the function f on [0, 2] defined by
g(x) =
x
if 0 ≤ x < 1
3−x
if 1 ≤ x ≤ 2
it holds that
g
−1
(y) =
y
3−y
if 0 ≤ y < 1
if 1 ≤ y ≤ 2.
So g −1 = g.
EXAMPLE 25
In Example 22 we proved that the function g on the interval I = (0, ∞),
defined by
g(x) =
1
,
1 + x2
is invertible.
In order to determine the inverse function g −1 , we first will prove that g(I) = (0, 1).
1
< 1 for all x > 0, g(I) ⊂ (0, 1).
Since 0 <
1 + x2
In order to prove the reverse inclusion, let y ∈ (0, 1). We will show that the equation
g(x) = y ⇐⇒
1
=y
1 + x2
has at least one solution.
Indeed, for 0 < y < 1 and x > 0,
1−y
1
⇐⇒ y + x2 y = 1 ⇐⇒ x2 =
⇐⇒ x =
y = g(x) ⇐⇒ y =
1 + x2
y
Hence, g(I) = (0, 1) and g −1 is the function on (0, 1), defined by
g −1 (y) =
EXERCISE 33
r
1
− 1.
y
Prove that the function f on [0, ∞), defined by
f (x) = x2 + 1
is invertible and find f −1 .
EXERCISE 34
Consider the function f on [0, 1), defined by
x
.
f (x) = √
1 − x2
(a) Prove that the function f is invertible.
(b) Prove that Rf = [0, ∞) and find f −1 .
r
1−y
.
y
1
30
Preliminaries
If f is an invertible function on an interval I, then the graph of f −1 is the reflection of the
graph of f in the line y = x.
y
f −1
y=x
(b, a)
f
(a, b)
x
FIGURE 10
The graph of the inverse function
For suppose that (a, b) is a point on the graph of f . Then b = f (a), or equivalently,
a = f −1 (b). In other words: (b, a) – which is the reflection of the point (a, b) in the line
y = x – is on the graph of f −1 .
In the next example we determine the inverse function of an exponential one.
The function f defined by
EXAMPLE 26
f (x) = 3 (1.8)x
is invertible as it is strictly increasing. Note that Df = (0, ∞). Now for y > 0
y = 3 (1.8)x ⇐⇒ ln y = ln 3 (1.8)x ⇐⇒ ln y = ln 3 + ln(1.8)x ⇐⇒ ln y = ln 3 + x ln 1.8
⇐⇒ x =
ln y
ln 3
−
.
ln 1.8 ln 1.8
Hence, for y > 0,
f −1 (y) =
EXERCISE 35
ln y
ln 3
−
.
ln 1.8 ln 1.8
Explain why the following functions are invertible and find their inverse
function.
(a) x → (0.9)x
EXERCISE 36
(b) x → 4 e0.5x
(c) x → 1 + ln x
Put the following exponential functions in the form t → b ekt .
(a) t → 3 (5.5)t
(b) t → 8 (0.4)t .
1
31
Preliminaries
Mixed exercises
EXERCISE 37
Write the set of all values of x satisfying 1 < |2 − x| ≤ 6 as a union of two
intervals.
EXERCISE 38
Write the following expression as one fraction
ax . ax2 + 4ax
.
4(x + 1)
x2 − 1
EXERCISE 39
Simplify, if possible, the following expression
r
EXERCISE 40
Simplify, if possible, the following expression
2
EXERCISE 41
5 2 5/2
.
·
8
3
log 3 +4 log 3.
Factor
x3 − 6x2 y + 9xy 2 − 16xz 2 .
EXERCISE 42
Simplify, if possible, the following expression
4x2 + 2x − 12
.
2x2 − x − 3
EXERCISE 43
Simplify, if possible, the following expression
3a3 + 5a2 b − 12ab2 − 20b3 .
EXERCISE 44
Solve the following equation for x:
(3b − 2x)4 = 256.
EXERCISE 45
Solve the following inequality
1
2
≤
.
2x + 1
x+3
EXERCISE 46
Use a long division to solve the following equation
x4 − 4x3 − 5x2 + 4x + 4 = 0.
EXERCISE 47
Draw in one picture the graphs of the functions f : x → 3x , g: x →3log(−x),
1
h: x → −3 log x and k: x → 3 log x.
1
32
EXERCISE 48
Solve the following equation for x ∈ [0, 2π]:
sin2
EXERCISE 49
Preliminaries
x
2
+ cos x = 1.
The function f defined by
2
f (x) =
ex + e−x
x2 + 1
2
is the composition of two functions g and h. Find the functions g and h.
EXERCISE 50
Find the derivative of the following functions:
(a) f : x →
EXERCISE 51
x2/3 − x3/2
x
(b) g: x →
q
ln 4x4 + 9 .
Write the following algebraic expression as the product of two or more nontrivial (i.e. not equal to 1) factors.
(a) 2x4 − 32
(b) x2 − y 2 + 5x + 5y
(c) x3 y − 4x2 y 2 + 4xy 3
(d) x2 + 2xy − 3y 2
(e) x3 + y 3 + xy 2 + x2 y.
EXERCISE 52
Simplify the following expressions.
2y
2+x 1−y
− 2 2
+
2
2
x y
xy
x y
√
x x+1
√
(c)
x
(a)
EXERCISE 53
x−y
x
3xy
−
+ 2
x + y x − y x − y2
√
√
x y−y x
√ .
(d) √
x y+y x
(b)
Solve the following inequalities.
x
4
≥1+
2
x
(a) x4 − 5x2 + 5 ≥ 1
(b)
(c) (x − 1)100 (x + 1) > 0
(d) 3x − 1 < 5x + 3 ≤ 2x + 15.
2
33
The Principle of mathematical induction
2
THE PRINCIPLE OF MATHEMATICAL
INDUCTION
A proof by induction is an important technique for verifying formulas involving natural
numbers. Examples of such formulas are
1 + 2 + · · · + n = 21 n(n + 1),
the number 9n − 5 is a multiple of 4,
and
n
X
i=1
i3 =
n 2
X
i .
i=1
The general structure of an induction proof can be explained as follows. We want to prove
that a statement P(n), which depends on n, is true for all natural numbers n. For instance,
in the first example the relevant statement is
P(n) : 1 + 2 + · · · + n = 21 n(n + 1).
The steps required in the proof are as follows. First, verify that the statement P(1) is true
(which means that the formula is correct for n = 1). Then prove that for each natural
number k, if statement P(k) is true, it follows that P(k + 1) must be true. Here assuming
that P(k) is true is called the induction hypothesis, and the step from P(k) to P(k + 1) is
called the induction step in the proof.
The principle of induction seems intuitively clear. If for each k the truth of P(k) implies
the truth of P(k + 1), then because P(1) is true, P(2) must be true, which, in turn, means
that P(3) is true, and so on.
Note that the algorithmic-like structure of such a proof resembles the ’structure’ of the
natural numbers: there is a starting number, the number one, and each number is directly
followed by precisely one other number. An analogy: consider a ladder with an infinite number of steps, numbered 1, 2, . . . Suppose you can climb the first step and suppose, moreover,
that after each step, you can always climb the next. Then you are able to climb up to any
step.
In the second section, the details of an induction proof are discussed. Furthermore, in the
third section of this chapter, some notations will be introduced.
2
34
1
The Principle of mathematical induction
STATEMENTS ABOUT NATURAL NUMBERS
We give some examples. In the first one a formula involving natural numbers will be proved
in a straightforward way.
EXAMPLE 1
We will show that
n3 − 3n2 + 3n > 0
for all n ∈ IN.
As
n3 − 3n2 + 3n = n n2 − 3n + 3 ,
and the discriminant of the equation x2 − 3x + 3 = 0 is negative, n2 − 3n + 3 > 0 for all n.
Since n > 0 the proof is complete.
EXAMPLE 2
Suppose we want to investigate whether the number
9n − 5
is a multiple of 4 for all n ∈ IN.
We observe that
91 − 5 = 4 = 1 · 4
92 − 5 = 76 = 19 · 4
93 − 5 = 724 = 181 · 4
94 − 5 = 6556 = 1639 · 4.
As we can go on in this way, we believe that 9n − 5 is a multiple of 4 for all n. In the
next example, however, we will show that it is not sufficient to check a statement for some
natural numbers and to think that ’one can go on in this way’.
EXAMPLE 3
Consider the expression
f (n) = n2 + n + 11.
If we evaluate this expression for various natural numbers, we observe that we always seem
to obtain a prime number. (Recall that a natural number n is prime if n > 1 and its only
positive divisors are 1 and n.) For example,
f (1) = 13
f (2) = 17
f (3) = 23
..
.
f (8) = 83
f (9) = 101
and all these numbers (as well as the ones skipped over) are prime.
2
35
The Principle of mathematical induction
On the basis of this experience we might conjecture that the expression n2 + n + 11 will
always produce prime numbers when n is a natural number.
However, for n = 10 we obtain the number 121 = 112 , which is definitely not a prime.
inductive reasoning
In the foregoing example we came to a conjecture by what is called inductive reasoning: on
the basis of looking at individual cases we make a general conclusion. It appeared to be of
little value in the example, because in the end there is no absolute causal logical relation
between a prime number and a number being expressible by the formula n2 +n+11. Although
this type of reasoning is not always successful, as seen above, sometimes it may lead to a
correct statement.
2
MATHEMATICAL INDUCTION
Next we discuss an example in which essentially a mathematical inductive proof is found.
Actually the causal logic relation between some steps is demonstrated which then can be
extended to a formal proof.
EXAMPLE 4
In Example 2 we investigated whether the number
9n − 5
is a multiple of 4 for all n ∈ IN. We observed that
91 − 5 = 4 = 1 · 4
92 − 5 = 76 = 19 · 4
93 − 5 = 724 = 181 · 4
94 − 5 = 6556 = 1639 · 4.
Due to Example 3 we know that going on in this way isn’t very useful. So we take a more
systematic approach.
95 − 5 = 9 · 94 − 5 =
96 − 5 = 9 · 95 − 5 =
· 9}4
|8 {z
+
is a multiple of 4
· 95}
|8 {z
is a multiple of 4
94 − 5
| {z }
is a multiple of 4
is a multiple of 4
+
95 − 5
| {z }
is a multiple of 4.
is a multiple of 4
In the first line we used our last calculation: 94 − 5 is a multiple of 4. In the second line we
used the result of the preceding line: 95 − 5 is a multiple of 4.
It is quite obvious that we can continue in this way: in each line we use the result obtained
in the preceding line. Intuitively it is clear that this must imply that the number 9n − 5 is
a multiple of 4 for any n ∈ IN.
Although we are quite convinced now that the number 9n − 5 is a multiple of 4 for any
natural number, we are still not quite satisfied. The reason is that a proof that ends with
the observation that ’we can continue in this way’ is not precise enough. That is: in general,
2
36
The Principle of mathematical induction
it may not be clear when we can say that we can continue in that way. The notion of a proof
by Mathematical Induction pinpoints when we are allowed to complete a proof in that way.
This tool enables us to conclude that a given statement about natural numbers is true for
all the natural numbers. It is discussed in the following theorem. Its proof is based on the
intuitive idea that each nonempty subset of IN must have a smallest element.
THEOREM 1
THE PRINCIPLE OF MATHEMATICAL INDUCTION
Let, for each natural number n, P(n) be a statement which is either true or
false.
Then P(n) is true for all n ∈ IN, provided that
(a) P(1) is true
(basis for induction)
(b) for every k ∈ IN, if P(k) is true, then P(k + 1) is true. (induction step)
PROOF
Let S = {m ∈ IN| the statement P(m) is false}. It is sufficient to prove that if
S is nonempty, then (a) or (b) is violated. Let S be nonempty and let s be the smallest
number in S. Then (a) is clearly violated if s = 1. If s > 1, then s − 1 is a natural number
which is not in S. Now P(s − 1) holds and P(s) does not. This violates (b) and completes
the proof.
If you consider the above proof just like a logical play not adding much to the insight,
then this may stem from your conception of the natural numbers. If you think of these
as the number one and that each number say n has precisely one consecutive follower in
which case n + 1, then the Principle of Mathematical Induction makes sense by itself. Note
furthermore that at (b) there are actually an infinite number of implications. Writing these
out separately is impossible for obvious reasons and likewise this holds for proving them
separately. Precisely this was expressed by ’we can continue in this way’ in Example 2.
That is to present the proof more like an algorithm: if the theorem holds for any number,
say 4711, then it also holds for its successor, in this case 4712.
It might be clear that reasoning by induction i.e. on the basis of individual cases is different
from reasoning by Mathematical Induction. Theorem 1 shows that a Mathematical Inductive
reasoning leads to valid conclusions where we have seen earlier that this might not be the
case for an arbitrary inductive reasoning. In the sequel, unless otherwise stated, we will
refer to Mathematical Induction just by the word Induction.
In the next example we will use the Principle of Induction to reconsider the statement in
Examples 2 and 4.
2
37
The Principle of mathematical induction
EXAMPLE 5
In order to prove that, for all n ∈ IN, the number
9n − 5
is a multiple of 4, we introduce, for n ∈ IN, the statement P(n): 9n − 5 is a multiple of 4.
(1) First we show that the statement P(1) is true: 91 − 5 = 4 is a multiple of 4.
(2) Let k ∈ IN and assume that the statement P(k) is true: that is 9k − 5 is a multiple of 4.
Then
9k+1 − 5 = 9 · 9k − 5 =
· 9k}
|8 {z
is a multiple of 4
+
9k − 5
| {z }
is a multiple of 4.
is a multiple of 4
This proves that the statement P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
EXERCISE 1
Consider the expression 9n − 5 for natural numbers n. Define Q(n) as the
following statement
9n − 5 is not a multiple of 8.
(a) Prove by Induction that for all natural numbers n the number 9n − 5 is
not a multiple of 8.
(b) Find an alternative proof of the statement
for any natural number n the number 9n − 5 is a multiple of 4.
[Clue: Note that
9n − 5 = 9n − 1 − 4 = (9 − 1)(9n−1 + 9n−2 + · · · + 1) − 4.]
induction hypothesis
The assumption, in step (2), that P(k) is true is known as the induction hypothesis.
EXAMPLE 6
We will show that, for every natural number n,
1 + 2 + 3 + · · · + n = 21 n(n + 1).
For n ∈ IN we introduce the statement P(n): 1 + 2 + · · · + n = 12 n(n + 1).
(1) First we show that the statement P(1) is true: 1 =
1
2
· 1 · (1 + 1).
(2) Let k ∈ IN and assume that the statement P(k) is true. In other words, we assume that
1 + 2 + · · · + k = 12 k(k + 1).
Then
1
1
1
|1 + 2 +{z· · · + k} +(k + 1) = |2 k(k{z+ 1)} +k + 1 = 2 (k + 1)[k + 2] = 2 (k + 1)[(k + 1) + 1].
=
This proves that the statement P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
2
38
EXAMPLE 7
The Principle of mathematical induction
Although the above example shows a nice application of a proof with
Induction, for the sake of history as well as simplicity we like to present a nice geometrical
proof originating at least back to the Pythagorean school. The summation
n
X
k
k=1
can be represented by the following n × (n + 1) matrix


1
1
1 1
1 1
1 1
1
.
.
.
1 1
0
1


1

1


1

1


1

1


1

.


.

.


1
1
1
1
1 1
1 1
1 1
1 1
1 1
1 .
.
.
.
.
. 1
1 1
1 0
0 0
1
1 1
1 1
.
.
.
1
1 0
0 0
1
1
1 1
1 1
1
.
.
.
.
.
.
1
1 1
1 0
0 0
0 0
0 0
0 0
1
1
1
.
.
.
.
.
.
1
1 1
1 0
0 0
0 0
0 0
0 0
0 0
0 0
.
.
.
.
.
.
1 1
. .
0 0
. .
0 0
. .
0 0
. .
0 0
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
0
0 0
0 0
0
.
.
.
.
.
.
0
0 0
0 0
0


0

0


0

0


0

0


0

.


.

.


0
.
.
.
.
0 0
0 0
.
.
0 0
0 0
0
The number of cells containing 1 equals the number of cells containing 0 which is equal to
n
X
k.
k=1
Hence,
n
X
k = 21 n(n + 1).
k=1
This proof shows that modelling your problem in the right way may lead you to an easy
solution. Modelling problems is one of the main activities which you are going to encounter
in the rest of your Econometrics and Operations Research study.
EXERCISE 2
Prove that for all n ∈ IN,
1
1
1
1
n
(a)
+
+
+ ···+
=
.
1·2 2·3 3·4
n(n + 1)
n+1
(b) n < 2n .
EXAMPLE 8
We will prove that for any natural number n
n
X
i3 =
i=1
For n ∈ IN we introduce the statement P (n):
n 2
X
i .
i=1
n
X
i=1
i3 =
n 2
X
i .
i=1
2
39
The Principle of mathematical induction
(1) First we show that the statement P(1) is true: 13 = 12 .
(2) Let k ∈ IN and assume that the statement P(k) is true:
Then, using Example 6 twice leads to
k+1
X
i3 =
k
X
i3 + (k + 1)3 =
=
i=1
k 2
X
i .
i=1
k 2
X
i + (k + 1)3
2
1
+ (k + 1)3 = 41 (k + 1)2 k 2 + 4k + 4
2 k(k + 1)
= 14 (k + 1)2 (k + 2)2 =
=
i3 =
i=1
i=1
i=1
k
X
k+1
X 2
i .
1
2 (k
2
+ 1)(k + 2)
i=1
This proves that the statement P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
In the next theorem we consider a statement depending on n and a second variable x. We
will prove by induction that this statement is true for all n ∈ IN and all x > −1.
THEOREM 2
BERNOULLI’S INEQUALITY
If x > −1, then for every n ∈ IN
(1 + x)n ≥ 1 + nx.
PROOF
Let x > −1. For n ∈ IN we introduce the statement P(n): (1 + x)n ≥ 1 + nx.
We will prove by induction that P(n) is true for every n ∈ IN.
(1) First we show that the statement P(1) is true: (1 + x)1 ≥ 1 + 1 · x. Note that in fact
an equality holds.
(2) Let k ∈ IN and assume that the statement P(k) is true, in other words: (1+x)k ≥ 1+kx.
Then this assumption and the fact that 1 + x > 0 imply that
(1 + x)k+1 = (1 + x)k (1 + x) ≥ (1 + kx)(1 + x)
≥
= 1 + (k + 1)x + |{z}
kx2 ≥ 1 + (k + 1)x.
≥0
This proves that the statement P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN. Since the
x > −1 was arbitrarily chosen, the statement is true for all x > −1 and all n ∈ IN.
EXERCISE 3
Prove that, for all r 6= 1 and all n ∈ IN,
1 + r + r2 + · · · + rn =
1 − rn+1
.
1−r
2
40
The Principle of mathematical induction
There is a generalization of the Principle of Induction that enables us to conclude that a
given statement is true for all natural numbers sufficiently large.
THEOREM 3
Let n0 ∈ IN and let, for each natural number n ≥ n0 , P(n) be a statement
which is either true or false.
Then P(n) is true for all natural numbers n ≥ n0 , provided that
(a) P(n0 ) is true
(b) for every k ≥ n0 , if P(k) is true, then P(k + 1) is true.
A proof can be based on the original Principle of Induction by introducing, for each n ∈ IN,
the statement Q(n): ’P(n + n0 − 1) is true’.
EXERCISE 4
Prove that, for every natural number n ≥ 2,
√
1
1
1
√ + √ + · · · + √ > n.
n
1
2
3
SOME NOTATION
Closely related to proofs by induction are ’recursive definitions’. For example, for a natural
number n, the number n! (read: ’n factorial’) is defined as the product of the natural
numbers less than or equal to n:
n! = 1 · 2 · . . . · (n − 1) · n.
This can be expressed more precisely as follows:
1! = 1,
and for n > 1
n! = n · (n − 1)! .
It will be convenient to define 0! = 1.
EXERCISE 5
Prove that, for all n ∈ IN,
2n ≤ (n + 1)! .
EXERCISE 6
binomial coefficient
n
Let n ∈ IN. For k ∈ {0, 1, 2, . . . , n} the binomial coefficient
(read: ’n
k
choose k’) is defined by
n
n!
.
=
(n − k)!k!
k
n
n
n
n
(a) Find
,
,
and
.
0
1
n−1
n
(b) Prove that for n ∈ IN and for i ∈ {1, 2, . . . , n}
n+1
n
n
=
+
.
i
i−1
i
2
41
The Principle of mathematical induction
The foregoing relation gives rise to the following
configuration, known as ’Pascal’s Triangle’.
n
If we write down the binomial coefficient
as the (k + 1)st number in the nth row, we
k
obtain
1
1
1
1
1
1
1
..
.
3
4
5
6
2
3
1
6
10
15
1
4
10
20
1
5
15
1
6
1
..
.
As a consequence of the property described in part (b), a number not on one of the sides of
this triangle is the sum of the two numbers above it.
The following theorem shows that the numbers in (the nth row of) ’Pascal’s Triangle’ appear
as the coefficients in the expansion of (a + b)n . For instance, the expansion
(a + b)6 = 1 · a6 + 6 · a5 b + 15 · a4 b2 + 20 · a3 b3 + 15 · a2 b4 + 6 · ab5 + 1 · b6
corresponds with the numbers in the six row of the Triangle.
THEOREM 4
BINOMIUM OF NEWTON
For every a, b ∈ IR and for every n ∈ IN,
(a + b)n =
Binomial Formula
PROOF
n n
n n−1
n
n n
a +
a
b + ···+
abn−1 +
b .
0
1
n−1
n
Let a, b ∈ IR. We will prove this result by induction on n and introduce, for
n ∈ IN, the statement P(n):
(a + b)n =
n n
n n−1
n
n n
a +
a
b + ··· +
abn−1 +
b .
0
1
n−1
n
(1) The statement P(1) is true, because for n = 1 the both expressions on the left- and
right-hand side of the equality are equal to a + b.
(2) Let k ∈ IN and assume that the statement P(k) is true, in other words:
(a + b)k =
k k
k k−1
k
k k
b .
a +
a
b + ··· +
abk−1 +
k
0
1
k−1
Then (a + b)k+1 = (a + b)(a + b)k = a(a + b)k + b(a + b)k . By multiplying the formula
for (a + b)k by a and by b respectively, one obtains that a(a + b)k + b(a + b)k equals
2
42
The Principle of mathematical induction
k k+1
k k
k
k
a
+
a b + ··············· +
a2 bk−1 +
abk
0
1
k−1
k
k k
k k−1 2
k
k k+1
k
a b+
a
b + ··············· +
ab +
b
0
1
k−1
k
+
k k+1
k
k
k
k k+1
k
a
+
+
ak b + · · · · · · · · · +
abk +
+
b
0
1
0
k
k
k−1
|
{z
}
{z
}
|
k+1
k+1
=
=
1
k
By using Exercise 6 (b) and the fact that
formula can be written as
k
k+1
k
k+1
=
and
=
, this
0
0
k
k+1
k + 1 k+1
k+1 k
k+1
k + 1 k+1
a
+
a b + ·········+
abk +
b
.
0
1
k
k+1
This proves that the statement P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
Using the sigma notation, the Binomial Formula can be summarized as follows
n X
n n−i i
a b.
(a + b) =
i
i=0
n
EXERCISE 7
Let n ∈ IN. Prove that
n X
n
n
n
=
+ ··· +
= 2n ,
(a)
i
0
n
i=0
n
X
n
n
n
=
+
+ · · · = 2n−1 .
(b)
i
1
3
i=1
i odd
EXERCISE 8
Prove that, for every n ∈ IN,
n Y
2
< (n + 1)2 .
1+
i
i=1
2
43
The Principle of mathematical induction
Mixed exercises
EXERCISE 9
Prove that, for every n ∈ IN,
n
X
i2 = 61 n(n + 1)(2n + 1).
i=1
EXERCISE 10
Let x be a real number between −1 and 1. Prove that |xn | ≤ |x| for all
n ∈ IN.
EXERCISE 11
Prove that, for every n ∈ IN,
n
X
1
1
≤ 2 − n−1 .
i!
2
i=1
[Clue: use Exercise 5.]
EXERCISE 12
Prove that, for every n ∈ IN and for every k ∈ {1, . . . , n},
n
n−1
k
=n
.
k
k−1
EXERCISE 13
Find two formula’s for the sum of all elements in the triangle array
a11
a21
a31
..
.
am1
a22
a32
..
.
am2
a33
..
.
am3
..
.
· · · amm
[Clue: what is the formula for the sum of the k th row or the k th column?]
EXERCISE 14
A sequence of numbers t1 , t2 , t3 , . . . satisfies
t1 = 2
tn+1 = 2tn
for all n ∈ IN.
Prove that tn = 2n for all n ∈ IN.
EXERCISE 15
Find the mistake (if any) in the proof of the following statement: all natural
numbers are equal.
Proof: For n ∈ IN we introduce the statement P(n):
if a and b are natural numbers satisfying a ≤ n and b ≤ n, then a = b.
We will prove P(n) by induction on n.
(a) The basis is obvious.
(b) (induction step) Suppose P(k) holds for some natural number k. It is
sufficient to prove P(k +1). To do so let a and b be two natural numbers
both smaller than or equal to k + 1. It is sufficient to prove a = b. As a
and b are both not greater than k + 1 it follows that a − 1 and b − 1 are
numbers which are smaller than or equal to k. Hence, by the induction
hypothesis P(k) we have a − 1 = b − 1. This however yields the desired
result a = b.
2
44
EXERCISE 16
The Principle of mathematical induction
Assume that 0 < a < b. Prove that for all n ∈ IN,
an < b n .
EXERCISE 17
Find the mistake (if any) in the proof of the following statement: all students
have the same gender.
Proof: For n ∈ IN we introduce the statement P(n):
in a row of n students all have the same gender.
We will prove P(n) by induction on n.
(a) The basis is obvious.
(b) (induction step) Let k be a natural number. In order to prove P(k + 1),
consider a row of k + 1 students. Here is an abstract picture of this row
σ1 , σ2 , σ3 , σ4 , . . . , σk , σk+1 .
Now by the induction hypothesis the k first students have the same
gender. Without loss of generality suppose these are female. So σ1 , σ2 ,
σ3 , up to σk are female. Similarly the k last students σ2 , σ3 , σ4 up to
σk+1 have the same gender. As the k first are female the k last must be
female as well. This proves the statement.
3
3
(Infinite) sequences
45
(INFINITE) SEQUENCES
Sequences report values on discrete moments. Examples of these are common, for instance
the daily stock exchange indices such as Dow Jones, AEX or DAX or the daily temperature
reports. These sequences are registered mainly because we want to study their behavior. The
stock exchange indices may indicate how economy is doing and these temperature reports
may tell you that you only need light clothing when you visit central China in summer. This
behavior is mainly found by statistical tools which are not our focus here. We will study
infinite sequences in an abstract way and zoom in on the behavior of the tails of such a
sequence. If these tails are arbitrarily close to a certain value, which we call the limit of the
sequence, then we call the sequence convergent. Questions like does it converge and if so,
what is its limit are of our main interest here.
’Analysis’ is that part of mathematics in which ’limits’ play a central role. That is why
concepts like the continuity, the derivative and the integral of a function are in the core of
this book. Roughly speaking, a function is continuous if, on its graph, a small change in
the horizontal direction corresponds with a small change in the vertical direction. Before
we are going to discuss – in Chapter 5 – the continuity of functions, we first concentrate on
limits of sequences. The most important reason to do so is the fact that in the definition of
continuity of a function sequences play a central role.
Further, sequences are a relevant topic apart from continuity. In applied mathematics all
kinds of problems (such as finding roots of equations) are solved by using a so-called iterative
algorithm that generates, step by step, a sequence of approximations of the solution of the
problem. In defining such an algorithm and in investigating its properties, the knowledge of
(the theory of) sequences cannot be missed.
In Section 1 we introduce the formal definition of a sequence, whereas Section 2 dwells on
recursive sequences. In Section 3 convergency of a sequence will be introduced. As the
application of the definition of convergency is not easy, some general results on convergency
3
46
(Infinite) sequences
are discussed in Sections 4 and 5. For instance in the former section we show that a convergent sequence is bounded and herewith provide a quick and easy check for non-convergent
sequences. In the latter section we develop so-called arithmetic rules like the sum of two
convergent sequences is convergent. Similar results hold for all basic arithmetic operations
by which we are able to decompose some sequences in (sums of) other sequences for which
the definition of convergency might more easy to apply.
1
SEQUENCES
We begin with a number of examples. The first example is taken from the mathematics
of finance, whereas the second example is concerned with a simple macroeconomic model.
In the third example we consider an iterative algorithm generating approximations of the
inverse of a given positive real number.
EXAMPLE 1
(COMPOUND INTEREST)
Suppose you deposit K Euros in a bank account and the interest rate is p% per year. By
the end of the first year the amount (in Euros) will be
K +K ·
p p
.
=K 1+
100
100
Suppose this new amount is left in the bank for another year. Then after a second year the
total amount (in Euros) will have grown to
K 1+
p p p
p 2
+K 1+
·
=K 1+
.
100
100 100
100
Obviously, this computation can be continued to obtain the total amount after the third,
p
fourth, . . . year: each year the principal increases by the (growth) factor 1 +
. So, given
100
the constant interest rate of p%, the yearly growth of the original investment of K Euros
can be described by the sequence
K 1+
p p 2
p 3
,K 1 +
,K 1+
,...
100
100
100
Note that each number in this sequence is formed by multiplying its predecessor by the same
number.
EXAMPLE 2
(AN INVESTMENT)
Assume that consumers and producers spend a fixed part q ∈ (0, 1) of their income on
consuming. Here the parameter q is called the marginal propensity to consume.
Now an entrepreneur wants to invest K Euros in building a factory, buying machines and so
on. Those who get payed for their services (bricklayers, installers, engineers, . . .) are going
to spend the q part of their salary. So they spend qK Euros. According to our assumption,
those who receive this money are going to spend q · (qK) = q 2 K Euros. Because this process
continues, the original investment leads to a total spending (in Euros) of
K + qK + q 2 K + · · · ,
3
47
(Infinite) sequences
that is the sum of the elements of the sequence
K, qK, q 2 K, . . . .
According to Exercise 2.3, the sum of the first n terms of this sequence is equal to
K
1 − q n+1
.
1−q
K
.
1−q
If for instance q = 53 , an investment of one million Euros leads to an increase of the income
Later on in this chapter (Lemma 1), we will show that the ’infinite’ sum is equal to
of 2.5 million Euros (explaining the importance of investment).
EXAMPLE 3
(INVERTING A REAL NUMBER)
In this example we describe how some pocket calculators determine for a real number a > 0
(an approximation of) its inverse a−1 .
Obviously, a−1 is the (unique) solution of the equation ax = 1, so it is the nonzero solution
of the equation
ax2 = x ⇐⇒ x + ax2 = 2x ⇐⇒ x = 2x − ax2 .
Geometrically, this means that we have to determine the nonzero first coordinate of the
intersection of the line y = x and the parabola y = 2x − ax2 , which is the graph of the
function f : x → 2x − ax2 . Now a sequence of successive approximations of a−1 can be
generated as is suggested in the following figure:
y
y=x
t3 = f (t2 )
y = 2x − ax2
t2 = f (t1 )
t1
FIGURE 1
recursion formula
t2
t3
x
a−1
An iterative algorithm
Formally, this sequence is obtained as follows.
• First choose some (starting) point t1 > 0 satisfying at1 < 1.
• Then the numbers t2 , t3 , . . . are determined by using the recursion formula
tn+1 = f (tn ) ⇐⇒ tn+1 = 2tn − at2n
(n ∈ IN).
If we apply this algorithm for a = 0.25, Excel generates the following 10 approximations:
3
48
n
tn
1
2
1, 000000000000
1, 750000000000
3
4
2, 734375000000
3, 599548339844
5
6
3, 959909616970
3, 999598190297
7
8
3, 999999959637
4, 000000000000
(Infinite) sequences
9 4, 000000000000
10 4, 000000000000
TABLE 1
Ten approximations generated by Excel
Apparently, after 7 steps the approximations are correct to 12 decimal places.
If we want to prove that this iterative algorithm ’works’ for any a > 0, if we want to know
how ’fast’ the algorithm leads to an approximation which is ’close enough’ to a−1 , or if
we want to compare this algorithm with another one, in all these situations the (infinite)
sequence t1 , t2 , . . . of approximations must be considered.
Of course it is also possible to write down some examples of sequences without referring to
some application.
EXAMPLE 4
Some examples are:
• the sequence of the natural numbers:
1, 2, 3, 4, . . . ,
• the sequence consisting of the squared inverses of the natural numbers:
1
1, 14 , 19 , 16
,... ,
• the sequence of prime numbers
2, 3, 5, 7, 11, 13, . . . ,
• a so-called alternating sequence:
−1, 1, −1, 1, −1, . . . ,
• and a so-called constant sequence:
2, 2, 2, . . . .
term
The elements of a sequence are called the terms of the sequence. If we take t as the symbol
for the terms of a sequence, then t1 , t2 and, more generally, tn (n ∈ IN) represent the first,
second and nth term of the sequence, respectively.
3
49
(Infinite) sequences
The sequence
t1 , t2 , t3 , . . .
is also denoted as
tn
∞
n=1
.
Sometimes a sequence is defined by giving a formula for tn in terms of n. Thus the second
∞
sequence from Example 4 can also be described as the sequence tn n=1 , given by
tn =
or, shortly, as the sequence
EXERCISE 1
1
n2
1 ∞
.
n2 n=1
Represent the nth term of the first, fourth and fifth sequence from Example
4 by means of a formula.
So far we didn’t try to give a rigorous definition of a sequence. Such a definition is not
hard to give. The important point about a sequence is that for each natural number n
there is a real number tn . The concept of a function can be used to formalize this kind of
correspondence.
sequence
DEFINITION
A sequence is a function whose domain is the set IN of the natural numbers.
From the point of view of this definition, the nth term of a sequence should be designated
by t(n). Instead we will use – as observed before – the subscript notation tn .
Once we know that a sequence can be seen as a function, it is not surprising that the set of
all terms of the sequence,
{tn | n ∈ IN}
range of a sequence
is called the range of the sequence. The range of a sequence can be finite or infinite. The
ranges of the first three sequences from Example 4 are infinite, whereas the ranges of the
other two sequence are finite.
EXERCISE 2
Determine the range of the first, second, fourth and fifth sequence from Example 4.
A sequence, like any function, can be graphed. For a sequence tn
the points n, tn represented in a coordinate plane.
∞
n=1
its graph consists of
3
50
(Infinite) sequences
tn
4
the sequence 1, 2, 3, . . .
3
the sequence −1, 1, −1, . . .
2
1
1
2
3
4
n
5
−1
FIGURE 2
The graph of two sequences
EXERCISE 3
Calculate the first five terms of the sequence tn

1


tn = n

−1
n
∞
n=1
, given by
if n is even
if n is odd.
Also sketch (a part of) the graph of this sequence.
Note that the graph of a sequence can reveal all kind of characteristics of the sequence.
Sometimes the graph can be used to get an idea of how to prove some property of the
sequence. However it is not possible to give a proof by just referring to the graph. So the
graph is no more than just an aid.
EXERCISE 4
Let the sequence tn
∞
n=1
be defined by
2
2−
2(n − 1)
n
tn =
=
.
1
3n − 1
3−
n
We see that, for large n, tn is close to 23 . However, tn will be different from
2
3
for any n. Determine the number, say N , such that from this number on your
pocket calculator does not distinguish between tn , tn+1 and .6666666667.
∞
∞
Repeat the question for the sequences vn n=1 and wn n=1 defined by vn =
2 n + (−1)n
2(n2 − 1)
and
w
=
, respectively.
n
3n2 − 1
3n − 1
Knowing that one can build new functions from old ones by adding them, multiply them
(with a scalar), and so on, the same operations can be performed on sequences to obtain
new sequences from old ones.
3
51
(Infinite) sequences
DEFINITION
sum
scalar product
product
quotient
If un
and vn
n=1
∞
un + vn n=1
∞
c un n=1
∞
un · vn n=1
u ∞
n
vn
EXAMPLE 5
∞
n=1
∞
are sequences, then the sequence
∞
∞
is the sum of the sequences un n=1 and vn n=1
n=1
∞
is the scalar product of the scalar c and the sequence un n=1
∞
∞
is the product of the sequences un n=1 and vn n=1
∞
∞
is the quotient of the sequences un n=1 and vn n=1 provided
that vn 6= 0 for all n ∈ IN.
The sequence tn
∞
n=1
given by
tn =
2n + 3 · 5n
,
5n − 4
is in fact the result of manipulating a number of more elementary sequences.
If we divide all terms in the fraction by 5n , we obtain
2 n
+3
2n + 3 · 5n
5
n .
=
5n − 4
1 − 4 · 15
∞
Apparently, the sequence tn n=1 is the result of applying some of the above mentioned
∞
∞
operations to the sequences 15 n=1 and 52 n=1 and the constant sequences 3, 3, 3, . . . and
1, 1, 1, . . . .
EXERCISE 5
2
Determine the sum, the difference, the product and the quotient of the se∞
∞
quences (−1)n n=1 and (−1)n+1 n=1 .
RECURSIVELY DEFINED SEQUENCES
In Example 3 we discussed the sequence tn
∞
n=1
where the first term t1 was given and where
the terms t2 , t3 , . . . were (successively) determined by means of the recursion formula
tn+1 = 2tn − at2n .
We say that this sequence is recursively defined.
Typically, by giving an equation each term of such a sequence (except the first one) is
related to the foregoing one. Examples are the yearly values of a savings account with a
fixed rent and a fixed yearly deposit and the growth of a population of rabbits or other
species. Assuming that offspring is a fixed proportion of the mature part of the population,
the increase of the population measured at discrete moments can be described by means of
a recursion formula.
In mathematics recursive sequences play an important role approximating irrational numbers
√
like π or 2 (by rational numbers) or approximating the roots of an equation. By defining the
sequence of approximations recursively, one can not only show that this sequence converges
to the relevant number but also the speed of convergence can be determined. This latter
topic is not addressed to in this chapter.
3
52
recursively defined
DEFINITION
A sequence sn
∞
n=1
(Infinite) sequences
is recursively defined if the first term s1 is given and
if a function f exists such that
sequence
sn+1 = f (sn ) for n ∈ IN.
The terms of a recursively defined sequence (except the first one) can be generated by
applying the function f to the first term, then applying the function f to the result, etc.
For the sequence in Example 3 this function f is given by f (x) = 2x − ax2 .
EXERCISE 6
You are asked to construct a sequence of approximations of
√
2. In order
to describe the algorithm generating the sequence we consider the parabola
y = x2 , the line y = 2 and the point P on the parabola with coordinates
(2, 4).
y
4
P
y = x2
S
2
Q
tk+1
tk
FIGURE 3
√
2
2
x
A sequence of approximations of
√
2
For the first term of the sequence we choose some number smaller than 1,
say t1 = 21 . If for some k ∈ IN, tk is given, then the next term tk+1 must be
determined as follows. Let Q be the point on the parabola corresponding to
tk , so Q has coordinates (tk , (tk )2 ). Then the term tk+1 is the first coordinate
of the intersection S of the line segment joining P and Q with the line y = 2
(as indicated in Figure 3).
(a) Show that the recursive formula of the sequence constructed in this way
is given by
tk+1 =
2 + 2tk
.
2 + tk
(b) Explain (by using Mathematical Induction) why all terms of the sequence
are positive.
3
53
(Infinite) sequences
REMARK
In Example 16 we will prove that for all k
√
√
√
2 − 2
2 − tk+1 = ( 2 − tk
2 + tk
and that
√
2− 2
< 31 .
2 + tk
√
This means that the distance between 2 and tk+1 is less than a third of the
√
distance between 2 and tk . So, each term of the sequence brings us three
√
times closer to 2 than its previous term. Therefore the behavior of this
√
sequence is such that the larger the index k the more tk resembles 2.
Exercise 6 shows that an irrational number can be approached by – more formally: is the
limit of – a sequence of rational numbers. In fact, each irrational number can be approached
by such a sequence. In Chapter 4 such a sequence is constructed for the number e.
EXERCISE 7
Consider the sequence of approximations t1 , t2 , t3 , . . . of
1
3
defined by
t1 = 1
tn+1 =
tn + 1
4
for n ∈ IN.
The following table has been generated by using Excel.
n
tn
1
1, 000000000000
10 0, 333335876465
2
3
0, 500000000000
0, 375000000000
11 0, 333333969116
12 0, 333333492279
4
5
0, 343750000000
0, 335937500000
13 0, 333333373070
14 0, 333333343267
6
0, 333984375000
15 0, 333333335817
7
8
0, 333496093750
0, 333374023438
16 0, 333333333954
17 0, 333333333489
9
0, 333343505859
18 0, 333333333343
TABLE 2
n
tn
The first 18 approximations of the number
1
3
The error of an approximation tn is equal to the difference tn − 13 .
If, for instance, this error is smaller than 10−4 , then we know that the first 4
nonzero digits of the approximation are correct. In the table above you can
observe that this is true for n ≥ 8.
Determine for each k ∈ {1, 2, 3, 4, 5, 6, 7, 8} a natural number N such that the
error of the approximation tn is smaller than 10−k , whenever n ≥ N .
3
54
(Infinite) sequences
3
THE LIMIT OF A SEQUENCE
1 ∞
the first 6 terms
If we write down for the sequence 1 − 2
n n=1
24 35
0, 43 , 89 , 15
16 , 25 , 36 ,
then we observe that these terms get closer and closer to 1. If we consider the graph of this
sequence, then it is striking that the distance between the horizontal line at level 1 and the
(points corresponding with the) terms of the sequence becomes smaller and smaller.
tn
1
1
FIGURE 4
2
3
4
5
n
6
1 ∞
The graph of the sequence 1 − 2
n n=1
These observations are summarized by saying that the sequence has limit 1.
The purpose of this section is to formalize this concept.
1
as close to 1 as we please.
n2
If n ≥ 4, for instance, the difference between the nth term of the sequence and 1 is smaller
Note that, by choosing n sufficiently large, we can make 1 −
than 0.1. For n > 10 the difference is smaller than 0.01. This brings us to the following
provisional definition.
PROVISIONAL DEFINITION 1
We call a number ℓ the limit of a sequence tn
∞
n=1
if we can get the nth term tn of the
sequence as close to ℓ as we please by choosing n sufficiently large.
This definition is far from perfect because it leaves open the question of what the meaning
is of ’close to’. As a first step in improving this definition, let us observe that the distance
|tn − ℓ| measures how close tn is to ℓ. So we are going to modify the former definition in the
following way.
PROVISIONAL DEFINITION 2
We call a number ℓ the limit of a sequence tn
∞
n=1
small as we please by choosing n sufficiently large.
if we can make the distance |tn − ℓ| as
3
55
(Infinite) sequences
Next we note that ’as small as we please’ in fact means that we can make the distance
smaller than any given number. So our third attempt to get a proper definition will be.
PROVISIONAL DEFINITION 3
We call a number ℓ the limit of a sequence tn
∞
n=1
if we can make the distance |tn − ℓ|
smaller than any arbitrary number ε > 0 by choosing n sufficiently large.
Finally, we will try to make precise what is meant by the phrase ’by choosing n sufficiently
∞
1
large’. In order to do so we consider once more the sequence tn n=1 given by tn = 1 − 2
n
for n ∈ IN.
We can make the distance |tn − 1| smaller than 0.01 by choosing n in such a way that
|tn − 1| < 0.01 ⇐⇒ −
1
1
< 0.01 ⇐⇒ 2 < 0.01 ⇐⇒ n2 > 100 ⇐⇒ n > 10.
n2
n
So in this case ’sufficiently large’ means larger than 10.
Similarly, we can make the distance |tn − 1| smaller than 0.0001 by choosing n larger than
100.
More generally, we can make the distance |tn − 1| smaller than an arbitrary number ε > 0
by choosing n in such a way that
1
1
1
1
< ε ⇐⇒ 2 < ε ⇐⇒ n2 > ⇐⇒ n > √ .
n2
n
ε
ε
√
So, given an arbitrary number ε > 0, ’sufficiently large’ means larger than 1/ ε.
|tn − 1| < ε ⇐⇒ −
Apparently, ’the inequality |tn − 1| < ε is satisfied for n sufficiently large’ means that there
exists a number, say N , such that the inequality is satisfied whenever n > N .
Now we can give the formal definition of a limit.
limit
DEFINITION
A number ℓ is called the limit of a sequence tn
number N exists such that
∞
n=1
if for every ε > 0 a
|tn − ℓ| < ε,
whenever n > N .
∞
If a sequence tn n=1 has the limit ℓ, then we say that the sequence converges
to ℓ. We write this as
lim tn = ℓ or as ‘tn → ℓ as n → ∞’.
n→∞
convergent/divergent
A sequence which has a limit is called convergent. Otherwise the sequence
is called divergent.
We will make some observations concerning this definition.
3
56
(Infinite) sequences
In general, the number N as described in the definition of the limit will
REMARK 1
depend on ε.
In the definition we have written down ’the limit of the sequence’. This
REMARK 2
already suggests that a convergent sequence has precisely one limit. For a proof that the
limit of a convergent sequence is unique, we refer to Exercise 34.
REMARK 3
We may assume that the number N , which plays such a prominent role in
the definition of the limit, is a natural number. If computations lead to an N ∈
/ IN, then
one can replace this N by any natural number larger than N .
A GRAPHICAL REPRESENTATION OF THE LIMIT
A number ℓ is the limit of a sequence tn
∞
n=1
if for every ε > 0 a number N can be found
such that for all terms tn with n > N the following holds:
|tn − ℓ| < ε
or equivalently
ℓ − ε < tn < ℓ + ε
or equivalently tn ∈ (ℓ − ε, ℓ + ε).
Graphically this can be explained as follows. For an arbitrarily positive number ε we draw
two horizontal lines in a coordinate system, one at the level ℓ − ε and one at the level ℓ + ε.
The sequence is convergent with limit ℓ if it is possible to find a vertical line at level N
such that, beyond this line, all points of the graph lie between the horizontal lines at the
indicated levels (as represented in Figure 5).
tn
ℓ+ε
ℓ
ℓ−ε
n
FIGURE 5
EXAMPLE 6
N
A graphical explanation of the limit of a sequence
Now we can formally prove that
1
lim 1 − 2 = 1.
n→∞
n
Given a number ε > 0 we should produce a number N such that
whenever n > N .
1−
1
− 1 < ε,
n2
3
57
(Infinite) sequences
As was shown previously, this inequality holds if
1
n> √ .
ε
√
Thus it suffices to let N = 1/ ε. We can organize this in a formal proof as follows.
√
Let ε > 0. If we choose N = 1/ ε, then for all n > N
1−
1 This proves that lim 1 − 2 = 1.
n→∞
n
1
1
− 1 = 2 < ε.
n2
n
In the foregoing example we first explained how we determined the number N given ε > 0.
Then we summarized our conclusions in a formal proof. This formal proof however has the
drawback that we just write down how to choose the number N without explaining how we
did find this value for N . Therefore, from now on, we first determine the number N ; then
we show that this number ’works’.
EXAMPLE 7
We will prove, by using the definition, that the sequence
n − 1 ∞
2n
is convergent with limit 12 .
n=1
According to the definition, we must find for any positive number ε a number N such that
n−1
−
2n
1
2
< ε,
whenever n > N .
step 1
We start with choosing an arbitrary positive number ε
Let ε > 0.
step 2
We determine N
Next we consider the left-hand side of the foregoing inequality more closely:
n−1
−
2n
1
2
=
n
1
1
n−1
−
.
= −
=
2n
2n
2n
2n
Finally, we want to know for which values of n the fraction
1
is smaller than ε. Since
2n
1
1
1
< ε ⇐⇒ 2n > ⇐⇒ n > ,
2n
ε
2ε
1
.
2ε
We show that the number N ’works’
we choose N =
step 3
Then, for all n > N ,
n−1
−
2n
step 4
We draw the conclusion
n−1
This proves that lim
= 21 .
n→∞ 2n
1
2
=
1
< ε.
2n
3
58
EXERCISE 8
(Infinite) sequences
Consider the sequence
1 ∞
√
.
n n=1
(a) Solve the following inequalities
1
√ <
n
EXERCISE 9
1
20 ,
1
√ <
n
1
101
1
1
and √ < 4 .
n
10
1
(b) Let ε > 0. Find a number N such that √ < ε, whenever n > N .
n
1 ∞
(c) Investigate the convergence of the sequence √
.
n n=1
∞
Consider the convergent sequence tn n=1 defined by
1
n+7
1
(d) tn =
n!
(g) tn = (−1)n
(−1)n+1
n+7
n2
(f) tn = 2
n +7
sin(n)
(i) tn =
.
n
n
n+7
2n
(e) tn = 3
n +7
ln(n)
(h) tn =
n
(b) tn =
(a) tn =
9 n
11
(c) tn =
First try to find out what the limits of these sequences must be.
Then determine, for each sequence, a suitable N , where ε takes the values 0.1
and 0.01. Here N and ε are as in the definition of the limit of a sequence.
In addition you may derive a suitable N for arbitrary ε and herewith prove
that the sequence at hand converges to the found limit.
EXERCISE 10
Prove, by using the definition, that
1
(a) lim
=0
(b)
n→∞ n
lim 1 +
n→∞
(−1)n = 1.
n
In the following examples (and exercises) we are going to discuss more complicated limits.
EXAMPLE 8
We will prove, by using the definition, that
lim
n→∞
n
= 0.
n3 + 1
Let ε > 0. Note that, for all n ∈ IN,
n3
n
n
.
= 3
+1
n +1
(1)
n
is smaller than ε.
n3 + 1
One could try to solve this problem as in the foregoing example:
Now we want to know for which values of n the fraction
n
< ε ⇐⇒ n < εn3 + ε ⇐⇒ εn3 − n + ε > 0.
n3 + 1
Since we don’t know how to solve this inequality analytically, we should come out with
something different. So we reconsider the expression (1) more closely and observe that for
all n
n
1
n
n
= 3
< 3 = 2.
n3 + 1
n +1
n
n
(2)
3
59
(Infinite) sequences
Obviously
1
1
1
< ε ⇐⇒ n2 > ⇐⇒ n > √ .
n2
ε
ε
1
So for n > N = √ , the right-hand side of inequality (2) is smaller than ε. Since the
ε
left-hand side of the equality is smaller than the right-hand side, also this left-hand side is
smaller than ε (for n > N ).
More formally: for all n > N ,
1
n
< 2 < ε.
n3 + 1
n
This proves that lim
n→∞
EXAMPLE 9
n
= 0.
n3 + 1
We will prove that
n2 + 2n
= 0.
n→∞ n3 − 5
lim
Let ε > 0. The inequality we have to consider is
n2 + 2n
n2 + 2n
< ε.
−
0
<
ε
⇐⇒
n3 − 5
n3 − 5
By considering only n ≥ 2, we can skip the absolute value signs, since n3 − 5 will be positive.
Note that by choosing in advance N ≥ 2, all numbers n > N are at least 3.
Observe that
n2 + 2n
<ε
n3 − 5
leads to a very messy inequality. Instead of solving this inequality, we try to find an upper
bound for the relevant fraction by seeking for an upper bound for the numerator and a lower
bound for the denominator.
Indeed,
n2 + 2n ≤ n2 + n2 = 2n2 when n ≥ 2
and
n3 − 5 > 21 n3 when n ≥ 3.
Note that by choosing in advance N ≥ 2, all numbers n > N are at least 3.
Thus for n ≥ 3,
2n2
n2 + 2n
4
<
= .
1
3
3
n −5
n
2n
To make this at most equal to ε, we want n ≥ 4/ε. Thus there are two conditions to be
satisfied: n ≥ 3 and n ≥ 4/ε. We can accomplish this by letting
Then for all n > N ,
4
N = max 2, .
ε
4
4
4
n2 + 2n
−0 < <
≤ 4 = ε.
n3 − 5
n
N
ε
n2 + 2n
= 0.
n→∞ n3 − 5
This proves that lim
3
60
(Infinite) sequences
The way above to find N is more or less along a straight forward path. Using arithmetic rules
as depicted in Chapter 1 we can also find such N . Note that for n > 2, n3 − 5 > n3 − 8 6= 0
and that n3 − 8 = (n − 2)(n2 + 2n + 4) (use a long division). Now you may already have
the clue because for n > 2,
n2 + 2n
n2 + 2n
n2 + 2n
1
n2 + 2n
< 3
=
<
.
= 3
3
n −5
n −5
n −8
(n − 2)(n2 + 2n + 4)
n−2
So, taking N = 2 +
EXAMPLE 10
1
ε
will do. Please check this!!
(THE ROOT METHOD)
We investigate the convergence of the sequence tn
tn =
∞
n=1
, given by
√
√
n + 1 − n.
Note that
√
√
√
√ n+1+ n
n+1−n
1
√
n+1− n
tn =
√ =√
√ =√
√ .
n+1+ n
n+1+ n
n+1+ n
Apparently, lim tn = 0.
n→∞
In order to prove this formally, observe that for all n
|tn | = √
1
1
√ ≤√ .
n
n+1+ n
1
1
1
Let ε > 0. Since √ < ε ⇐⇒ n > 2 , we choose N = 2 . Then for all n > N ,
n
ε
ε
1
|tn | ≤ √ < ε.
n
This proves that lim tn = 0.
n→∞
EXERCISE 11
The sequence an
∞
n=1
is given by

1


3n
an =

3
n
if n is even
if n is odd.
Prove, by using the definition of a limit, that lim an = 0.
EXERCISE 12
Use the definition of a limit to prove that
2n − 1
=2
(a) lim
n→∞ n + 2
EXERCISE 13
n→∞
1 − n2
(b) lim
= − 12
n→∞ 2n2 + 1
(c)
lim
n→∞
r
n+1
= 0.
n2
Prove, by using the definition, that
2 n
3
n→∞
lim
= 0.
[Clue: an option could be to write
(1 + 12 )n ≥ 1 + 21 n.]
3
2
=1+
1
2
and use Bernoulli’s Inequality
3
61
(Infinite) sequences
Let tn
EXERCISE 14
∞
n=1
be a convergent sequence with limit ℓ. Let an
∞
n=1
and bn
∞
n=1
be sequences defined by an = tn − ℓ and bn = tn + ℓ, respectively. Prove that
the first sequence converges to 0 and that the second one converges to 2ℓ.
geometric sequence
In Exercise 13 we considered an example of a so-called geometric sequence. More generally,
this is a sequence of the form
r, r2 , r3 , . . . ,
where r is some number called the ratio of the sequence. It appears that the convergence
of a geometric sequence depends on the value of r. As in Exercise 13 one can prove the
following result.
LEMMA 1
The geometric sequence r, r2 , r3 , . . . converges (to zero) if |r| < 1.
EXERCISE 15
In a Dutch secondary school book a number ℓ is called the limit of a sequence
∞
tn n=1 if the following property is satisfied.
For any k ∈ IN a number Nk ∈ IR can be found such that
|tn − ℓ| < 10−k ,
whenever n > Nk .
Explain why this definition of convergence is equivalent with ours.
EXERCISE 16
∞
be a sequence and let ℓ be a number.
∞
Prove that the sequence tn n=1 converges to ℓ if and only if the sequence
Let tn
n=1
t437 , t438 , t439 , . . .
converges to ℓ.
tail of a sequence
Let tn
∞
n=1
be a sequence. The (N -)tail of this sequence is tN +n
tN +1 , tN +2 , tN +3 , . . .
∞
n=1
i.e. the sequence
Clearly in Exercise 16 we introduced the 436-tail of that sequence. Actually in that exercise
the number 436 was chosen arbitrarily and therefore the exercise shows that a sequence
∞
converges tn n=1 to limit value ℓ if, and only if, a tail of this sequence converges to ℓ.
There is an other way to describe convergence of a sequence which is based on this tail
notion and the distance between a set S and a real number x.
Let the distance between S and x be smaller than ε if for all s in S we have that |s − x| < ε.
∞
The following exercise now say that sequence tn n=1 converges to real number ℓ if, and
∞
only if, for every ε > 0 there are N -tails of tn n=1 whose distance to ℓ is smaller than ε.
Loosely speaking this means that convergence of a sequences boils down to being able to
find tails which are arbitrary close to the limit value.
3
62
EXERCISE 17
(Infinite) sequences
∞
converges to the number ℓ if and only if,
∞
for every ε > 0 there are N -tails of tn n=1 whose distance to ℓ is smaller
Check that the sequence tn
n=1
than ε.
4
PROPERTIES OF CONVERGENT SEQUENCES
In this paragraph we will discuss two of the many properties of convergent sequences. We
will show that a convergent sequence is bounded and that a convergent sequence with nonnegative terms has a non-negative limit.
THEOREM 1
Let tn
ℓ ≥ 0.
PROOF
Let tn
∞
∞
n=1
n=1
be a sequence which converges to ℓ. If tn ≥ 0 for all n ∈ IN, then
be a non-negative sequence converging to ℓ.
By means of a proof by contradiction we will prove that ℓ ≥ 0. So we assume that ℓ < 0.
∞
Since tn n=1 is convergent with limit ℓ, for any ε > 0 a number N exists such that
|tn − ℓ| < ε ⇐⇒ ℓ − ε < tn < ℓ + ε,
whenever n > N .
tn
N
n
ℓ+ε
ℓ
ℓ−ε
FIGURE 6
A negative limit
If we choose an ε > 0 as suggested in the figure, we obtain a contradiction because the terms
at the right-hand side of the vertical line at level N are negative!
So we choose ε = − 21 ℓ (which is positive!). Then a number N exists such that
|tn − ℓ| < − 12 ℓ =⇒ tn < ℓ + (− 21 ℓ) < 0,
whenever n > N . This however implies that tn < 0 for n > N , which contradicts the fact
that the terms of the sequence are non-negative. The statement ℓ < 0 is therefore not true,
in other words: ℓ ≥ 0.
Note that if all terms of a convergent sequence are positive, the limit of this sequence may
be zero.
3
63
(Infinite) sequences
EXERCISE 18
Give an example of a convergent sequence with positive terms of which the
limit is positive and an example of a convergent sequence with positive terms
of which the limit is zero.
EXERCISE 19
Let tn
∞
n=1
be a convergent sequence with limit ℓ.
(a) Assume that tn ≤ b for all n ∈ IN. Prove that ℓ ≤ b.
(b) Assume that all the terms of the sequence are contained in a closed
interval I. Prove that ℓ ∈ I.
Before we can prove that each convergent sequence is bounded, we need to introduce some
definitions.
DEFINITION
A sequence tn
∞
n=1
is called bounded above if a number u exists such that
tn ≤ u for all n ∈ IN. In this situation the number u is an upper bound of
the sequence.
We call a sequence tn
∞
n=1
bounded below if a number l exists such that
tn ≥ l for all n ∈ IN. In this situation the number l is a lower bound of the
sequence.
A sequence tn
bounded sequence
bounded.
∞
n=1
which is bounded above and bounded below is called
In Exercise 6, the sequence tn
EXAMPLE 11
t1 =
1
2
∞
n=1
(recursively) defined by
2 + 2tn
for n ∈ IN
2 + tn
was introduced. Figure 3 suggests that the number 0 is a lower bound of this sequence,
√
whereas 2 is an upper bound of the sequence. In the next chapter this will be proved
tn+1 =
formally.
EXERCISE 20
Prove that a sequence tn
exists such that
∞
n=1
is bounded if and only if a number m > 0
|tn | ≤ m,
for all n ∈ IN.
THEOREM 2
PROOF
Every convergent sequence is bounded.
Let tn
∞
n=1
be a convergent sequence with limit ℓ. Then for any ε > 0 a number
N exists such that
|tn − ℓ| < ε,
whenever n > N .
3
64
(Infinite) sequences
We choose ε = 1. Then a natural number N exists such that
|tn − ℓ| < 1,
whenever n > N . According to the Triangle Inequality, for all n > N ,
|tn | = |tn − ℓ + ℓ| ≤ |tn − ℓ| + |ℓ| < 1 + |ℓ|.
Now we choose
m = max{|t1 |, |t2 |, . . . , |tN |, 1 + |ℓ|}.
Then |tn | ≤ m for all n ∈ IN.
Note that a bounded sequence is not necessarily convergent. This will be illustrated in the
following example.
EXAMPLE 12
(THE ALTERNATING SEQUENCE)
We will prove that the alternating sequence (−1)n
tn
∞
n=1
, which is bounded, diverges.
1
n
−1
FIGURE 7
The alternating sequence is bounded but not convergent
Suppose that the sequence converges, say to ℓ. Then for any ε > 0 there exists a number N
such that
(−1)n − ℓ < ε,
whenever n > N . So for ε = 1 a number N exists such that
(−1)n − ℓ < 1,
whenever n > N . As a consequence, for odd n > N ,
| − 1 − ℓ| < 1 ⇐⇒ −1 < −1 − ℓ < 1 =⇒ ℓ < 0,
whereas for even n > N ,
|1 − ℓ| < 1 ⇐⇒ −1 < 1 − ℓ < 1 =⇒ ℓ > 0.
Since this is impossible, the sequence doesn’t have a limit.
3
65
(Infinite) sequences
An important consequence of Theorem 2 is, that a sequence which is not bounded cannot be
convergent. This property may be helpful if you want to show that a sequence is divergent.
This is illustrated in the following example.
The (geometric) sequence tn
EXAMPLE 13
∞
n=1
defined by
tn = 2 n
is divergent because it is not bounded above.
In fact we will show that no real number u can be an upper bound of the sequence.
This will be accomplished by proving that for any (positive) real number u there exists a(t
least one) term, say tk , such that tk > u.
So let u > 0. As
2n > u ⇐⇒ ln 2n > ln u ⇐⇒ n ln 2 > ln u ⇐⇒ n >
we choose a natural number k such that k >
ln u
,
ln 2
ln u
, then
ln 2
tk = 2k > u.
Hence, the sequence 2, 4, 8, 16, . . . is not bounded and divergent.
By using a similar approach as in the foregoing example, one can prove, more generally, that
∞
the sequence rn n=1 is divergent if |r| > 1.
EXERCISE 21
Prove that the following sequences are divergent:
1 − n2 ∞
n + 1 ∞
√
(a)
(b)
.
n
n n=1
n=1
EXERCISE 22
Let tn
EXERCISE 23
∞
be a sequence with positive terms which converges to 0.
1 ∞
is not bounded.
Prove that the sequence
tn n=1
∞
We consider a sequence an n=1 satisfying a1 = 2 and an+1 ≥ 3an for all
n=1
n ∈ IN.
5
(a) Prove that an ≥ 2 · 3n−1 for all n ∈ IN.
∞
(b) Prove that the sequence an n=1 is not bounded.
ELEMENTARY ARITHMETIC RULES FOR LIMITS OF SEQUENCES
We can formulate some arithmetic rules for convergent sequences. For example, the sum of
two convergent sequences is itself a convergent sequence, and the limit of the sum of the
sequences is the sum of the limits. Similar statements can be made about the product and
the quotient of convergent sequences. As a result, the arithmetic rules allow us to reduce
the study of the convergence of a complicated sequence to the study of the convergence of
less complicated sequences.
3
66
(Infinite) sequences
We want to prove that the sequence
EXAMPLE 14
n + √n ∞
√
n n n=1
converges to 0. We could use the definition of the limit, but instead we break down the
sequence into more simple sequences.
Since the nth term of the sequence can be written as
√
n+ n
1
1
√
= √ + ,
n n
n n
1 ∞
1 ∞
√
.
and the sequence
n n=1
n n=1
Since both sequences converge to zero (according to Exercises 8 and 10 (a)), one expects that
the given sequence is in fact the sum of the sequence
the original sequence converges to zero too. This is established in the following theorem.
THEOREM 3
ARITHMETIC RULES FOR LIMITS OF SEQUENCES
If a sequence sn
then
∞
n=1
converges to s and a sequence tn
∞
n=1
converges to t,
∞
(a) the sequence sn + tn n=1 converges to s + t,
∞
(b) the sequence sn tn n=1 converges to st,
s ∞
s
n
converges to (provided that tn 6= 0 for all n ∈ IN
(c) the sequence
tn n=1
t
and t 6= 0).
Sum Rule
Product Rule
Quotient Rule
PROOF
(a) In order to show that sn + tn → s + t as n → ∞, we need to make the
difference |(sn + tn ) − (s + t)| small. Using the Triangle Inequality, we get
|(sn + tn ) − (s + t)| = |(sn − s) + (tn − t)| ≤ |sn − s| + |tn − t|.
Now given any ε > 0, since sn → s as n → ∞, there exists a number N ′ such that
|sn − s| <
ε
,
2
whenever n > N ′ . Similarly, since tn → t as n → ∞, there exists a number N ′′ such that
|tn − t| <
ε
,
2
whenever n > N ′′ .
Thus if we choose N = max{N ′ , N ′′ }, then
|(sn + tn ) − (s + t)| ≤ |sn − s| + |tn − t| <
whenever n > N .
This proves that lim (sn + tn ) = s + t.
n→∞
ε ε
+ = ε,
2 2
3
67
(Infinite) sequences
(b) Using the Telescope Trick, we obtain the inequality
|sn tn − st| = |(sn tn − sn t) + (sn t − st)| ≤ |sn tn − sn t| + |sn t − st|
= |sn | · |tn − t| + |t| · |sn − s|.
According to Theorem 2, as the sequence sn
exists such that |sn | < m for all n ∈ IN.
∞
n=1
is convergent, it is bounded. So an m > 0
Letting M = max{m, |t|} > 0, we obtain the inequality
|sn tn − st| ≤ M |tn − t| + M |sn − s|.
Now given any ε > 0, there exist numbers N ′ and N ′′ such that
|tn − t| <
ε
,
2M
|sn − s| <
ε
,
2M
whenever n > N ′ , and
whenever n > N ′′ .
If we choose N = max{N ′ , N ′′ }, then
|sn tn − st| ≤ M |tn − t| + M |sn − s| < M
ε ε ε
ε +M
= + = ε,
2M
2M
2 2
whenever n > N .
This proves that lim sn tn = st.
n→∞
The proof of part (c) is (partly) discussed in Exercise 33.
EXERCISE 24
∞
∞
be convergent sequences with limit s and t, respec∞
tively. We consider for numbers a and b the sequence a sn + b tn n=1 .
Let sn
n=1
and tn
n=1
Prove that this sequence converges to as + bt.
EXAMPLE 15
In Example 5 we observed that the sequence tn
tn =
2n + 3 · 5n
,
5n − 4
∞
n=1
, given by
is the result of manipulating a number of more elementary sequences.
In fact, by writing the nth term as
2 n
+3
2n + 3 · 5n
5
n
=
n
5 −4
1 − 4 · 51
it appeared that the sequence is composed of the sequences
1 ∞
5 n=1
and
2 ∞
5 n=1
and the
constant sequences 3, 3, 3, . . . and 1, 1, 1, . . . . Note that the first two sequences converge to
0 and that the other two converge to 3 and 1, respectively.
3
68
According to the foregoing exercise,
2 n
+3→ 0+3= 3
5
and
1−4·
1 n
5
(Infinite) sequences
as n → ∞
→ 1 − 4 · 0 = 1 as n → ∞.
So, the application of part (c) of Theorem 3 leads to the conclusion that the sequence tn
converges to 3.
EXERCISE 25
6
∞
n=1
Prove that the following sequences are convergent and determine their limits.
2n − 1 ∞
1 − n2 ∞
(−1)n ∞
(a)
(b)
(c)
.
n + 2 n=1
2n2 + 1 n=1
2n − 1 n=1
OTHER ARITHMETIC RULES FOR LIMITS OF SEQUENCES
If we investigate the convergence of the sequence
sin n ∞
n
n=1
,
1 ∞
the first idea might be to write the sequence as the product of the sequence
and
n n=1
∞
the sequence sin n n=1 . However if we consider the second sequence for a few moments one
might suspect that this sequence is divergent. This makes it impossible to apply Theorem
3 (b).
Now, for all n ∈ IN,
−
sin n
1
1
≤
≤ .
n
n
n
1 ∞
and
One could say that the given sequence is ’sandwiched’ between the sequences −
n n=1
1 ∞
. Since both sequences converge to zero one might expect that the given sequence
n n=1
converges to zero too. This is established by the following lemma.
LEMMA 2
THE SANDWICH LEMMA
Let an
all n,
∞
n=1
and bn
∞
n=1
be sequences converging to ℓ and suppose that for
a n ≤ tn ≤ b n .
∞
Then the sequence tn n=1 converges to ℓ.
EXERCISE 26
Prove the Sandwich Lemma.
In the next example, the Sandwich Lemma will be used to show that the recursively defined
sequence introduced in Exercise 6 is convergent.
EXAMPLE 16
In Exercise 6 we investigated the (convergence of the) sequence tn
defined by
t1 =
tn+1 =
1
2
2 + 2tn
2 + tn
for n ∈ IN.
∞
n=1
,
3
69
(Infinite) sequences
The terms of this sequence can be seen as successive approximations of the number
√
2 as
was suggested in Figure 3.
√
√
2, by showing that the ’error’ 2 − tn we
√
make if we consider tn as an approximation of 2, converges to zero.
We will prove that this sequence converges to
Note that for n ∈ IN,
√
√
√
√
2 2 + 2 tn − 2 − 2tn
2 + 2tn
=
2 − tn+1 = 2 −
2 + tn
2 + tn
√
√ √
√
2 2 − tn + 2 tn − 2
2− 2 √
=
2 − tn .
=
2 + tn
2 + tn
As, for all n,
√
√
2− 2
2− 2
0<
<
< 31 ,
2 + tn
2
this implies that
√
2 − tn+1
| {z }
0<
error in step n+1
√
2− 2 √
=
2 − tn <
2 + tn
1
3
√
2 − tn ,
| {z }
error in step n
so that in each step of the iteration process the error is multiplied by a factor smaller than
1
3
∈ (0, 1). It follows that for all n
0<
√
2 − tn+1 <
1
3
√
2 − tn <
1 2
3
√
2 − tn−1 < · · · <
1 n
3
√
2 − t1 .
[To be perfectly correct, one should prove this relation by induction.]
n
√
2 − tn+1 = 0 or
By Lemma 1, lim 13 = 0. So the Sandwich Lemma implies that lim
n→∞
n→∞
√
lim tn = 2.
n→∞
Several operations on a sequence preserve the convergence of the sequence. In the next
exercises two examples are discussed.
EXERCISE 27
Let tn
∞
n=1
be a sequence with non-negative terms which converges to ℓ.
Then according to Theorem 1, ℓ ≥ 0.
√
√ ∞
tn n=1 converges to ℓ if
Prove that the sequence
(a) ℓ = 0
(b) ℓ > 0.
EXERCISE 28
∞
be a convergent sequence with limit ℓ.
∞
Prove that the sequence |tn | n=1 converges to |ℓ|.
Let tn
n=1
3
70
(Infinite) sequences
Mixed exercises
EXERCISE 29
EXERCISE 30
Use the definition to prove that the following sequences are convergent.
∞
n2 − 1
(a) an n=1 with an = 2
n +1
√
∞
n+1
(b) bn n=1 with bn =
n
∞
n+1
(c) cn n=1 with cn = √ .
n n
∞
Let xn n=1 be a convergent sequence with limit ℓ.
∞
Prove, by using the definition, that the sequence yn n=1 , given by
yn = xn +
xn
,
n
converges to ℓ.
EXERCISE 31
EXERCISE 32
Consider the sequences an
∞
n=1
and bn
∞
n=1
.
Give a proof or a counterexample for the following statements:
∞
∞
(a) If the sequence an n=1 converges and the sequence bn n=1 diverges,
∞
then the sequence an + bn n=1 diverges.
∞
∞
(b) If the sequence an bn n=1 converges, then the sequences an n=1 and
∞
bn n=1 are convergent.
Investigate the convergence of the sequence
p
∞
n2 + n − n n=1 .
EXERCISE 33
Let tn
∞
n=1
be a convergent sequence with nonzero terms and limit ℓ > 0.
(a) Prove that a number K exists such that
tn ≥
ℓ
,
2
whenever n > K.
1 ∞
1
converges to .
tn n=1
ℓ
1
2
[Clue: according to part (a),
≤ , for every n > K.]
tn
ℓ
(c) Prove part (c) of Theorem 3 for the case t > 0.
(b) Prove that the sequence
EXERCISE 34
Let tn
∞
n=1
be a sequence which converges to both ℓ and ℓ′ .
Prove, with the definition of limit, that ℓ = ℓ′ .
[Clue: Assume that ℓ 6= ℓ′ , choose ε = |ℓ − ℓ′ | and find a contradiction.]
3
71
(Infinite) sequences
EXERCISE 35
Consider the sequence tn
t1 = 1
tn+1 =
∞
n=1
defined by
3 + tn
1 + tn
(a) Prove that for any n, tn >
for n ∈ IN.
√
√
3 if and only if tn+1 < 3.
(b) For all n show that
√
√
√
tn+1 − 3
3−1
1− 3
√
≤
<
.
1 + tn
2
tn − 3
(c) Prove that tn
EXERCISE 36
∞
n=1
converges to
Prove that the sequence tn
to zero.
∞
n=1
√
3.
defined by tn =
n2 + 7n + π
converges
n3 + nπ + ln 7
∞
be a sequence with positive terms converging to 0.
∞
Prove that the sequence ln(xn ) n=1 diverges.
EXERCISE 37
Let xn
EXERCISE 38
Babs has invested α Euro at the stock exchange market. A friend of her, who
n=1
follows a bachelor in Econometrics and Operations Research (one should not
be surprised about this coincidence), has figured out that after n months the
expected value of her investment equals
cos xn i
2n2 h
1−
α Euro.
+π
n
n2
Here xn
∞
n=1
is an unknown sequence which mimics the hazarding fluctua-
tions in the stock exchanges. When may Babs expect that the value of her
investment is at least 1.99α Euro?
EXERCISE 39
Let tn
∞
n=1
be a sequence converging to ℓ.
Prove that for any k ∈ IN, the sequence tkn
∞
n=1
converges to ℓk .
72
3
(Infinite) sequences
4
4
73
Bounded sequences
BOUNDED SEQUENCES
In Chapter 3, we have discussed a number of results that can be used to prove that a sequence
converges. Unfortunately, most of these techniques depend on knowing (or guessing) what
the limit of the sequence is before we begin. Often it is desirable to show that a given
sequence is convergent without knowing the precise value of the limit. In this chapter we
obtain an important result that enables us to do so.
Furthermore, we will prove that it is always possible to extract from an arbitrary bounded
sequence another sequence which is convergent. In Chapter 6, this result plays an essential
role in proving some of the properties of continuous functions.
1
MONOTONE SEQUENCES
In this section we will consider sequences that are increasing or decreasing.
increasing sequence
DEFINITION
A sequence tn
A sequence tn
decreasing sequence
∞
n=1
∞
is called increasing if tn ≤ tn+1 for all n, that is:
t1 ≤ t2 ≤ t3 ≤ . . . ≤ tk ≤ tk+1 ≤ . . .
n=1
is called decreasing if tn ≥ tn+1 for all n, that is:
t1 ≥ t2 ≥ t3 ≥ . . . ≥ tk ≥ tk+1 ≥ . . .
A sequence is called monotone if it is an increasing or decreasing sequence.
monotone sequence
If all relevant inequalities in the foregoing definition are strict, we say that the sequence is
strictly increasing, strictly decreasing or strictly monotone.
In proving the monotonicity of a sequence one often considers the difference or the quotient
of two subsequent terms of that sequence.
EXAMPLE 1
is increasing.
The sequence an
∞
n=1
with for n ∈ IN
an =
n−1
n+1
4
74
Bounded sequences
Let n ∈ IN. We will show that an+1 ≥ an . Indeed,
an+1 − an =
n
n−1
n(n + 1) − (n − 1)(n + 2)
−
=
n+2 n+1
(n + 1)(n + 2)
=
2
n2 + n − n2 − n + 2
=
> 0.
(n + 1)(n + 2)
(n + 1)(n + 2)
The sequence bn
EXAMPLE 2
∞
n=1
with for n ∈ IN
bn =
2n
n!
is decreasing. Note that the terms of this sequence are positive.
bn+1
≤ 1. Indeed,
Let n ∈ IN. We will show that
bn
bn+1
bn
2n+1
2n+1
2
n!
(n + 1)!
=
=
=
×
≤ 1.
n
2
(n + 1)n! 2n
n+1
n!
The geometric sequence rn
EXAMPLE 3
∞
n=1
is monotone if r > 0 and r 6= 1. If
0 < r < 1, then the sequence is decreasing, if r > 1, then the sequence is increasing.
If r < 0, then the sequence is not monotone.
EXERCISE 1
Prove that the geometric sequence rn
∞
n=1
is increasing if r > 1.
If an increasing sequence is bounded above by u ∈ IR, then its graph is below the horizontal
line at level u. If we move down that horizontal line ’as far as possible’ we get a situation
as suggested in the following figure.
y
u
1
FIGURE 1
2
3
4
5
6
7
x
An increasing sequence which is bounded above
The convergence of the sequence seems to be ’inevitable’ in this situation.
THEOREM 1
MONOTONE SEQUENCE PROPERTY
(a) An increasing sequence which is bounded above, is convergent.
(b) A decreasing sequence which is bounded below, is convergent.
A proof of this theorem is beyond the basic level considered sufficient for this course.
4
75
Bounded sequences
EXERCISE 2
∞
(a) Prove that the sequence bn n=1 introduced in Example 2 is convergent.
2n
1
(b) Prove that
< for all n ≥ 6.
n!
n
(c) Prove that lim bn = 0.
n→∞
In Example 16 of Chapter 3, we showed how the Sandwich Lemma can be used to prove that
a recursively defined sequence is convergent. In the next example we discuss an alternative
method which is based on the Monotone Sequence Property. The method consists of four
steps.
(1) First, one shows that the sequence is bounded below and/or bounded above.
(2) Secondly, the monotonicity of the sequence is demonstrated.
(3) Then the convergence of the sequence is established.
(4) Finally, the existence of the limit is used to actually determine the limit.
EXAMPLE 4
In Example 16 of Chapter 3, we considered the sequence tn
by
t1 =
1
2
∞
n=1
defined
2 + 2tn
for n ∈ IN.
2 + tn
√
We proved that this sequence converges to 2. Here we will prove the same result by using
tn+1 =
the Monotone Sequence Property.
(a) We will show that the sequence is bounded. In fact we will prove by induction that
√
0 < tn < 2 for all n ∈ IN.
√
(1) Clearly, t1 = 21 ∈ (0, 2).
√
(2) Let k ∈ IN and suppose that 0 < tk < 2. Then
>0
tk+1
z }| {
2 + 2tk
=
>0
2 + tk
| {z }
>0
and
√
√
√
2 + 2tk √
2 + 2tk − 2 2 − tk 2
− 2=
2=
2 + tk
2 + tk
√
√
√
√
√
2− 2
2(tk − 2) − 2(tk − 2)
=
(tk − 2) < 0.
=
2 + tk
2 + tk
√
So 0 < tk+1 < 2.
√
According to the Principle of Induction, 0 < tn < 2 for all n ∈ IN.
∞
(b) We will prove that the sequence tn n=1 is increasing by showing that tn+1 − tn > 0
tk+1 −
for all n ∈ IN. Let n ∈ IN. Then
2 + 2tn − 2tn − t2n
2 − t2n
2 + 2tn
− tn =
=
> 0,
2 + tn
2 + tn
2 + tn
√
where the inequality follows from the relations tn > 0 and tn < 2.
√
∞
(c) Since the sequence tn n=1 is increasing and bounded above (by 2), the Monotone
tn+1 − tn =
Sequence Property implies that it is convergent.
4
76
Bounded sequences
(d) Let ℓ be the limit of the sequence: lim tn = ℓ. Then lim tn+1 = ℓ.
n→∞
n→∞
Since tn > 0 for all n, Theorem 3.1 implies that ℓ ≥ 0.
So, according to the Arithmetic Rules for limits of sequences,
2 + 2 lim tn
lim tn+1 =
n→∞
n→∞
2 + lim tn
=⇒ ℓ =
n→∞
2 + 2ℓ
=⇒ 2ℓ + ℓ2 = 2 + 2ℓ =⇒ ℓ2 = 2.
2+ℓ
√
Hence, ℓ = 2.
This proves that the given sequence converges to
√
2.
In the foregoing example we first proved in step (c) that the relevant sequence was convergent. After that we determined in step (4) its limit. Using the technique described in step
(d) to find the limit without proving its existence first can lead to problems. Consider
∞
for example the sequence tn n=1 defined by t1 = 2 and tn+1 = 2tn for n ∈ IN. If one
(illegally) applies the idea used in step (d), then ℓ = lim tn satisfies ℓ = 2ℓ, or ℓ = 0.
n→∞
However, according to Exercise 2.14, tn = 2n for n ∈ IN and in Example 3.13 we showed
that this sequence is divergent.
EXERCISE 3
The sequence tn
∞
n=1
is defined by
t1 = 4
tn+1 =
1
2
tn +
2
tn
for n ∈ IN.
Note that for n ∈ IN, tn+1 = f (tn ), where f is the function on (0, ∞) defined
2
by f (x) = 12 x +
. In Figure 2 it is shown how the first terms of this
x
sequence can be constructed by using the graph of this function f .
y
y=x
3
f
2
1
1
2
t3
FIGURE 2
3
t2
4
= t1
x
The first terms of the sequence
√
2 for every n ∈ IN.
∞
(b) Prove that the sequence tn n=1 is decreasing.
∞
(c) Prove that the sequence tn n=1 converges and determine its limit.
(a) Prove that tn >
4
2
77
Bounded sequences
THE NUMBER
e
In order to prove that the limit
1 n
lim 1 +
n→∞
n
∞
∞
exists, we introduce the sequences an n=1 and bn n=2 , defined by
1 n
an = 1 +
n
and
bn = 1 +
1 n
,
n−1
respectively. These sequences appear to be monotone. We will prove that the sequence
∞
bn n=2 is decreasing. Because the terms of this sequence are positive, the proof is complete
bn
if we can show that the quotient
of two subsequent terms is at least equal to 1. Indeed,
bn+1
for all natural numbers n > 1,
bn
bn+1
n n+1 n −1
1 n
1+
n+1 n − 1
n2
1
n−1
n−1
=
=
= n−
n + 1 n+1
1 n+1
(n − 1)(n + 1)
n
1+
n
n
n2 n+1 n − 1 n + 1 n − 1
1 n+1 n − 1
1+ 2
= 1+ 2
≥
=
2
n −1
n
n −1
n Bernoulli’s
n −1
n
Inequality
= 1+
1
n−1
n − 1
n
=
n n−1
= 1.
n−1 n
In the next exercise we ask you to prove that the sequence an
EXERCISE 4
∞
n=1
is increasing.
an
≥ 1 for all n > 1.
an−1
∞
Explain why this implies that the sequence an n=1 is increasing.
Prove, by using Bernouilli’s Inequality, that
In order to prove that the increasing sequence an
∞
n=1
is bounded above, we ’compare’ the
two sequences introduced at the beginning of this section.
For n > 1 we have
bn = 1 +
1 n
1 n > 1+
= an .
n−1
n
Since one can easily prove by induction that for all n > 1
an < b n ≤ b 2 ,
an
∞
n=1
is an increasing sequence which is bounded above by b2 = 4. So according to the
Monotone Sequence Property, the sequence converges. Its limit is denoted by e:
1 n
lim 1 +
= e.
n→∞
n
EXERCISE 5
(a) Prove that for all n ∈ IN,
1
bn+1 = 1 +
an .
n
∞
(b) Prove that the sequence bn n=2 is convergent and find its limit.
4
78
EXAMPLE 5
Bounded sequences
We will prove that
1 n
lim 1 −
= e−1 .
n→∞
n
Note that for any natural number n larger than 1,
As
1−
1
1
1
1 n h n − 1 in
1
= h
=
= h n in = h
1 in
1
1 in−1
n
n
1+
1+
1+
n−1
n−1
n−1
n−1
1
n−1
.
= h
1 in−1 n
1+
n−1
h
lim 1 +
n→∞
1 in−1
= e,
n−1
the Quotient Rule for limits of sequences implies that
1
= e−1 .
1 in−1
n−1
n − 1 ∞
Obviously, since the limit of the sequence
is 1, the Product Rule for limits of
n
n=1
sequences establishes that the limit we are investigating is equal to e−1 .
lim h
1+
n→∞
EXERCISE 6
(a) Prove that, for all n ∈ IN,
1+
2
1 1 1+
.
= 1+
n
n
n+1
(b) Evaluate the limit
2 n
.
lim 1 +
n→∞
n
3
SUBSEQUENCES
If we are given a sequence t1 , t2 , t3 , . . ., then sequences such as
t1 , t3 , t5 , t7 , . . .
and t1 , t4 , t9 , t16 , . . .
which can be obtained by deleting terms from the original sequence, are called subsequences
of the original sequence. Formally,
DEFINITION
Let tn
∞
n=1
be a sequence and let n1 , n2 , n3 , . . . be any sequence. of natural
numbers such that n1 < n2 < n3 < · · ·. The sequence
tn1 , tn2 , tn3 , . . .
subsequence
is called a subsequence of tn
∞
n=1
.
or
tnk
∞
k=1
4
79
Bounded sequences
EXAMPLE 6
The sequence t1 , t3 , t5 , t7 , . . . is a subsequence of the sequence tn
∞
n=1
. Take
n1 = 1, n2 = 3, n3 = 5, . . ., or more generally, nk = 2k − 1 for k ∈ IN. This sequence of
odd-numbered terms can be represented as
t2k−1
∞
k=1
.
The sequence t1 , t4 , t9 , t16 , . . . is a subsequence of the sequence tn
2
∞
n=1
too. Take n1 = 1,
n2 = 4, n3 = 9, . . ., or more generally, nk = k for k ∈ IN. This sequence can be represented
as
tk 2
EXERCISE 7
∞
∞
k=1
.
Consider the sequence (tn )n=1 and write down its subsequence consisting of
(a) the odd-numbered terms starting with t5 ,
(b) the terms numbered 2, 4, 8, 16, 32, . . ..
EXERCISE 8
Prove that the sequence 1, 12 , 13 , . . . is a subsequence of the sequence
1
1
1
1, √ , √ , √ , . . .
2
3
4
If we consider a subsequence of a convergent sequence, then this subsequence is convergent
with the same limit as the original sequence.
THEOREM 2
∞
If a sequence (tn )n=1 converges to ℓ, then any subsequence also converges to
ℓ.
PROOF
∞
Let (tnk )k=1 be a subsequence of the sequence xn
∞
n=1
.
Since n1 , n2 , n3 , . . . is a strictly increasing sequence of natural numbers, one can easily prove
by induction that nk ≥ k for all k ∈ IN.
Let ε > 0. Since lim tn = ℓ, there exists an N ∈ IR such that |tn − ℓ| < ε, whenever n > N .
n→∞
According to the observation made at the beginning of this proof, for all k > N
nk ≥ k > N,
so that |tnk − ℓ| < ε for all k > N .
This proves that lim tnk = ℓ.
k→∞
This theorem can be used effectively to prove that certain sequences are divergent. One
consequence of the theorem is the following result:
if a sequence has two subsequences which converge to different limits, then the
original sequence is divergent.
∞
For example, the alternating sequence (−1)n n=1 is divergent, since the subsequence
∞
(−1)2k−1 k=1 consisting of the odd-numbered terms converges to −1, while the subsequence
∞
(−1)2k k=1 consisting of the even-numbered terms converges to 1.
4
80
Bounded sequences
Another consequence of the theorem is the following result:
if a sequence has a divergent subsequence, then the original sequence is divergent.
∞
For example, the sequence n 2 − (−1)n
is divergent, since the subsequence
n=1
∞
2k 2 − (−1)2k
k=1
consisting of the even-numbered terms is in fact the divergent sequence 2k
EXERCISE 9
Consider the sequence
∞
k=1
.
1, 12 , 22 , 13 , 23 , 33 , 14 , 42 , 34 , 44 , . . .
(a) Determine a subsequence which converges to 1.
(b) Prove that the given sequence is divergent.
EXERCISE 10
∞
be a sequence such that
∞
(1) the sequence (−1)n bn n=1 converges and
Let bn
n=1
(2) bn ≥ 0 for every n ∈ IN.
∞
Prove that the sequence bn n=1 converges to 0.
[Clue: consider Theorem 1 in Chapter 3.]
4
THE THEOREM OF BOLZANO-WEIERSTRASS
A given sequence may or may not have a convergent subsequence. For instance, the sequence
1, 2, 4, 8, . . . does not have any convergent subsequence (every subsequence is unbounded
and hence divergent), while the alternating sequence has an infinite number of convergent
subsequences.
There are several conditions that guarantee the existence of a convergent subsequence.
For example, if the range of a sequence is finite, then this sequence has a convergent subsequence. This can be seen as follows.
∞
If the range of a sequence (tn )n=1 consists of a finite number of elements, then at least
one element in that range, say t123 , occurs infinitely many times as term of the sequence.
Consequently, by eliminating all terms not equal to the number t123 from the sequence, a
convergent subsequence can be obtained.
The next theorem gives another, more general, condition that guarantees the existence of
a convergent subsequence. The result plays a role in the proof of a number of important
results to be discussed in Chapter 6.
THEOREM 3
BOLZANO-WEIERSTRASS
Every bounded sequence has a convergent subsequence.
4
81
Bounded sequences
PROOF
Let xn
∞
n=1
be a bounded sequence. Then we can find an interval [a, b] containing
all the terms of that sequence.
bisection method
To extract a convergent subsequence, we shall use the so-called bisection method. First we
bisect the interval [a, b]. Then (at least) one of the two closed subintervals obtained in
this way, must contain an infinite number of terms of the sequence. Let [a1 , b1 ] be such an
interval. Note that the length of this interval is
b1 − a1 = 12 (b − a).
We next bisect the interval [a1 , b1 ] and we let [a2 , b2 ] be a closed subinterval containing an
infinite number of terms of the sequence. Then
b2 − a2 = 12 (b1 − a1 ) = 41 (b − a).
By continuing this process, we obtain a sequence of intervals [a, b] ⊃ [a1 , b1 ] ⊃ [a2 , b2 ] ⊃ · · ·
such that each subinterval [an , bn ] contains an infinite number of terms of the sequence
∞
xn n=1 and
n
bn − an = 12 (b − a).
Observe that the left-hand endpoints of these intervals, a1 , a2 , . . ., form an increasing se-
quence which is bounded above by b. Hence, by the Monotone Sequence Property, this
sequence converges, say to ℓ ∈ [a, b].
Since for all n
b n = an +
1 n
(b
2
− a),
the Arithmetic Rules for limits of sequences imply that lim bn = ℓ.
n→∞
∞
Finally, we shall prove that the sequence xn n=1 has a subsequence which converges to ℓ.
Since the interval [a1 , b1 ] contains an infinite number of terms of the sequence, we may choose
an n1 ∈ IN such that xn1 ∈ [a1 , b1 ]. Since the interval [a2 , b2 ] contains an infinite number
of terms of the sequence, there exists a natural number n2 > n1 such that xn2 ∈ [a2 , b2 ].
∞
Continuing in this fashion, we get a subsequence xnk k=1 such that xnk ∈ [ak , bk ] for all
k ∈ IN. Hence, for all k ∈ IN,
ak ≤ xnk ≤ bk ,
and it follows from the Sandwich Lemma that lim xnk = ℓ.
k→∞
compact interval
If we call an interval compact if it is both bounded and closed, then the foregoing result can
be formulated as follows:
if all terms of a sequence belong to a compact interval, say [a, b], then the sequence
has a convergent subsequence with a limit which also belongs to that interval [a, b].
EXAMPLE 7
A sequence may have a convergent subsequence although the sequence
is unbounded. The sequence 0, 1, 0, 2, 0, 3, . . . has the convergent subsequence 0, 0, 0, . . .
(consisting of the odd-numbered terms).
4
82
EXERCISE 11
Consider the sequence tn
∞
n=1
Bounded sequences
defined by
tn = n + (−1)n n.
(a) Prove that this sequence has a convergent subsequence.
(b) Write down two different strictly increasing subsequences of the se∞
quence tn n=1 .
4
83
Bounded sequences
Mixed exercises
EXERCISE 12
The sequence tn
∞
n=1
is defined by
t1 = 2
tn+1 =
EXERCISE 13
√
3 + 2tn
for n ∈ IN.
(a) Prove that 2 ≤ tn ≤ 3 for every n ∈ IN.
∞
(b) Prove that the sequence tn n=1 is increasing.
∞
(c) Prove that the sequence tn n=1 converges and find its limit.
Evaluate the limit
1 n
lim 1 +
.
n→∞
2n
EXERCISE 14
∞
be a sequence such that the subsequence a2n n=1 of the even∞
numbered terms and the subsequence a2n−1 n=1 of the odd-numbered terms
Let an
∞
n=1
both converge to ℓ.
Prove that the sequence an
∞
n=1
converges to ℓ.
(This exercise shows that when you weave together two convergent sequences
with the same limit, a sequence results also converging to this limit.)
EXERCISE 15
Let an
∞
n=1
be the sequence defined by
an =
n
X
(−1)k−1
k=1
k
.
∞
(a) Prove that the subsequence a2n n=1 is increasing and that the subse∞
quence a2n−1 n=1 is decreasing.
(b) Prove that a2 ≤ a2n < a2n+1 ≤ a1 for every n ∈ IN.
∞
∞
(c) Prove that the subsequences a2n n=1 and a2n−1 n=1 converge.
EXERCISE 16
(d) Prove that lim a2n = lim a2n−1 .
n→∞
n→∞
∞
(e) Prove that the sequence an n=1 converges.
The sequence tn
∞
n=1
is defined by
t1 = 0
tn+1 = αtn + β
for n ∈ IN,
where β > 0 and 0 ≤ α < 1.
β
for every n ∈ IN.
(a) Prove that 0 ≤ tn ≤
1−α ∞
(b) Prove that the sequence tn n=1 is monotone.
∞
(c) Prove that the sequence tn n=1 converges and determine its limit.
84
4
Bounded sequences
5
5
85
Continuity of a function
CONTINUITY OF A FUNCTION
Continuity of a function is a notion that pervades all techniques you will encounter in the
remainder of your studies and afterwards. Virtually all optimization techniques are based
either directly on continuity of the objective function, or on its differentiability, and hence
implicitly on continuity. Also approximation techniques in statistics and probability theory,
such as the central limit theorem, and the analysis of dynamical systems in growth theory
and macro economics require a good understanding of the notion of continuity.
In this chapter we focus explicitly on the definition of continuity of a function, whereas
in the next chapter a few applications are central. This gives you the opportunity to get
acquainted with the ‘official’ formulation of continuity in terms of sequences and to develop
your intuition for this definition. Furthermore, it gives you the opportunity to practice the
computational skills that you need to develop to prove the (dis)continuity of a function. Like
we just said, these skills are basic for a good understanding of for example approximation
techniques in non-linear optimization and the workings of statistical tests in econometric
models.
In this chapter we first present the formal definition of continuity of a function. Then we
discuss which basic functions are continuous and how we can construct new functions from
old ones (the Arithmetic Rules). In Section 3, two tests for continuity – the Sandwich
Lemma and the Glue Lemma – are central. Finally, we deal with an important alternative
definition of continuity that is often used in approximation techniques.
1
CONTINUITY OF A FUNCTION, A FORMAL DEFINITION
Let’s consider the functions g and h on [0, ∞) defined by

 1−x
√
g(x) = 1 − x

1
if x 6= 1
if x = 1

 1 −√x
and h(x) = 1 − x

1
Their graphs are represented in the following figure.
if x < 1
if x ≥ 1.
5
86
y
Continuity of a function
y
g
2
2
1
1
x
1
FIGURE 1
h
1
x
The graphs of the functions g and h
It appears that the graph of the first function has a ’hole’ at 1, whereas the graph of the
second one makes a ’jump’ at 1.
The function f on [0, ∞) defined by

 1−x
√
f (x) = 1 − x

2
if x 6= 1
if x = 1
and whose graph is given below
y
f
2
1
x
1
FIGURE 2
The graph of the function f
is ’well-behaved’ at 1: for values of x close to 1, the corresponding values f (x) are close to
f (1). For the function g however, we can find values of x arbitrarily close to 1 such that
the distance between g(x) and g(1) is at least 1. Similarly, for values of x close to 1 (but
smaller than 1) the distance between h(x) and h(1) is at least 21 . We say that the function
f is continuous at x = 1, whereas the functions g and h are discontinuous at x = 1. Note
well that in case of the discontinuous function h the values h(x) are even equal to h(1) for
values of x larger than 1. However, not all values h(x) are close to h(1), no matter how
close we choose x to be to 1.
EXERCISE 1
Give a sketch of the graph of the function k on [0, ∞) defined by

1−x


 1 − √x if x < 1
k(x) =

11
if x = 1

 2
1
if x > 1.
Is this function continuous at x = 1?
5
87
Continuity of a function
As a first intuition this geometric interpretation of continuity is fine: when a function f is
discontinuous at c, its graph makes a ’jump’ at c or it has a ’hole’ at c, continuity means
that no such a jump or hole occurs. However, we need a more precise (if you wish: more
fundamental) definition if you want to investigate the continuity at x = 0 of functions like
(
(
1
1
sin
if
x
=
6
0
x sin
if x 6= 0
g(x) =
and
h(x)
=
x
x
0
if x = 0
0
if x = 0.
We use our knowledge on the convergence of sequences to construct a more precise definition.
We present two examples to put the geometric intuition on its more formal footing, and to
show how a formal definition in terms of sequences should work. For the sake of the argument
we start with a simple function of which you already may suspect it is continuous.
EXAMPLE 1
We will show that the function f defined by
f (x) = x2
is continuous at x = 2.
So, we have to prove that, for points x close to 2, the corresponding values f (x) are close to
f (2) = 4. We can make this precise by using limits of sequences. To get an idea, consider
∞
an arbitrary sequence xn n=1 which converges to 2. Then the terms of this sequence get
closer and closer to 2. Hence the values f (xn ) should get closer an closer to f (2), that
is: we must show that the corresponding ’sequence of images’ f (x1 ), f (x2 ), . . . converges to
f (2) = 4. In other words, we should check that x21 , x22 , . . . converges to 4, This however is
an immediate consequence of the Product Rule for limits of sequences.
It is good to notice that we constructed a proof for an arbitrary sequence x1 , x2 , . . . converging to 2. This means that the values f (x1 ), f (x2 ), . . . converge to f (2) no matter how we
approach 2! The importance of this subtlety becomes evident when you take a look at the
discontinuity of the function h in Figure 1: when you approach 1 from the right, the values
h(xn ) do converge to h(1). However, there is a way, namely from the left, to approach 1
such that the corresponding values h(xn ) do not converge to h(1).
EXAMPLE 2
We will show that the function g on [0, ∞) defined by
g(x) =
is continuous at c, where c ≥ 0.
Consider an arbitrary sequence xn
∞
n=1
√
x
with nonnegative terms which converges to c.
We must show that the corresponding ’sequence of images’ g(x1 ), g(x2 ), . . . converges to
√
g(c) = c, i.e. we have to prove that
lim
n→∞
√
xn =
√
c.
According to Exercise 3.27 this is true. Hence, the function g is continuous at c.
5
88
Continuity of a function
Hopefully the above examples convinced you that a decent formal definition of continuity
ought to look as follows.
continuity of a function
DEFINITION
Let f be a function on an interval I. We say that f is continuous at c ∈ I if
∞
for any sequence xn n=1 in I converging to c,
lim f (xn ) = f (c).
n→∞
This is denoted by lim f (x) = f (c). When f is continuous at every c ∈ I we
x→c
say that f is a continuous function.
EXERCISE 2
Prove that the function f (x) = x2 + 2x + 6 is continuous at x = 3.
Although the line of reasoning in Examples 1 and 2 can be used frequently, it is only valid
for those cases in which all the hard work already has been done for sequences. In the next
example we will show you a more fundamental approach which is necessary because the
problem cannot be solved by simply using the Arithmetic Rules for limits of sequences.
The function f on the interval [0, ∞) defined by

 1−x
√
if x 6= 1
f (x) = 1 − x

2
if x = 1
is continuous at x = 1.
∞
Consider an arbitrary sequence xn n=1 with positive terms converging to 1. We assume
EXAMPLE 3
that the terms of the sequence are not equal to 1 (can you explain why this is allowed?).
We must show that the corresponding ’sequence of images’ f (x1 ), f (x2 ), . . . converges to
f (1) = 2, i.e. we have to prove that
lim
n→∞
Since lim 1 −
n→∞
1 − xn
= 2.
√
1 − xn
√ xn = 0, we cannot apply the Quotient Rule for limits of sequences.
However, after the following simplification (here we use that xn 6= 1)
√ √ 1 − xn 1 + xn
√
1 − xn
=
= 1 + xn ,
√
√
1 − xn
1 − xn
we obtain
√ 1 − xn
= lim 1 + xn = 2.
√
n→∞
n→∞ 1 −
xn
lim
As the sequence xn
at x = 1.
EXERCISE 3
∞
n=1
was arbitrarily chosen, this proves that the function f is continuous
Prove that the function h on I = [0, ∞) defined by
 √
 2 x − 4 if x 6= 4
h(x) =
x−4
1
if x = 4.
2
is continuous at x = 4.
5
89
Continuity of a function
A few remarks are in order. First of all, it is virtually impossible to underestimate the
importance and significance of a thorough understanding of the above definition for the
next few years of your studies. Continuity of a function is such a basic notion that it
pervades practically all areas of scientific activity, be it in the natural sciences, chemistry,
econometrics, finance or engineering. This is one of the reasons why we give you ample room
to practise with this definition (another one being that continuity is an excellent opportunity
to improve your computing skills).
Let us first get better acquainted with this formal definition in the next few examples and
exercises.
EXERCISE 4
Prove that the function q defined by
(
(2 + 3t)2 − 4
q(t) =
2t
6
if t 6= 0
if t = 0
is continuous at t = 0.
Of course we also want to prove that certain functions are continuous at any point c in their
domain. The same techniques we practised just now can be used to construct such proofs.
We first show in an example how this can be done.
EXAMPLE 4
Define the function f on IR by
f (x) =
x
.
x2 + 1
We prove that f is a continuous function. Take an arbitrary c ∈ IR. We have to prove that
∞
f is continuous at x = c. Let xn n=1 be a sequence converging to c. We have to show that
xn
c
lim
= 2
.
n→∞ x2
+
1
c
+1
n
Since xn → c as n → ∞, the Arithmetic Rules for limits of sequences imply that
lim
n→∞
As the sequence xn
∞
n=1
c
xn
= 2
.
x2n + 1
c +1
was arbitrarily chosen, this proves that f is continuous at c. Since
c was arbitrarily chosen, we may conclude that f is continuous.
EXERCISE 5
Define the function f on (0, ∞) by f (x) =
EXERCISE 6
Define the function t on IR by
t(y) =
1
. Prove that f is continuous.
x
5y 7 − y 2 + 15
.
y2 + 7
Prove that the function t is continuous.
EXERCISE 7
Let c be a point in an interval I. Let f be a function defined on I and suppose
that f is continuous at c. Suppose further that f (x) ≥ 0 for all x 6= c in I.
Prove that f (c) ≥ 0.
5
90
EXERCISE 8
Continuity of a function
√
3
x.
Consider the function f on (1, ∞) defined by f (x) =
Prove, for arbitrary c > 1, that f is continuous at x = c.
[Clue:
√
√
√
√ √
√
x − c = ( 3 x − 3 c) · (( 3 x)2 + 3 x 3 c + ( 3 c)2 ).]
EXERCISE 9
Consider the function f on IR defined by f (x) =
√
3
x.
Prove that f is continuous at x = 0. Prove that f is continuous.
2
ARITHMETIC RULES FOR CONTINUOUS FUNCTIONS
Now that we developed a good understanding of continuity of a function, we continue with
a discussion of continuous functions: which basic functions are continuous, and how can we
construct new continuous functions from old ones (the arithmetic rules).
The next theorem enables you to construct new continuous functions out of basic ones.
The continuity of these basic functions – the constant function c 7→ a, the power function
x 7→ xa , the goniometric functions sin, cos and tan, the exponential function x 7→ bx and
the logarithmic function x 7→ b log x (with a ∈ IR and b > 0) – will be discussed in this book.
THEOREM 1
ARITHMETIC RULES FOR CONTINUOUS FUNCTIONS
Let f and g be functions on an interval I containing c. If f and g are
continuous at c, then
(a) the sum function f + g is continuous at c,
Sum Rule
(b) the product function f · g is continuous at c,
f
is continuous at c
(c) the quotient function
g
for all x ∈ I).
Product Rule
Quotient Rule
(provided that g(x) 6= 0
We only prove part (a) and leave the proofs of the other parts to the reader.
∞
Assume that f and g are continuous at c ∈ I. Take an arbitrary sequence xn n=1 in I
PROOF
that converges to c. Then lim f (xn ) = f (c) and lim g(xn ) = g(c). So, according to the
n→∞
n→∞
Arithmetic Rules for sequences,
lim (f + g)(xn ) = lim f (xn ) + g(xn ) = lim f (xn ) + lim g(xn ) = f (c) + g(c)
n→∞
n→∞
n→∞
n→∞
= (f + g)(c).
This proves that the sum function f + g is continuous at c.
EXAMPLE 5
We will prove by induction that the basic function
gn : x → xn
is continuous for all n ∈ IN.
5
91
Continuity of a function
Let c ∈ IR. For n ∈ IN we introduce the statement P(n): the function gn : x → xn is
continuous at c.
(1) Obviously, the statement P(1) is true: the function g1 : x → x is continuous at c.
(2) Let k ∈ IN and assume that P(k) is true, that is: the function gk : x → xk is continuous
at c.
Then, for all x,
gk+1 (x) = xk+1 = xk · x = gk (x) · g1 (x),
that is: gk+1 is the product of the continuous functions gk and g1 . So, according to the
Product Rule, the function gk+1 is continuous at c.
This proves that P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
As c was arbitrarily chosen, this proves that the function gn is continuous for all n.
EXAMPLE 6
Let p be a polynomial function, that is
p(x) = a0 + a1 x + a2 x2 + · · · + an xn ,
for some n ∈ IN and a0 , a1 , . . . , an ∈ IR.
Then, according to the foregoing example and the Arithmetic Rules for continuous functions,
the polynomial function p is continuous.
EXAMPLE 7
We show that the function f on [0, ∞), defined by

 1−x
√
if x 6= 1
f (x) = 1 − x

2
if x = 1
is continuous. In Example 3 we already showed that f is continuous at x = 1.
Now let c ≥ 0 and c 6= 1. We show the continuity of f at c. According to Example 2,
√
the function x 7→ x is continuous at c. So the Sum Rule for continuous functions implies
√
that the function x 7→ 1 − x is continuous at c (here we use the fact that the constant
function x 7→ 1 is continuous). According to the foregoing Example, the function x 7→ 1 − x
is continuous at c. Finally, the Quotient Rule for continuous functions implies that the
function f is continuous at c.
Since c was arbitrarily chosen, the function f is continuous.
EXERCISE 10
(a) Prove that the function h introduced in Exercise 3 is continuous.
(b) Prove that the function q introduced in Exercise 4 is continuous.
EXERCISE 11
Let n ∈ IN. Prove that the function f defined on IR \ {0} by f (x) =
1
, is
xn
continuous.
The following theorem shows that we also construct new continuous functions when we take
composition of continuous functions.
5
92
THEOREM 2
Continuity of a function
CONTINUITY OF A COMPOSITION OF FUNCTIONS
Let f be a function on an interval I, and let g be a function on an interval J
containing c, such that g(J) ⊂ I. Suppose that g is continuous at c and that
f is continuous at g(c). Then the composite function f ◦ g is continuous at
c.
PROOF
Assume that g is continuous at c and that f is continuous at g(c). We prove that
the composite function f ◦ g is also continuous at c.
∞
Let xn n=1 be a sequence in J converging to c. As g is continuous at c,
lim g(xn ) = g(c).
n→∞
Because f is continuous at g(c) ∈ I and lim g(xn ) = g(c),
n→∞
lim f ◦ g (xn ) = lim f (g(xn )) = f (g(c)) = f ◦ g (c).
n→∞
As the sequence xn
at c.
∞
n=1
n→∞
was arbitrarily chosen, the composite function f ◦ g is continuous
Let g be a continuous function on an interval J, which is non-negative, in
√
other words, g(x) ≥ 0 for every x ∈ J. We prove that the function g on J, defined by
EXAMPLE 8
p
√
( g)(x) = g(x),
is also continuous.
The function f on [0, ∞), defined by f (x) =
√
x, is continuous. Furthermore, g(J) ⊂ [0, ∞).
By Theorem 2, applied to the functions f and g, the composite function f ◦ g on J, given
by
(f ◦ g)(x) = f (g(x)) =
is continuous.
EXERCISE 12
p
g(x),
Let g be a continuous function on an interval J. Define the function |g| on
J by |g|(x) = |g(x)|. Prove that the function |g| is continuous.
Another way to construct new continuous functions from old ones is by taking the inverse.
This will be discussed in the next chapter.
3
TWO OTHER TOOLS TO PROVE CONTINUITY
In this section we discuss two methods to prove the continuity of a function known under
the name of Sandwich Lemma and Glue Lemma.
5
93
Continuity of a function
THEOREM 3
SANDWICH LEMMA FOR FUNCTIONS
Let f be a function defined on an interval I containing c. Suppose that g
and h are functions defined on I such that
• g(c) = h(c),
• g and h are continuous at c, and
• g(x) ≤ f (x) ≤ h(x), for all x ∈ I.
Then f is also continuous at c.
The Sandwich Lemma states that, when a function f is ’sandwiched’ between two functions
g and h – in the sense that g(x) ≤ f (x) ≤ h(x) holds for all x close to the point c – with
g(c) = h(c), then continuity of g and h at c implies continuity of f at c.
y
h
f
g
x
c
FIGURE 3
The Sandwich Lemma
The strength of this Lemma lies in the fact that, even though f may be a relatively complicated function to analyze, we can (try to) choose g and h to be rather simple functions of
which we already know that they are continuous.
EXERCISE 13
Prove the Sandwich Lemma for functions.
EXERCISE 14
Prove that the function f on [0, ∞), defined by
 2 √
x + x
√
f (x) =
 x+ x
1
if x > 0
if x = 0
is continuous at x = 0.
EXAMPLE 9
We will prove that the function f defined by f (x) = sin x is continuous at
x = 0. The proof consists of two parts.
(a) First we show that for each real number x
| sin x| ≤ |x|.
We consider several cases. First suppose that 0 ≤ x ≤ π/2.
5
94
Continuity of a function
We construct an angle of x radians, as shown in the following figure.
y
1
P
x
x
O
S
FIGURE 4
A part of the unit circle
Then the length of the chord P S is 2 sin x, whereas the length of the arc P S is 2x. Since
a straight line is the shortest distance between two points, it follows that 0 ≤ 2 sin x ≤ 2x,
which implies that | sin x| ≤ |x|.
The inequality is trivial if x ≥ π/2, since in this case
| sin x| ≤ 1 < π/2 ≤ x = |x|.
Thus | sin x| ≤ |x| whenever x ≥ 0. If x ≤ 0, then −x ≥ 0 and
| sin x| = | − sin x| = | sin(−x)| ≤ | − x| = |x|.
(b) Now take g(x) = −|x| and h(x) = |x|. We know from (a) that g(x) ≤ f (x) ≤ h(x).
According to Exercise 12, g and h are continuous at c = 0, and g(0) = h(0) = 0. Hence, by
the Sandwich Lemma, f is continuous at x = 0.
EXERCISE 15
(a) Prove, by using Theorem 1.6, that for all numbers a and b,
| sin a − sin b | ≤ | a − b |.
(b) Prove that the sine function is continuous.
(c) Prove that the cosine function is continuous.
EXERCISE 16
The purpose of this exercise is to prove that the function f defined by
(
sin x
if x 6= 0
f (x) =
x
1
if x = 0
is continuous at x = 0.
(a) Let x ∈ (0, π/2). By comparing, in the figure below, the area of the
triangle OP Q, the area of the circular sector OP Q and the area of the
triangle ORQ, prove that
sin x ≤ x ≤ tan x.
5
95
Continuity of a function
y
R
1
P
x
O
x
Q
S
(b) Let x ∈ (0, π/2). Prove that
cos x ≤
sin x
≤ 1.
x
(c) Prove that the inequalities in part (b) also hold for x ∈ (−π/2, 0).
(d) Prove that f is continuous at x = 0.
EXERCISE 17
Consider the function f defined by
f (x) =
(
x sin
1
x
0
if x 6= 0
if x = 0.
(a) Make a sketch of the graph of f . By the way: ’sketch’ means that you
do not need to execute a full-blown analysis. It does mean though that
you are required to give a decent visual impression of the problematic
parts of the graph (in this case close to x = 0).
(b) Show that f is continuous at x = 0.
Another useful method for proving continuity of functions is the Glue Lemma. The Glue
Lemma states that, when we glue two continuous functions – whose respective domains
overlap at a common boundary point – together, the result is again continuous. For a proof
we refer to the Appendix.
THEOREM 4
GLUE LEMMA
Let g be a function defined on an interval [a, c] that is continuous at c, and let
h be a function defined on the interval [c, b] that is continuous at c. Suppose
further that g(c) = h(c). Then the function f defined by, for all x in [a, b],
f (x) =
is continuous at c.
g(x)
h(x)
if x ∈ [a, c]
if x ∈ (c, b]
5
96
Continuity of a function
Obviously the Glue Lemma derives its name from its content.
the function g
the glued function f
the function h
y
y
y
g(c) = h(c)
a
x
c
FIGURE 5
c
b
x
a
c
b
x
The Glue Lemma
We briefly discuss the use of this Lemma in a few applications which are presented in the
exercises below.
EXERCISE 18
Show that the function g defined by
3
x +2
if x < 1
g(x) =
x(x + 2) if x ≥ 1
is continuous at x = 1.
EXERCISE 19
Consider the function p defined by
t
if t > 1
p(t) =
2
2t − 1 if t ≤ 1.
Prove that p is continuous at t = 1.
EXERCISE 20
Consider the function u defined by
u(x) = max{−5x, 3x}.
Prove that u is continuous at x = 0.
4
CONTINUITY OF A FUNCTION, AN ALTERNATIVE DEFINITION
We now have found a workable definition of continuity of a function that quite well matches
our intuitive feeling of what continuity means. However, as you may have noticed, this
definition does not give any information about how close f (x) is to f (c).
In this section we derive an alternative definition of continuity that is more geared towards
usage in such cases. It is typically used to give hard bounds on how close x should be to c
in order to ensure that f (x) is within a certain pre-specified distance of f (c). In that sense
the alternative definition does more than just checking continuity. It is for example used to
give accurate approximations of values of a function such as square roots or logarithms, or
to estimate the maximum value of a function in optimization theory.
5
97
Continuity of a function
We first state the alternative definition and explain its meaning.
A proof can be found in the Appendix.
THEOREM 5
LINKING LIMIT LEMMA
Let f be a function on an open interval I containing c. Then f is continuous
at c if and only if for every ε > 0 there is an interval (c − δ, c + δ) around c
such that
|f (x) − f (c)| < ε,
for all x in the interval (c − δ, c + δ).
The Linking Limit Lemma states that, when a function f is continuous at a point c in its
domain, for any possible ε > 0 we can find an interval (c − δ, c + δ) around c such that for
every point x in this interval (c − δ, c + δ) the value f (x) is in the interval (f (c) − ε, f (c) + ε).
So indeed, points close to c get mapped to points close to f (c).
y
f (c) + ε
f (c)
f (c) − ε
c−δ
FIGURE 6
c c+δ
b
x
The alternative definition of continuity
Figure 6 displays the meaning of the alternative definition in detail. Given is a point c on
the horizontal axis, and its image f (c) on the vertical axis. Given a small real number ε > 0,
we can build the interval (f (c) − ε, f (c) + ε) around f (c) on the vertical axis. The Linking
Limit Lemma states that, given this interval on the vertical axis, we can find an interval
(c − δ, c + δ) around c on the horizontal axis such that, given any x in this interval, its image
f (x) is between the values f (c) − ε and f (c) + ε. This fact is visualized by the blue strip
in the figure. On the horizontal axis all values of x within the blue strip are in the interval
(c − δ, c + δ). These points x are mapped via the blue strip to values on the vertical axis that
are indeed between f (c) − ε and f (c) + ε. Note that δ has to be sufficiently small. If δ were
bigger, for example large enough to include b in the interval (c − δ, c + δ), then the images
of points x are no longer mapped within the interval (f (c) − ε, f (c) + ε). For example, the
image f (b) of b is clearly not within this interval (f (c) − ε, f (c) + ε) on the vertical axis.
5
98
Continuity of a function
You might have wondered why the Linking Limit Lemma is only formulated for open intervals. The only reason why we did this really is to make the formulation a bit easier. An
appropriately adjusted version of the Linking Limit Lemma can be formulated for other intervals than open intervals. Exercise 24 tells you how to adjust the statement of the Linking
Limit Lemma accordingly.
How does the alternative definition help us to prove continuity? Let us see how it works in
an easy example.
EXAMPLE 10
Consider again the quadratic function f : x → x2 , and take c = 2.
Let us first take for example ε =
1
10 .
Since we already know that f is continuous at x = 2,
according to the Linking Limit Lemma we must be able to find an interval (2 − δ, 2 + δ) such
that
|f (x) − 4| <
1
10 ,
for all x in the interval (2 − δ, 2 + δ). How do we compute this interval?
Well,
|f (x) − 4| <
1
10
⇐⇒ x2 − 4 <
1
10
⇐⇒ |x + 2| |x − 2| <
1
10 .
Now notice that, when we manage to find the interval (2 − δ, 2 + δ), we automatically have
|x+ 2| ≤ |x− 2|+ 4 < 4 + δ and |x− 2| < δ. So, when we take δ < 1, we know that |x+ 2| < 5.
Thus, the above inequality is satisfied when
|x − 2| <
This is it. We can now take δ =
1
50 .
1
50 .
Then |x + 2| < 5 and |x − 2| <
|f (x) − 4| = |x + 2| |x − 2| < 5 ·
1
50
=
1
50 .
Hence,
1
10 ,
and we have our desired level of approximation for the values of the function close to 2.
Of course there is nothing special about ε =
for an arbitrary ε > 0. Take δ = min{1,
1
5 ε}.
1
10 .
We can perform the same computations
Then for any x in the interval (2 − δ, 2 + δ) we
know that |x + 2| < 5 (because δ ≤ 1), and that |x − 2| < 51 ε (because δ ≤ 51 ε). Hence,
|f (x) − 4| = |x + 2| |x − 2| < 5 · 15 ε = ε,
and f is continuous at x = 2 by the Linking Limit Lemma.
EXERCISE 21
Consider the function g defined by
g(x) = x2 + 2x + 1.
Take ε =
1
1000 .
Construct an interval (2 − δ, 2 + δ) such that
|g(x) − g(2)| < ε,
for all x ∈ (2 − δ, 2 + δ). In general, use the Linking Limit Lemma to prove
that g is continuous at x = 2.
5
99
Continuity of a function
EXERCISE 22
Consider the function k defined by k(x) = x(x − 1)(x + 2).
Construct an interval (1 − δ, 1 + δ) such that
|k(x)| <
1
1000 ,
for all x ∈ (1 − δ, 1 + δ).
EXERCISE 23
Consider the function h on [0, ∞) defined by h(z) = z 3 + z. Take ε =
1
1000 .
Construct an interval (2 − δ, 2 + δ) such that
|h(z) − h(2)| < ε,
for all z ∈ (2 − δ, 2 + δ). Use the Linking Limit Lemma to prove that h is
continuous at z = 2.
EXERCISE 24
Let f be a function on an interval [a, b). Show the following variant on the
Linking Limit Lemma. The function f is continuous at a if and only if for
every ε > 0 there is an interval [a, a + δ) around a such that
|f (x) − f (a)| < ε,
for all x in the interval [a, a + δ).
EXAMPLE 11
Define the function r on the interval [0, ∞) by
 2√
 x x
√
if x > 0
r(x) = x + x

0
if x = 0.
We prove that r is continuous at x = 0. Note that, for x > 0,
√
x2
x2 x
√ =
√ < x2 .
|r(x) − r(0)| =
x+ x
1+ x
√
Let ε > 0. Take δ = ε. Then, according to the inequality displayed above, for any
x ∈ [0, δ),
√
x2 x
√ < x2 < δ 2 = ε.
|r(x) − r(0)| =
x+ x
Hence, by the Linking Limit Lemma, r is continuous at x = 0.
EXERCISE 25
Define the function h on [0, ∞) by h(x) =
√
x. Use the Linking Limit Lemma
to prove that h is continuous.
EXERCISE 26
√
10 up to 2 decimals.
√
Again consider the function h on [0, ∞) defined by h(x) = x. Show that,
We construct a rational approximation of
√
√
| x − 10| <
1
1000 ,
3
3
whenever x ∈ 10 − 1000
, 10 + 1000
. Use this information to argue that
√
√
3.162 approximates 10 up to two decimals, that is: 10 = 3.16 . . .
5
100
Continuity of a function
Mixed exercises
EXERCISE 27
Let f be a function defined on IR that is continuous at c. Assume that
f (c) > 0.
EXERCISE 28
Prove that there exists a δ > 0 such that f (x) > 0 for all x in the interval
c − δ, c + δ .
Construct a rational approximation of
√
4
3 up to 5 decimals. Argue that your
answer is indeed a good approximation.
[Clue 1: take a look at the hint at Exercise 8.
Clue 2: we can of course not forbid you to use a pocket calculator to aid you
to do your computations. All we ask you to do is to motivate your answer.]
EXERCISE 29
Consider the function m defined by
m(z) = min{z, z 3 − 3z 2 + 3z}.
Sketch the graph of m.
Show that m is a continuous function.
EXERCISE 30
Prove that the function f defined by
1
f (x) = √
x2 + 1
is continuous.
EXERCISE 31
Prove that the function z on (−1, ∞) defined by
p
z(y) = y 3 + 1,
is continuous at y = 2.
EXERCISE 32
We show that there exists an x ∈ IR such that x2 = −2.
Define the function f by
f (x) =
6
.
x2 + 4
6
= 3. So, 6 = 3(x2 + 4).
x2 + 4
This implies that −6 = 3x2 , and hence x2 = −2. Clearly the conclusion is
Take an x such that f (x) = 3. Then
nonsensical. Where is the mistake? Motivate your answer.
EXERCISE 33
The aim of this exercise is to produce an alternative proof of the continuity
√
at x = 1 of the function g on [0, ∞) defined by g(x) = 2 x.
√
(a) Prove that 2 x ≤ 1 + x for all x ≥ 0.
√
(b) Prove that x(3 − x) ≤ 2 x for all x ≥ 0.
√
√
[Clue: observe that x = 1 is a solution to x(3 − x) = 2, and use the
√
technique of long division to factorize the equation x(3 − x) − 2 = 0.]
(c) Use the Sandwich Lemma to prove that g is continuous at x = 1.
5
101
Continuity of a function
Appendix.
Proof of the Linking Limit Lemma.
A.
Suppose that for every ε > 0 there is an interval (c − δ, c + δ) around c such that
|f (x) − f (c)| < ε,
whenever x in the interval (c − δ, c + δ). Let xn
∞
n=1
be a sequence in the interval (a, b)
that converges to c. We show that lim f (xn ) = f (c).
n→∞
Let ε > 0. We have to show that there is an N such that, for all n > N ,
|f (xn ) − f (c)| < ε.
We construct this N as follows. Because of our assumption, we can obtain an interval
(c − δ, c + δ) around c such that |f (x) − f (c)| < ε for all x in this interval. As the sequence
∞
xn n=1 converges to c, we can take an N such that for all n > N ,
|xn − c| < δ.
Now take an n > N . Then |xn − c| < δ. Hence, by the choice of the interval (c − δ, c + δ),
|f (xn ) − f (c)| < ε.
Since n > N was chosen arbitrarily, this implies that lim f (xn ) = f (c).
n→∞
B.
We prove the contraposition. Suppose that the statement ‘for every ε > 0 there is an
interval (c − δ, c + δ) around c such that
|f (x) − f (c)| < ε
for all x in the interval (c − δ, c + δ)’ is not true. Then (verify this!!) there is a number
larger than zero, let us say ∆ > 0, such that, for any interval (c − δ, c + δ) around c there is
a point x in the interval (c − δ, c + δ) such that
|f (x) − f (c)| ≥ ∆.
Now we use this observation to construct a sequence xn
∞
n=1
as follows. First take δ = 1.
Then there is a point x1 in the interval (c − 1, c + 1) such that
|f (x1 ) − f (c)| ≥ ∆.
Next, take δ = 12 . Then there is a point x2 in the interval (c − 12 , c + 12 ) such that
|f (x2 ) − f (c)| ≥ ∆.
5
102
Et cetera. In general we take δ =
1
n.
Continuity of a function
Then there is a point xn in the interval (c − n1 , c + n1 )
such that
|f (xn ) − f (c)| ≥ ∆.
Thus we get a sequence x1 , x2 , . . . in the interval (a, b). Since xn ∈ (c − n1 , c + n1 ) for all n,
it is clear that this sequence converges to c. However, since for any n,
|f (xn ) − f (c)| ≥ ∆,
it is also clear that the sequence f (x1 ), f (x2 ), . . . of images does not converge to f (c). Hence,
the function f is not continuous at c.
Proof of the Glue Lemma.
We only consider the case where a < c < b (the other cases
are trivial, but obnoxious).
Let ε > 0. Since g is continuous at c, we know – by Exercise 24 – that there is an interval
(c − δg , c] around c such that
|g(x) − g(c)| < ε,
for all x in (c − δg , c]. Similarly we can take an interval [c, c + δh ) around c such that
|h(x) − h(c)| < ε,
for all x in [c, c + δh ). Take δ = min{δg , δh }. Take an arbitrary x in the interval (c − δ, c + δ).
If x < c. Then x ∈ (c − δg , c], and
|f (x) − f (c)| = |g(x) − g(c)| < ε.
If x ≥ c. Then x ∈ [c, c + δh ), and
|f (x) − f (c)| = |h(x) − h(c)| < ε.
In both cases we find that |f (x) − f (c)| < ε. Hence, since x was chosen arbitrarily in the
interval (c − δ, c + δ), we know by the Linking Limit Lemma that f is continuous at c.
6
6
Continuous functions
103
CONTINUOUS FUNCTIONS
In this chapter, some important properties of continuous functions defined on a compact
(that is: closed and bounded) interval are developed. We show for example that each such
function
– has a global maximum and global minimum, and
– attains every value between the minimal and maximal value.
These properties are based on the Bolzano-Weierstrass Theorem.
Further, we consider continuous functions on a compact interval which are invertible.
1
PROPERTIES OF A CONTINUOUS FUNCTION ON A COMPACT INTERVAL
The main objective of this section is to explore some of the ’global’ properties of continuous
functions that are defined on a closed and bounded interval. Such intervals are called
compact. Specifically, we shall prove two important theorems for such functions:
(a) they assume both a minimum and a maximum value on the interval, i.e. there exists
numbers c and d in that interval such that f (c) ≤ f (x) for all x in that interval and
f (d) ≥ f (x) for all x in that interval.
(b) they assume every value between the minimal and maximal value of the function.
The geometrical intuition of continuity is that the graph of a continuous function on a
bounded interval must be an unbroken curve with no jumps. The following result gives a
kind of justification for this intuition.
THEOREM 1
INTERMEDIATE VALUE THEOREM
Let f be a continuous function on an interval [a, b] and suppose that f (a) 6=
f (b). Then for any number t between f (a) and f (b) there exists a τ ∈ [a, b]
such that f (τ ) = t.
PROOF
We assume that f (a) < f (b).
Our proof is based on the bisection method as applied in the proof of the Bolzano-Weierstrass
Theorem. So first we bisect the interval [a, b] and we consider the value of the function f at
6
104
the midpoint 21 (a + b) of this interval. If f
1
2 (a + b)
Continuous functions
= t, then we stop and take τ = 21 (a + b).
Otherwise we consider one of the two closed half-intervals obtained by the bisection:
– if f 21 (a + b) > t, we consider the interval [a1 , b1 ] at the left-hand side of the midpoint;
– if f 12 (a + b) < t, we consider the interval [a1 , b1 ] at the right-hand side of the midpoint.
Note that in both cases f (a1 ) < t and f (b1 ) > t.
Now in the next step we repeat the foregoing with the interval [a1 , b1 ] instead of [a, b].
By continuing in this fashion, we obtain either a solution of the equation f (x) = t with
∞
∞
a < x < b or we find sequences an n=1 and bn n=1 such that for all n ∈ IN
(1)
an ≤ an+1 < bn+1 ≤ bn ,
n
bn − an = 12 (b − a), and
(2)
(3)
f (an ) < t < f (bn ).
As in the proof of the Bolzano-Weierstrass one can show that these conditions imply that
∞
the sequence an n=1 converges to a point in the interval [a, b], say to τ . Similarly, the
∞
sequence bn n=1 converges to a point in the interval [a, b], say to σ. Then, according to the
Arithmetic Rules for limits of sequences,
1 n
(b
n→∞ 2
σ − τ = lim bn − lim an = lim (bn − an ) = lim
n→∞
n→∞
n→∞
Hence, σ = τ . Since the function is continuous at τ ,
− a) = 0.
f (τ ) = lim f (an ) ≤ t ≤ lim f (bn ) = f (σ) = f (τ ).
n→∞
n→∞
So f (τ ) = t, which completes the proof.
Figure 1 shows a typical situation. The points a, f (a) and b, f (b) are on opposite sides
of the horizontal line at level t. The graph of the function f must cross this line in order to
go from one point to the other.
y
f (b)
f
y=t
t
f (a)
a
FIGURE 1
τ
b
x
A continuous function takes on all values between any two of its values
The fact that a continuous function on a compact interval attains a global maximum and a
global minimum is the content of the following Theorem of Weierstrass. Its proof is based
on the Bolzano-Weierstrass Theorem.
6
105
Continuous functions
THEOREM 2
WEIERSTRASS
Let f be a continuous function on an interval [a, b]. Then the function f
assumes both a maximum and a minimum value on the interval. That is,
there exist points c and d in the interval [a, b] such that
f (c) ≤ f (x) ≤ f (d)
for all x ∈ [a, b].
PROOF
We only prove the existence of a point d as described in the theorem.
(a) First we will prove that the range
S = {f (x)| x ∈ [a, b]}
of the function f is bounded above.
If not, then no natural number is an upper bound of the set S. Hence, for each n ∈ IN,
we can find an element xn ∈ [a, b] such that f (xn ) > n. In this way we obtain a sequence
∞
xn n=1 that is contained in the interval [a, b]. Since the sequence is bounded, the Bolzano∞
Weierstrass Theorem implies the existence of a convergent subsequence, say xnk k=1 . Then,
∞
by the continuity of the function f , the sequence of images f (xnk ) k=1 is convergent too.
∞
However, it is quite clear that every subsequence of f (xn ) n=1 is unbounded and hence
cannot converge to anything. This contradiction shows that the set S is bounded above.
(b) Let U be the set of upper bounds of the range S of f , that is:
u ∈ U ⇐⇒ u ≥ f (x) for all x ∈ [a, b].
Claim: there exist sequences y1 , y2 , y3 , . . . in S and u1 , u2 , . . . in U such that for every n,
u n − yn =
1
.
n
In order to construct such sequences, choose some y ∈ S and consider the unbounded
sequence
y, y + 1, y + 2, . . .
As S is bounded there is a k ∈ {0, 1, . . .} such that
y1 = y + k ∈ S
u1 = y + k + 1 ∈
/ S.
Then, according to the Intermediate Value Theorem, u1 ∈ U. Furthermore, u1 − y1 = 1.
By considering the sequence y, y + 21 , y + 1, y + 1 21 , . . . one can find, in a similar way, a y2 ∈ S
and u2 ∈ U with u2 − y2 = 21 .
By continuing this process in the obvious way, one finds two sequences with the property
mentioned in the claim.
6
106
Continuous functions
(c) As y1 , y2 , . . . is a sequence in the range S of the function f , for any n there exists an
xn in [a, b] with f (xn ) = yn . By Bolzano Weierstrass we may assume that the sequence
x1 , x2 , . . . converges, say to d. Then, by continuity of f , the sequence y1 , y2 , . . . (of images)
1
converges to f (d). As un = yn + , the sequence u1 , u2 , . . . also converges to f (d).
n
Now let x ∈ [a, b]. As un is an upper bound of the range S of f , un ≥ f (x) for all n. So
f (d) = lim un ≥ f (x).
n→∞
The conditions closedness and boundedness of the interval are essential. For instance, the
1
function f : x → on the interval (0, 1] and the function g: x → x on the interval [0, ∞) are
x
continuous. Yet they do not attain a maximum value.
bounded function
Note that we proved in part (a) of the foregoing proof that the range of a continuous function
on a compact interval is bounded above. In a similar way one can prove that the range is
bounded below too. We say in this situation that the function is bounded.
EXERCISE 1
Let f be a function on an interval I.
Prove that the function is bounded if and only if an m > 0 exists such that
|f (x)| ≤ m for all x ∈ I.
EXERCISE 2
Let f be a positive, continuous function on an interval [a, b].
Prove that a number m > 0 exists such that f (x) ≥ m for every x ∈ [a, b].
If we combine the Theorem of Weierstrass and the Intermediate Value Theorem we obtain
the following result.
COROLLARY
The range of a continuous function defined on a compact interval is a compact interval.
PROOF
According to the Theorem of Weierstrass, for a continuous function f defined on
a compact interval I there exists numbers c and d in I such that
m = f (c) ≤ f (x) ≤ f (d) = M
for all x ∈ I. Since for every t ∈ [m, M ] an x ∈ I exists such that f (x) = t, the range of f
is f (I) = [m, M ].
One of the main applications of the Intermediate Value Theorem is the existence of solutions
of equations. This topic will be discussed in the following examples.
6
107
Continuous functions
EXAMPLE 1
We consider the equation
x3 − x2 + 2x + 6 = 0.
In order to prove that this equation has at least one solution, we introduce the function
f : x → x3 − x2 + 2x + 6. Then f (0) = 6 and f (−2) = −10. Further, by Example 5.6, the
restriction of the function f to the interval [−2, 0] is continuous. Finally, the number 0 lies
between f (−2) = −10 and f (0) = 6. Hence, according to the Intermediate Value Theorem,
there exists a number z ∈ (−2, 0) such that f (z) = 0, that is: z 3 − z 2 + 2z + 6 = 0.
EXAMPLE 2
Let t be any positive real number and let n ∈ IN. We will show that the
equation
xn = t
1
has a unique positive solution, which will be denoted by t n or
√
n
t. In order to do so we
introduce the function f : x → xn , where x ≥ 0.
(a) First we deal with the existence of a solution.
If 0 < t < 1, we consider the restriction of the function f to the interval [0, 1]. This
restriction of f is continuous and the number t lies between f (0) = 0 and f (1) = 1. Hence,
according to the Intermediate Value Theorem, there exists a number τ ∈ (0, 1) such that
f (τ ) = t ⇐⇒ τ n = t, that is: τ is a solution of the equation.
If t = 1, then 1n = t, that is: 1 is a solution of the equation.
1
If t > 1, then 0 < < 1. Hence, there exists a number τ > 0 such that
t
1 n
1
1
τ n = ⇐⇒ n = t ⇐⇒
= t,
t
τ
τ
that is: τ −1 is a solution of the equation.
(b) Now assume that a and b are different positive solutions of the equation xn = t, say
0 < a < b. Then, according to Exercise 2.16,
t = an < bn = t,
which is impossible.
EXERCISE 3
Let f be a continuous function on an interval [a, b] with f (a)f (b) < 0.
Show that f has a zero on (a, b), in other words, that there is a z ∈ (a, b) with
f (z) = 0.
EXERCISE 4
Let f be the function defined by
f (x) = x3 + x2 − 17x + 16.
Prove that the function f has at least one zero on (0,2), at least one on (2,4)
and at least on (−∞, 0).
Another famous result on continuous functions is the Brouwer Fixed Point Theorem. We
first state this theorem and then briefly discuss it.
6
108
THEOREM 3
Continuous functions
THEOREM OF BROUWER
Let f be a continuous function on an interval [a, b] such that f (x) ∈ [a, b] for
every x ∈ [a, b]. Then f has a fixed point. That is, there exists x∗ ∈ [a, b]
such that f (x∗ ) = x∗ .
The condition in Brouwer’s Theorem states that the graph of f lies inside the square [a, b] ×
[a, b]. This is represented in Figure 2. The point x∗ is called a fixed point of f . Note that
the intersection of the graph of the function f and the 45o line y = x intersect at x∗ .
y
b
f
a
a
b
x
FIGURE 2
The function f has a fixed point
EXERCISE 5
Use the Intermediate Value Theorem to prove the Theorem of Brouwer.
EXERCISE 6
Give an example of a continuous function f on (0, 1) without a fixed point,
whereas f (x) ∈ (0, 1) for every x ∈ (0, 1).
EXERCISE 7
Give an example of a discontinuous function f on [0, 1] without a fixed point,
while f (x) ∈ [0, 1] for every x ∈ [0, 1].
The Theorem of Weierstrass, the Intermediate Value Theorem and Brouwer’s Theorem are
examples of so-called existence theorems. Such theorems assert that something exists without
telling you how to find it.
Students prefer finding solutions instead of worrying about their existence. They argue: if
I can calculate a solution of a problem, then it isn’t necessary to worry about whether a
solution exists. This is, however, false logic.
Suppose you want to find the largest natural number. Of course this problem has no solution.
Suppose however that you try to calculate a solution by proceeding as follows:
Let N be the largest natural number. Since 1 is a natural number, we must have N ≥ 1.
Since N 2 is a natural number, it cannot exceed the largest natural number. Therefore,
N 2 ≤ N and so N (N − 1) ≤ 0. Thus N − 1 ≤ 0 and therefore N ≤ 1.
We also know that N ≥ 1, so we may conclude that N = 1. In other words: 1 is the
largest natural number.
6
109
Continuous functions
The only error we have made here is in the assumption (in the first line) that the problem
has a solution. This explains why existence theorems are needed!
2
CONTINUITY OF THE INVERSE FUNCTION
In this section we discuss the continuity of the inverse of a continuous function.
EXERCISE 8
Let f be an invertible, continuous function on the interval [a, b]. Assume that
f (a) < f (b).
(a) Prove that [f (a), f (b)] ⊂ Rf .
(b) Prove that Rf ⊂ [f (a), f (b)].
[Clue: assume that a c ∈ (a, b) exists such that f (c) > f (b), make a
picture and derive a contradiction.]
There is a simple relation between strict monotonicity and invertibility. For suppose that a
function f on an interval I is strictly monotone, say f is strictly increasing, i.e. if x, x′ are
two different points in I, say x < x′ , then f (x) < f (x′ ). Hence, the function f is invertible.
However an invertible function is not necessarily strictly monotone.
y
2
g
g(x) =
x
3−x
if 0 ≤ x < 1
if 1 ≤ x ≤ 2
1
1
2
x
A one-to-one function which isn’t strictly monotone
FIGURE 3
For continuous functions on an interval however being one-to-one implies strict monotonicity.
THEOREM 4
Let f be a continuous function on an interval I. Then f is one-to-one if and
only if f is strictly monotone.
PROOF
As we observed preceding this theorem, a strictly monotone function is one-to-
one. So we only need to prove the reverse statement.
Assume that the function f is one-to-one. We will prove that f is strictly monotone for the
case that I = [a, b]. Since f is one-to-one, f (a) 6= f (b), say f (a) < f (b). We will show that
f is strictly increasing.
6
110
Continuous functions
Let x, x′ ∈ [a, b] such that x < x′ . We have to prove that f (x) < f (x′ ).
By Exercise 8, f (x′ ) > f (a). The function f restricted to the interval [a, x′ ] is continuous
and one-to-one. So, according to Exercise 8, f (x) ∈ [f (a), f (x′ )]. Hence, f (x) < f (x′ ) [note
that f (x) 6= f (x′ ) because f is one-to-one].
THEOREM 5
Let f be a one-to-one and continuous function on an interval [a, b]. Then
the inverse function f −1 has the following properties:
(a) Df −1 is the closed interval determined by f (a) and f (b).
(b) the inverse function is strictly monotone
(c) the inverse function is continuous.
PROOF
In view of the foregoing theorem, the function f is strictly monotone. In this
proof we will assume that the function f is strictly increasing.
(a) This is in fact Exercise 8.
(b) Assume that f (a) ≤ y < y ′ ≤ f (b). Then x, x′ ∈ [a, b] exists such that y = f (x) and
y ′ = f (x′ ). Since f is strictly increasing x < x′ .
[If x > x′ , y = f (x) > f (x′ ) = y ′ > y, which is a contradiction.]
Hence, f −1 (y) = x < x′ = f −1 (y ′ ), that is: the inverse function f −1 is strictly increasing.
(c) Let d ∈ f (a), f (b) . We will prove that the inverse function f −1 is continuous at y = d.
∞
Let yn n=1 be a sequence in [f (a), f (b)] converging to d. According to the definition, the
function f −1 is continuous at y = d if we can prove that
f −1 (yn )
| {z }
=xn
∞
n=1
→ f −1 (d)
| {z }
as
n → ∞.
=c
Let ε > 0 such that (c − ε, c + ε) ⊂ [a, b].
y
f (c + ε)
d
f (c − ε)
f
a
c−ε
c
c+ε
b
x
Let δ be the distance from d to the closer of the two points f (c − ε) and f (c + ε). [Thus
δ = min{f (c + ε) − d, d − f (c − ε)}.]
6
111
Continuous functions
Since lim yn = d, there exists an N ∈ IR such that
n→∞
|yn − d| < δ,
whenever n > N .
Then for n > N
|yn − d| < δ =⇒ −δ < yn − d < δ =⇒ d − δ < yn < d + δ
=⇒ f (c − ε) ≤ d − δ < yn < d + δ ≤ f (c + ε)
=⇒ f (c − ε) < yn < f (c + ε).
Because the function f −1 is strictly increasing, these inequalities imply that, for n > N ,
c − ε < f −1 (yn ) < c + ε =⇒ f −1 (yn ) − c < ε.
This proves that lim f −1 (yn ) = d.
n→∞
With a little more work one can prove that the properties (b) and (c) of the theorem also
hold if the function is defined on a non-compact interval.
EXAMPLE 3
The function f , defined by
f (x) = x3 + x,
is strictly increasing. So the function is invertible. Since the range of the function f is IR,
we know that the equation x3 + x = y has a unique solution for any y ∈ IR. Although we
cannot find a nice formula (in terms of y) for this unique solution f −1 (y), the function f −1
is a strictly increasing continuous function.
EXERCISE 9
Prove that the range of the function f : x → x3 + x is IR.
6
112
Continuous functions
Mixed exercises
EXERCISE 10
Show that the equation
x7 + x5 = 4 − x
has a solution on the interval [1, 2]. Show that the equation has a unique
solution, say x∗ , on the interval [1, 2]. Show that x∗ < 1.1 (and hence,
x∗ = 1.0.....).
EXERCISE 11
Let f be a continuous function on an interval [a, b] with f (x) 6= 0 for every
x ∈ [a, b].
Prove that either f (x) > 0 for every x ∈ [a, b] or f (x) < 0 for every x ∈ [a, b].
EXERCISE 12
Let f be the function on the interval [0, 1] defined by f (x) = x2 and let
g: [0, 1] → [0, 1] be a continuous function.
Show that the equation f (x) = g(x) has a solution.
EXERCISE 13
Let f be a continuous function on IR which is periodical with period 2, which
means that f (x + 2) = f (x) for every x ∈ IR. Prove that f has a minimum
and maximum.
EXERCISE 14
Let f be a continuous function on the interval [0, 1] with f (0) = f (1).
Prove that a c ∈ [0, 12 ] exists such that f (c) = f (c + 12 ).
[Clue: Introduce a function h on the interval [0, 12 ].]
EXERCISE 15
Let f be a continuous function on an interval I. Suppose that f is strictly
increasing, that is, for every x, y ∈ I with x < y we have f (x) < f (y). Show
that for an arbitrary constant c ∈ IR, f (x) = c has at most one solution.
Suppose further that there are a and b in I with f (a) < c < f (b). Show that
f (x) = c has exactly one solution x∗ , and that a < x∗ < b.
EXERCISE 16
We approximate
√
5
10 as follows. Show that the equation x5 = 10 has a
unique solution. (Hint: use the previous exercise). Let us call this solution
x∗ . Show that x∗ = 1.584... Motivate your answer.
7
113
Limits of functions
7
LIMITS OF FUNCTIONS
We defined the continuity of a function using limits of sequences of images. However,
sometimes the limits of these images exist at a point c, while the function is not continuous
at c. An example of such a situation is represented in Figure 1 by the function g at the
right-hand side. Apparently, when a function is not continuous, or even not defined, at a
certain point x = c, the limit of the sequence of images can exist. In order to discuss such
situations more carefully, we introduce ‘the limit of a function’.
In the last two sections of this chapter, we discuss discontinuities, and behavior of functions
at infinity, in particular linear asymptotic behavior.
1
LIMITS OF FUNCTIONS
Consider the functions f on [0, 1) ∪ (1, ∞) and g on [0, ∞) defined by
f (x) =
1−x
√
1− x

 1 −√x
and g(x) = 1 − x

1
if x 6= 1
if x = 1,
respectively. Their graphs are represented in the following figure.
y
y
g
f
2
2
1
1
x
1
FIGURE 1
If xn
∞
n=1
1
x
The graphs of the functions f and g
is an arbitrary sequence in the interval [0, ∞) with limit 1 such that xn 6= 1 for
all n, then
√ 1 − xn
= lim 1 + xn = 2.
√
n→∞
n→∞ 1 −
xn
lim f (xn ) = lim
n→∞
Similarly, lim g(xn ) = 2.
n→∞
7
114
Limits of functions
Apparently, when a function is not continuous, or even not defined, at a certain point x = c,
the limit of the sequence of images can exist. In order to discuss such situations more
carefully, we introduce
limit of a function
DEFINITION
Let I be an interval containing c and let f be a function, defined at all x ∈ I,
except perhaps at c. We say that f has a limit at c if there is an ℓ ∈ IR such
∞
that for any sequence xn n=1 in I \ {c} converging to c,
lim f (xn ) = ℓ.
n→∞
This is denoted by lim f (x) = ℓ.
x→c
Observe that, in the foregoing definition, all the terms of the sequence xn
∞
n=1
should be
different from c. The function may not be defined at c and if it is, the value of the function
at c is, in this context, irrelevant.
EXAMPLE 1
According to this definition,
lim f (x) = lim g(x) = 2,
x→1
x→1
where f and g are the functions introduced at the beginning of this Section.
EXAMPLE 2
We show that
1
= 0.
x
be a sequence with nonzero terms which converges to 0. Then, for all n,
lim x sin
x→0
Let xn
∞
n=1
xn sin
1
1
= |xn | · sin
≤ |xn |,
xn
xn
1
≤ |xn |. As lim |xn | = 0, the Sandwich Lemma implies that
n→∞
xn
1
lim xn sin
= 0.
n→∞
xn
∞
1
As the sequence xn n=1 was chosen arbitrarily, this proves that lim x sin = 0.
x→0
x
so that −|xn | ≤ xn sin
By modifying the proof of Theorem 5.1 in the obvious way, one can prove the following
rules.
THEOREM 1
ARITHMETIC RULES FOR LIMITS OF FUNCTIONS
Let I be an interval containing c and let f and g be functions on I \ {c}. If
lim f (x) = ℓ and lim g(x) = m, then
x→c
Sum Rule
Product Rule
Quotient Rule
x→c
(a) lim (f + g)(x) = ℓ + m
x→c
(b) lim (f · g)(x) = ℓ · m
x→c
f
ℓ
(c) lim (x) =
(provided that g(x) 6= 0 for all x ∈ I \ {c} and that
x→c g
m
m 6= 0).
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115
Limits of functions
EXAMPLE 3
We will prove that
lim cos x = 1.
x→0
Observe that, for al x ∈ (−1, 1),
cos x =
p
1 − sin2 x.
According to Example 5.9, lim sin x = 0. Hence, by the Product Rule, lim sin2 x = 0,
x→0
x→0
whereas the Sum Rule implies that lim 1 − sin2 x = 1 − 0 = 1.
x→0
√
According to Example 2 in Chapter 5 the function g(x) = x is continuous at x = c for any
c ≥ 0. So
lim cos x = lim
n→∞
EXERCISE 1
n→∞
Prove that
p
√
1 − sin2 x = 1 = 1.
√
lim (x2 − 1) x + 1 = 0.
x→1
In the next example we consider a situation where a limit doesn’t exist.
EXAMPLE 4
We consider the function h on [0, ∞) defined by

 1 −√x
if x < 1
h(x) = 1 − x

1
if x ≥ 1.
Its graph is presented in the following figure.
y
2
h
1
x
1
FIGURE 2
If xn
∞
n=1
The graph of the function h
is an arbitrary sequence in the interval [0, ∞) such that xn < 1 for all n, then
√ 1 − xn
= lim 1 + xn = 2.
√
n→∞
n→∞ 1 −
xn
lim h(xn ) = lim
n→∞
However, if xn > 1 for all n, then lim h(xn ) = 1.
n→∞
This implies that the limit lim h(x) doesn’t exist:
x→1
∞
• lim h(xn ) = 2 for any sequence xn n=1 at the left-hand side of 1, and
n→∞
∞
• lim h(yn ) = 1 for any sequence yn n=1 at the right-hand side of 1.
n→∞
7
116
Limits of functions
In order to prove that a limit of a function doesn’t exist one can exploit, as before, the fact
that we defined the limit of a function in terms of limits of sequences:
a limit lim f (x) does not exist if one can either
x→c
(1) construct two sequences in I \ {c} both converging to c such that the two corresponding
’sequences of images’ have different limits
or
(2) construct a sequence xn
’sequence of images’
EXAMPLE 5
∞
in I \ {c} converging to c such that the corresponding
n=1
∞
f (xn ) n=1
is divergent.
Sequences of images with different limits.
Consider the function g on IR \ {0}, defined by
g(x) =
|x|
.
x
We investigate the existence of the limit lim g(x).
x→0
∞
∞
1
1
We choose the sequences xn n=1 and yn n=1 defined by xn =
and yn = − , respecn
n
tively. Then
1
n
= 1,
lim g(xn ) = lim
n→∞
n→∞ 1
n
while
1
−
n = −1.
lim g(yn ) = lim
1
n→∞
n→∞
−
n
This implies that the two sequences of images have different limits. So the limit of the
function at 0 does not exist.
EXAMPLE 6
Sequence of images is unbounded.
Consider the function f on IR \ {−2, 2}, defined by
f (x) =
1
.
x2 − 4
We investigate the existence of the limit lim f (x).
x→2
∞
1
We choose the sequence xn n=1 defined by xn = 2 + .
n
Then
1
1
1
n
f (xn ) =
>
=
=
2
2
1 2
1
2
(2 + ) − 4
+
n
n n2
n
So f (xn ) is clearly unbounded and the limit of the function at 2 does not exist.
EXERCISE 2
Prove that the following limit does not exist:
√
x
√
x→1 1 −
x
lim
7
2
117
Limits of functions
ONE-SIDED LIMITS
Although the limit lim h(x) doesn’t exist for the function h introduced in Example 4, the
x→1
∞
limit of the sequence of images h(x1 ), h(x2 ), . . . is equal to 2 for any sequence xn n=1 at
the left-hand side of 1 which converges to 1. In order to describe such a situation formally,
we introduce so-called one-sided limits.
one-sided limits
DEFINITION
Let I be an interval containing c and let f be a function, defined at all x ∈ I,
except perhaps at c. We say that f has a limit from the left at c if there is
∞
an ℓ ∈ IR such that for any sequence xn n=1 in I at the left-hand side of c
and converging to c,
lim f (xn ) = ℓ.
n→∞
This is denoted by lim f (x) = ℓ.
x↑c
Of course we can also define the limit from the right lim f (x) = ℓ in the obvious way.
x↓c
EXAMPLE 7
For the function h introduced in Example 4,
lim h(x) = 2
x↑1
EXAMPLE 8
and
lim h(x) = 1.
x↓1
We will prove that
√
x2 x
√ = 0.
lim
x↓0 x +
x
Observe that the function x 7→
√
x2 x
√ is defined on the interval I = (0, ∞) and that 0 is
x+ x
an endpoint of this interval.
∞
Let xn n=1 be a sequence with positive terms converging to 0. Then
√
x2n xn
x2
= lim √ n .
√
n→∞ xn +
n→∞
xn
xn + 1
lim
According to the Arithmetic Rules for limits of sequences, this limit is equal
√ to 0.
∞
x2 x
√ = 0.
As the sequence xn n=1 was chosen arbitrarily, this proves that lim
x↓0 x +
x
EXERCISE 3
Let I be an open interval containing c, let f be a function defined on I \ {c}
and let ℓ be some number.
Prove that lim f (x) = ℓ if and only if lim f (x) = ℓ and lim f (x) = ℓ.
x→c
EXERCISE 4
x↓c
Prove that the limit from the right
lim sin
x↓0
doesn’t exist.
1
x
x↑c
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118
3
Limits of functions
DISCONTINUITIES
The techniques we discussed at the end of Section 1 can also be used to show that certain
functions are not continuous. By means of several examples and exercises we explain how
this can be done.
In order to prove that a function f defined on an interval I is not continuous at c ∈ I,
∞
one has to construct a sequence xn n=1 in I converging to c such that the corresponding
∞
‘sequence of images’ f (xn ) n=1 does not converge to f (c).
In the next examples and exercises we illustrate this principle. First an easy example.
EXAMPLE 9
Consider the function f on IR defined by
f (x) =
x2 + 1
0
if x 6= 0
if x = 0.
This function is obviously not continuous at c = 0 (sketch the graph of this function). A
formal proof of the discontinuity of f at c = 0 runs as follows. Take (for example) the
∞
sequence xn n=1 defined by xn = n1 for all n. Then clearly lim xn = 0. But
n→∞
lim f (xn ) = lim
n→∞
h 1 2
n→∞
n
i
+ 1 = 1,
while f (0) = 0 6= 1. This shows that f is discontinuous at x = 0.
EXERCISE 5
Sketch the graph of the function r defined by
r(x) =
x2 + 1 if x ≥ 1
−x − 1 if x < 1.
Prove that r is not continuous.
Of course the discontinuity in Example 9 looks very artificial, because we could easily ’mend’
this discontinuity of f by taking f (0) = 1 instead of, as we did above, f (0) = 0. However,
there are more severe cases of discontinuity as you will see in the next example and exercise.
EXAMPLE 10
We will show that the function g defined by
g(x) =
(
|x|
x
1
if x 6= 0
if x = 0
is not continuous at x = 0. Choose the sequence xn
∞
n=1
1
defined by xn = − . Then
n
1
n = −1 6= 1 = g(0).
lim g(xn ) = lim
1
n→∞
n→∞
−
n
−
This shows that g is discontinuous at x = 0.
7
119
Limits of functions
So far the argument is not any different from the previous Example. However, the next
exercise shows that the discontinuity of g is of a more severe nature than the discontinuity
we created in Example 9.
EXERCISE 6
Let ℓ be an arbitrary real number. Define the function h by
h(x) =
(
|x|
x
ℓ
if x 6= 0
if x = 0.
Sketch the graph of h. Prove that h is not continuous.
Thus, h has a discontinuity at x = 0 no matter what value we choose h to have at 0. In fact,
the function f in Example 9 has a limit at x = 0, that is why the function can be ’repaired’.
However, as we have seen in Example 5, the function g in the previous example does not
have a limit at x = 0 (although both the limit from the left as the limit from the right exist),
and we cannot ’repair’ g, no matter what value we would assign to g at x = 0. From the
graphs of f and h it is clear that there really is a difference in type of discontinuity between
these two functions.
4
BEHAVIOR OF A FUNCTION AT INFINITY
Finally, we briefly discuss asymptotic behavior of the values f (x) when x goes to ∞ or −∞.
The techniques we present are used for example to find trends in statistical data, or to study
long-run behavior of dynamical systems in macro economics.
DEFINITION
Let f be a function defined on an interval (a, ∞) and let ℓ be some number.
We say that f (x) converges to ℓ when x becomes large, if for every ε > 0
there exists a number H such that
|f (x) − ℓ| < ε,
whenever x > H. This is denoted by lim f (x) = ℓ.
x→∞
horizontal asymptote
The line y = ℓ is called a horizontal asymptote (at infinity) of the function
f.
1
has a horizontal asymptote y = 0 at infinity. The next figure
x
visualizes this fact. As the figure suggests, the (vertical) distance between the graph of the
The function g(x) =
function g and the horizontal axis can be made arbitrarily small.
7
120
Limits of functions
y
1
x
ε
x
H
−ε
FIGURE 3
A horizontal asymptote
EXERCISE 7
Write down the definition of lim f (x) = ℓ.
EXAMPLE 11
x→−∞
We will prove that
√
x
= 0.
lim
x→∞ x2 − 1
Let ε > 0. We must find a number H such that
√
x
< ε,
x2 − 1
whenever x > H. Note that for x > 2,
√
√
√
√
1
x
x
x
x
=
≤
≤
=√ .
x2 − 1
(x − 1)(x + 1)
x+1
x
x
Since
√
1
1
1
√ < ε ⇐⇒ x > ⇐⇒ x > 2 ,
ε
ε
x
1
we choose H = max 2, 2 . Then, for x > H,
ε
√
x
1
≤ √ < ε.
x2 − 1
x
√
x
= 0.
This proves that lim 2
x→∞ x − 1
There is a heuristic technique to determine the horizontal asymptote of a fractional function
that works quite well. The idea is to divide all terms in the fraction by the highest degree
term of the denominator. For example take the function
s(x) =
2x3 + x + 1
.
15x3 + x2
The highest degree term in the denominator is 15x3 . Therefore we divide all terms in both
the numerator and the denominator by x3 (in other words, we multiply all terms by
and we get
1
1
+ 3
x2
x .
1
15 +
x
2+
s(x) =
1
x3 ),
7
121
Limits of functions
Thus, when x goes to infinity, the numerator converges to 2 while the denominator converges
to 15. Hence, the horizontal asymptote of s at infinity is y =
2
15 .
Of course this is only
a heuristic, so it is not a formal proof (yet). But it helps you to get a feeling for which
asymptote you are looking for. Still, a formal proof requires an argument along the lines of
the definition as in the above example.
The heuristic also works for non-existence of limits.
The function
EXAMPLE 12
x2 + 3
t(x) =
5x − 17
3
x
t(x) =
.
17
5−
x
x+
becomes
When x goes to infinity, the denominator converges to 5, while the numerator goes to infinity.
Hence, this function goes to infinity when x goes to infinity.
Also fractional functions that have root functions in their terms can be handled this way.
The function q on (0, ∞) defined by
EXAMPLE 13
has highest degree term
√
√
2x + 1 + 5
q(x) = √
5x + 7 − 4
5x + 7 in its denominator. Therefore we divide all terms by
and q(x) can be written as
√
x
r
1
5
2+ + √
x
x
.
q(x) = r
7
4
5+ − √
x
x
√
√
2
Hence, the horizontal asymptote of the function q at infinity is y = √ = 15 10.
5
√
x−x
has a horizontal
EXERCISE 8 Prove that the function g on (0, ∞) defined by g(x) = √
x+x
asymptote at infinity.
EXERCISE 9
Prove, by
√ using the definition, that the function h on (−∞, 0) defined by
1−x
h(x) =
has a horizontal asymptote y = 0 at minus infinity.
x
EXERCISE 10
Prove that the limit
lim
x→∞
3+x
√
x
doesn’t exist.
[ Clue: Give a proof by contradiction.]
Of course a function could have two different horizontal asymptotes, one at infinity and one
at minus infinity.
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122
EXERCISE 11
Limits of functions
Consider the function s on IR \ {−1} defined by
1
(1 + x)2
s(x) =
.
1
|x| +
(1 + x)2
x+
Find the horizontal asymptotes of s. Sketch the graph of s.
A more general form of convergence at infinity is the notion of a linear asymptote, in statistics
also known as the trend.
linear asymptote
Let f be a function defined on an interval (a, ∞). We say that f (x) converges
DEFINITION
to mx + b when x becomes large, if for every ε > 0 there exists a number H
such that
|f (x) − (mx + b)| < ε,
whenever x > H.
The line y = mx + b is called a linear asymptote (at infinity) of the function
linear asymptote
f.
EXAMPLE 14
Consider the function f on IR \ {0} defined by
f (x) = 1 +
x2 − 1
.
2x
We show that the line y = 1 + 12 x is a linear asymptote at infinity of the function f . Note
that for every x > 0,
1
x2 − 1 − x2
x
x2 − 1
f (x) − 1 + 21 x = 1 +
=
=
−1−
.
2x
2
2x
2x
Let ε > 0. Take H =
1
1
1
. Then for x > H we know that
<
= ε. Hence, for x > H,
2ε
2x
2H
1
< ε.
f (x) − 1 + 21 x =
2x
This shows that y = 1 + 21 x is a linear asymptote of f at infinity.
y
y = 1 + 12 x
f
1
x
FIGURE 4
A linear asymptote
7
123
Limits of functions
EXERCISE 12
Consider the function z defined by z(x) =
√
1 + x2 + 1.
Show that y = x + 1 is a linear asymptote of the function z at infinity. Also
show that y = 1 − x is a linear asymptote of the
[Clue: the ’root method’.]
Obviously, a horizontal asymptote can simply be viewed as a special case of a linear asymptote, namely the case where the line y = mx + b happens to have a horizontal slope, that
is, where m = 0.
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124
Limits of functions
Mixed exercises
EXERCISE 13
Let c be a point in an open interval I. Let f be a function defined on I \ {c}
and suppose that lim f (x) = ℓ. Suppose further that f (x) ≥ 0 for all x in
x→c
I \ {c}.
Prove that ℓ ≥ 0.
EXERCISE 14
Let d be an arbitrary real number. On the interval (−1, ∞), we define the
function g by
√
 1+x−1
g(x) =
|x|

d
if x 6= 0
if x = 0.
Prove that g is not continuous at x = 0.
EXERCISE 15
Consider the function h defined by
h(x) = 1 − x +
p
x2 − 2x + 3.
Show that h has a horizontal asymptote at infinity.
Show that y = 2 − 2x is a linear asymptote of h at minus infinity.
Sketch the graph of h.
EXERCISE 16
Find the horizontal asymptote at infinity of the function w on (0, ∞) defined
by
√
3 x3 + 5x − 12
w(x) = √
.
3x3 + 2x2 + 15
Motivate your answer.
EXERCISE 17
Does the function g on (0, ∞) defined by
√
4
x3 + 2
g(x) = √
2x3 + 3
have a horizontal asymptote at infinity? If so, prove this, if not, prove that
no asymptote exists.
EXERCISE 18
Let f be a continuous (nonzero) function on IR for which
lim f (x) = lim f (x) = 0.
x→−∞
x→∞
Assume that f (0) > 0.
(a) Prove that numbers H1 < 0 and H2 > 0 exist such that f (x) < f (0) for
all x ∈
/ [H1 , H2 ].
(b) Prove that f has a maximum.
8
8
125
Derivatives
DERIVATIVES
A fundamental problem in analysis is concerned with finding the slope of the tangent to a
curve at a given point on the curve. This problem is the subject of this chapter. As we will
see, it has important applications in economics.
In the first section we introduce the relevant concepts. In the second section higher-order
derivatives are discussed. The Arithmetic Rules for differentiation, like Product Rule and
Chain Rule, will be presented in section three. Section four deals with the derivative of the
inverse function. As an application of derivatives, marginality and elasticity are discussed
in the fifth section. The final section contains a summary of derivatives of basic functions
(standard derivatives).
1
DIFFERENTIABILITY
In the following figure we illustrate certain types of ’misbehavior’ which continuous functions
can display.
y
y = |x|
y = x2
y
y=x
x
(a)
y
x
(b)
y=
p
|x|
x
(c)
Some non-smooth graphs
FIGURE 1
The graphs of these functions are ’bent’ at the origin, unlike the graph of the next figure,
where it is possible to draw a ’tangent’ at each point.
y
x
FIGURE 2
A smooth graph
8
126
Derivatives
Note that we used quotation marks because we did not yet provide a suitable definition of
a tangent. In plane geometry a tangent to a circle is customarily defined to be a line which
intersects the circle exactly once. Obviously, it doesn’t make sense to adopt this as the
general definition of a tangent to a curve. With such a definition, each vertical line would
be a tangent to the parabola y = x2 . Furthermore, the three curves in Figure 1 would have
more than one tangent at the origin.
A more promising approach to the definition of a tangent (at a point c, f (c) on the graph
of a function f ) starts with secant lines and uses the notion of limits. If x 6= c, then, as
presented in Figure 3(a), the two points c, f (c) and x, f (x) on the graph of f determine
a straight line whose slope is
f (x) − f (c)
.
x−c
difference quotient
For obvious reasons, this slope is also called a difference quotient.
y
y
f (x)
f (x) − f (c)
f (c)
f (c)
x−c
c
x
x
EXAMPLE 1
x
(b)
(a)
FIGURE 3
c
Some secant lines
Let f be the quadratic function defined by
f (x) = x2 − 2x + 4.
The slope of a straight (non-vertical) line through the points (2, 4) and x, f (x) on the
graph of the function f is equal to
x2 − 2x
f (x) − 4
=
= x.
x−2
x−2
As Figure 3(b) suggests, the tangent at c, f (c) seems to be the limit – in some sense – of
the ’secant lines’ as x approaches c. Although we don’t have a formal definition of a ’limit’
of lines, we can talk about the limit of their slopes: the slope of the tangent to the graph of
f at c, f (c) should be
f (x) − f (c)
.
lim
x→c
x−c
8
Derivatives
127
EXAMPLE 2
For the function f introduced in Example 1 the slope of the tangent to the
graph of f at the point (2, 4) should be
lim
x→2
f (x) − 4
= lim x = 2.
x→2
x−2
This brings us to the following definition.
DEFINITION
Let f be a function defined on an interval I containing c.
We say that f is differentiable at c if the limit
differentiable
f (x) − f (c)
x→c
x−c
lim
exists. In this case this limit is called the derivative of f at c and denoted
df
(c).
by f ′ (c) or
dx
If the function is differentiable at every c in I, then f is said to be differen-
derivative
tiable (on I) and the function f ′ : I → IR is called the derivative of f .
tangent
Next we define the tangent to the graph of f at c, f (c) to be the line through c, f (c)
with slope f ′ (c). An equation of this tangent is
y = f ′ (c)(x − c) + f (c).
This means that this tangent is defined only if f is differentiable at c !
EXAMPLE 3
According to Example 2, the function f defined by
f (x) = x2 − 2x + 4
is differentiable at 2 and f ′ (2) = 2. Furthermore, an equation of the tangent to the graph
of f at the point (2, 4) is
y = 2(x − 2) + 4 ⇐⇒ y = 2x.
EXAMPLE 4
We will prove that the function f : x → x2 − 2x + 4 is differentiable.
Let c ∈ IR. Note that for any x 6= c,
(x2 − 2x + 4) − (c2 − 2c + 4)
f (x) − f (c)
=
x−c
x−c
=
=
x2 − c2 − 2(x − c)
x−c
(x − c)(x + c) − 2(x − c)
x−c
= x + c − 2,
so that
lim
x→c
f (x) − f (c)
= lim (x + c − 2) = 2c − 2.
x→c
x−c
So the function f is differentiable at c and f ′ (c) = 2c − 2. Since c was arbitrarily chosen,
the function is differentiable and its derivative is given by f ′ (x) = 2x − 2.
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128
EXAMPLE 5
Derivatives
We will prove that the function f on (0, ∞), defined by
1
f (x) = √ ,
x
is differentiable. Note that for c > 0 and a positive x 6= c
f (x) − f (c)
=
x−c
=
1
1
√
√
√
√ √
√
√ −√
1
1
c− x
c− x c+ x
x
c
√ √ =
√ √ √
√
=
x−c
x−c
x−c
x c
x c
c+ x
1 c−x
1
1
√ √ √
√ = −√ √ √
√ .
x−c x c c+ x
x c( c + x)
Since the limit as x → c exists for the expression at the right-hand side of the equality
sign, the function f is differentiable at c and, according to the Arithmetic Rules for limits
of functions,
f ′ (c) = lim
x→c
f (x) − f (c)
1
1
√ =− √ .
= lim − √ √ √
x→c
x−c
x c( c + x)
2c c
Hence, the function f is differentiable and its derivative is given by
f ′ (x) = −
EXAMPLE 6
1
√ .
2x x
The function f , defined by f (x) = sin x is differentiable at 0.
Since
f (x) − f (0)
sin x
=
,
x
x
Exercise 5.16 implies that the function f is differentiable at 0 and that
f ′ (0) = lim
x→0
EXERCISE 1
sin x
= 1.
x
Use the definition to prove that each of the following functions is differentiable
at every point in its domain. Find its derivative and give an equation of the
tangent to the graph of the function at 3.
(a) f : x → x2
(c) k: x → a
EXERCISE 2
(b) g: x → x3
(a ∈ IR)
(d) h: u →
2
+ A (A ∈ IR).
3u
Prove that the function g defined by
g(x) = x|x|
is differentiable at 0.
EXERCISE 3
Let f be the function on [0, ∞), defined by
f (x) =
√
x.
Prove that the function f is differentiable at c for all c > 0.
8
Derivatives
129
EXAMPLE 7
An example of a function that is not differentiable at every point of its
domain is the absolute value function g defined by g(x) = |x|.
Note that for any x 6= 0,
g(x) − g(0)
|x|
=
=
x
x
1
if x > 0
−1 if x < 0,
so that
lim
g(x) − g(0)
=1
x
lim
g(x) − g(0)
= −1.
x
x↓0
and
x↑0
g(x) − g(0)
doesn’t exist. Also consider Example 7.9 where we proved,
x→0
x
|x|
by using sequences, that the limit lim
doesn’t exist. So the function g is not differentiable
x→0 x
at 0. If we look at the graph of the function g – which is represented in Figure 1(a) – then
Hence, the limit lim
we observe that this graph is not smooth at the origin.
For obvious reasons, the two one-sided limits in the foregoing example are called the righthand derivative and the left-hand derivative, respectively, of g at 0. In general
DEFINITION
Let f be a function defined on an interval I containing c.
If the limit
lim
x↓c
f (x) − f (c)
x−c
exists, it is called the right-hand derivative of the function f at c.
right-hand derivative
The left-hand derivative of the function f at c is defined in a similar way.
left-hand derivative
Obviously, a function f defined on an open interval containing c is differentiable at c if and
only if right-hand and left-hand derivative at c exist and are equal.
EXAMPLE 8
Let f (x) =
p
|x|. We will prove that the left-hand derivative at 0 and the
right-hand-derivative at 0 do not exist.
Note that for x 6= 0,
√
1
x

p

=√
x
|x|  x
f (x) − f (0)
=
= √

x
x

 −x = − √1
x
−x
In this case the right-hand limit
lim
x↓0
doesn’t exist.
1
f (x) − f (0)
= lim √
x↓0
x
x
if x > 0
if x < 0.
8
130
In order to prove this, we consider the sequence xn
∞
n=1
, defined by xn =
Derivatives
1
. Then
n2
1
f (xn ) − f (0)
= n,
=r
xn
1
n2
f (xn ) − f (0)
doesn’t exist. Hence, the right-hand derivaxn
tive of f at 0 doesn’t exist. In a similar way one can prove that the left-hand derivative of
which implies that the limit lim
n→∞
f at 0 doesn’t exist.
cusp
We say that the graph of the function f , which has been represented in Figure 1(c), has a
cusp at the origin. If you were ’travelling along’ the graph, you would have to stop and turn
180◦ at the origin.
EXERCISE 4
Prove that the function f defined by
x if x ≥ 0
f (x) =
x2 if x < 0
is not differentiable at 0 (see Figure 1(b)).
The following theorem shows that differentiability is a stronger condition than continuity.
THEOREM 1
Let f be a function defined on an interval I containing c. If f is differentiable
at c, then f is continuous at c.
PROOF
For every x ∈ I \ {c},
f (x) = (x − c)
f (x) − f (c)
+ f (c).
x−c
So, according to the Arithmetic Rules for limits of functions,
h
i
f (x) − f (c)
+ lim f (c)
lim f (x) = lim (x − c) lim
x→c
x→c
x→c
x→c
x−c
= 0 · f ′ (c) + f (c) = f (c).
That is: the function f is continuous at c.
Note that the converse of this result is not true: a continuous function need not to be
differentiable. Keep in mind the absolute value function x → |x| we discussed before.
EXERCISE 5
Consider the function
x2 − 2 if x ≥ 1
x2
if x < 1.
(a) Prove that the function f is differentiable at c 6= 1.
f: x →
(b) Prove that lim f ′ (x) exists.
x→1
(c) What about the following proof? ”The derivative f ′ (x) exists for all
x 6= 1 and also its limit at 1 exists, so f is differentiable at 1.”
(d) Is the function f differentiable at 1?
(e) Draw the graph of the function.
8
131
Derivatives
EXERCISE 6
Let f be a function which is differentiable on an interval I.
Prove that the function 5f is differentiable on I and find (5f )′ .
2
HIGHER-ORDER DERIVATIVES
In general we obtain for a function f , by taking the derivative, a new function f ′ whose
domain may be properly contained in the domain of f . Of course we can investigate whether
the function f ′ is differentiable at c where c is in the domain of f ′ . If the derivative of f ′ at
c exists, then it is denoted as f ′′ (c) instead of (f ′ )′ (c). In this case we say that f is twicedifferentiable at c and the number f ′′ (c) is called the second derivative of f at c. Another
d2 f
(c).
notation for the second derivative of the function f at c is
dx2
′′′
′′ ′
Sometimes we can repeat this process and define f = (f ) . Because this notation rapidly
becomes unwieldy, the following notation will also be used: f (2) = f ′′ and for k ≥ 3
′
f (k) = f (k−1) .
higher-order derivatives
For k ≥ 2, the functions f (k) are called the higher-order derivatives of f .
The graph of the function
EXAMPLE 9
h: x →
x2
2
−x
if x ≥ 0
if x < 0
is presented in the following figure.
y
h
x
FIGURE 4
The graph of the function h
Obviously, Exercise 1 (a) implies that
h′ (x) =
while for x 6= 0
2x
if x > 0
−2x if x < 0,
 2

x
if x > 0
h(x) − h(0)  x = x
=
2

x
 − x = −x if x < 0.

x
8
132
Derivatives
Now
lim
h(x) − h(0)
= lim x = 0
x↓0
x
lim
h(x) − h(0)
= lim −x = 0.
x↑0
x
x↓0
and
x↑0
So h′ (0) = 0.
The foregoing can be summarized as follows
h′ (x) = 2|x|.
It follows that h′′ (0) doesn’t exist.
Existence of the second derivative is thus a rather strong criterion for a function to satisfy.
Even a ’smooth looking’ function like h reveals some irregularity when examined with the
second derivative.
3
ARITHMETIC RULES FOR DIFFERENTIABLE FUNCTIONS
In the foregoing sections the derivative has been determined by using the definition. Although this is sometimes the only possible approach, a large number of functions can be
differentiated without referring to the definition. In the following theorems a technique will
be provided for differentiating functions which are formed from a few basic functions by
the process of addition, multiplication, division and composition. Examples of such basic
functions are: the constant function x → c (where c ∈ IR), the power function x → xk ,
√
(where k ∈ IN), the root function x → x, the natural logarithm x → ln x, the exponential
function x → ex and the functions sin and cos. The derivatives of these basic functions are
summarized in Section 6.
The derivative of the sum of two differentiable functions is just what one would expect: the
sum of the derivatives.
THEOREM 2
SUM RULE
Let f and g be functions on an interval I containing c and assume that f
and g are differentiable at c. Then the function f + g is differentiable at c
and
(f + g)′ (c) = f ′ (c) + g ′ (c).
PROOF
For every x ∈ I with x 6= c, we have
(f + g)(x) − (f + g)(c)
f (x) − f (c) g(x) − g(c)
=
+
.
x−c
x−c
x−c
Since the limit as x → c exists for both terms in the expression at the right-hand side of
the equality sign, the Arithmetic Rules for limits of functions imply that this limit also
8
133
Derivatives
exists for the expression at the left-hand side of the equality sign. So the function f + g is
differentiable at c and, according to these rules,
(f + g)(x) − (f + g)(c)
f (x) − f (c)
g(x) − g(c)
= lim
+ lim
x→c
x→c
x→c
x−c
x−c
x−c
(f + g)′ (c) = lim
= f ′ (c) + g ′ (c).
The derivative of the product of two differentiable functions leads to a less straightforward
but symmetric formula. In the proof we apply the well-known ’telescope method’: a number
is not changed if the same quantity is added to and subtracted from it.
THEOREM 3
PRODUCT RULE
Let f and g be functions on an interval I containing c and assume that f
and g are diffe-rentiable at c. Then the function f · g is differentiable at c
and
(f · g)′ (c) = f ′ (c)g(c) + f (c)g ′ (c).
PROOF
For every x ∈ I with x 6= c, we have
(f · g)(x) − (f · g)(c)
f (x)g(x) − f (c)g(c)
=
x−c
x−c
=
=
f (x)g(x) − f (c)g(x) + f (c)g(x) − f (c)g(c)
x−c
g(x) − g(c)
f (x) − f (c)
g(x) + f (c)
.
x−c
x−c
Since the limit as x → c exists for both terms in the expression at the right-hand side of the
equality sign, the Arithmetic Rules for limits of functions imply that this limit also exists for
the expression at the left-hand side of the equality sign. So the function f · g is differentiable
at c and, according to these rules,
(f · g)′ (c) = lim
x→c
(f · g)(x) − (f · g)(c)
g(x) − g(c)
f (x) − f (c)
= lim
g(x) + lim f (c)
x→c
x→c
x−c
x−c
x−c
= f ′ (c)g(c) + f (c)g ′ (c).
Note that we have used Theorem 1 to conclude that lim g(x) = g(c).
x→c
An obvious generalization of Exercise 6 shows that (cf )′ = cf ′ , where f is a differentiable
function on an interval I and c is some real number.
EXAMPLE 10
We show that the function
f : x → xn
is differentiable for all n ∈ IN and that
f ′ (x) = nxn−1 .
8
134
Derivatives
For n ∈ IN we introduce the statement P(n): the function f : x → xn is differentiable and
f ′ (x) = nxn−1 .
(1) First we show that the statement P(1) is true:
for any c ∈ IR
lim
x→c
x−c
f (x) − f (c)
= lim
= lim 1 = 1,
x→c
x−c
x − c x→c
that is: the function f : x → x is differentiable at c and f ′ (c) = 1.
(2) Let k ∈ IN and assume that P(k) is true, that is: the function f : x → xk is differentiable
and f ′ (x) = kxk−1 .
Now let g: x → xk+1 . In order to prove that the function g is differentiable, we introduce the function h: x → x. Then g = f · h and, according to the Product Rule for
differentiable functions, g is differentiable and
g ′ (x) = f (x) · h′ (x) + f ′ (x) · h(x) = xk · 1 + kxk−1 · x = (k + 1)xk .
This proves that P(k + 1) is true.
According to the Principle of Induction, the statement P(n) is true for all n ∈ IN.
EXERCISE 7
EXERCISE 8
Find the derivative of the following functions.
(a) f (x) = x2 ln x + x ln x
√
(b) f (x) = x2 x
(c) f (x) = (1 − sin x) ex
(d) f (x) = sin x · cos x.
Let f be a differentiable function on an interval I.
Prove that for every n ∈ IN the function
fn = f · f · · · · · f
|
{z
}
n times
is differentiable and that for all x ∈ I
′
n−1 ′
f (x).
f n (x) = n f (x)
EXAMPLE 11
Once again we consider for n ∈ IN the function f defined by f (x) = xn .
According to the foregoing example, the function f is twice-differentiable and
f ′′ (x) = n(n − 1)xn−2 .
More generally, for k ∈ IN, the kth derivative f (k) exists and

n!

xn−k if k ≤ n
(k)
f (x) = (n − k)!

0
if k > n.
Putting together the theorems proved so far, we can now determine the derivative of the
polynomial function p, defined by
p(x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn .
8
135
Derivatives
By applying the Sum Rule and Product Rule and by referring to Example 10, we obtain
p′ (x) = a1 + 2a2 x + · · · + (n − 1)an−1 xn−2 + nan xn−1 .
This process can be continued easily: in each next step the highest power of x is reduced by
1 and one more of the coefficients ai is eliminated. Note that
p(n) (x) = n! an
and that p(k) (x) = 0 for k > n.
Clearly, the next step is to find the derivative of a quotient f /g of two functions. Because
it is a lot simpler and also sufficient (by the Product Rule), we will determine the derivative
of the function 1/g.
THEOREM 4
Let g be a function on an interval I containing c and assume that g is differentiable at c and that g(c) 6= 0. Then the function 1/g is differentiable at
c and
PROOF
′
1
g ′ (c)
(c) = −
.
g
[g(c)]2
Before we consider the difference quotient
1
1
1
1
(x)−
(c)
−
g
g
g(x) g(c)
=
,
x−c
x−c
we must check that g(x) 6= 0 for x sufficiently close to c.
Indeed, since g(c) 6= 0 and g is continuous at c, according to Exercise 5.37, an interval
(c − δ, c + δ) exists such that g(x) 6= 0 for all x ∈ I in that interval.
For every x ∈ I \ {c} contained in this interval, we have
1
1
1
1
(x)−
(c)
−
g(x) − g(c)
1
g
g
g(x) g(c)
=
=−
.
x−c
x−c
g(x)g(c)
x−c
g(x) − g(c)
exist, the Arithmetic Rules for limits of functions imply
x−c
that the limit of the expression at the right-hand side of the equality sign exists. So the
Since lim g(x) and lim
x→c
x→c
function 1/g is differentiable at c and, according to these rules,
1
1
′
(x)−
(c)
1
g ′ (c)
1
g(x) − g(c)
g
g
(c) = lim
.
= − lim
· lim
=−
x→c
x→c g(x)g(c) x→c
g
x−c
x−c
[g(c)]2
Note that we have used Theorem 1 to conclude that lim g(x) = g(c).
x→c
EXERCISE 9
Consider the function g on (0, ∞), defined by
g(x) =
1
,
xn
where n ∈ IN. Prove that the function g is differentiable and find its derivative.
8
136
Derivatives
The general formula for the derivative of a quotient is now easy to derive.
THEOREM 5
QUOTIENT RULE
Let f and g be functions on an interval I containing c and assume that f and
g are differentiable at c. Then, if g(c) 6= 0, the function f /g is differentiable
at c and
′
f
g(c)f ′ (c) − f (c)g ′ (c)
(c) =
.
g
[g(c)]2
EXERCISE 10
Find the derivative of the following functions.
√
2 x
x2 − 1
√
(a) f (x) =
(x ≥ 0)
(b) f (x) = 2
x +1
1+ x
(c) f (x) =
1 − sin x
2 + cos x
(d) f (x) =
ln x
.
ex
In Theorem 6.2 we proved that the composition of two continuous functions is continuous.
A similar result holds for the composition of differentiable functions, and is known as the
Chain Rule.
THEOREM 6
THE CHAIN RULE
Let f be a function on an interval I, and let g be a function on an interval
J containing c, such that g(J) ⊂ I. If g is differentiable at c and f is
differentiable at g(c), then the composite function f ◦ g is differentiable at c
and
(f ◦ g)′ (c) = f ′ g(c) · g ′ (c).
PROOF
We will prove this result only for the situation where g(x) 6= g(c) for all x ∈ J.
For all x ∈ J with x 6= c, we have
f g(x) − f g(c)
f g(x) − f g(c) g(x) − g(c)
(f ◦ g)(x) − (f ◦ g)(c)
=
=
·
.
x−c
x−c
g(x) − g(c)
x−c
Since g is differentiable at c,
lim
x→c
g(x) − g(c)
= g ′ (c).
x−c
Furthermore, as the function g is continuous at c, for a sequence xn
∞
n=1
in J converging to
c, g(xn ) → g(c) as n → ∞. In combination with the fact that the function f is differentiable
at g(c) this implies that
f g(xn ) − f g(c)
→ f ′ g(c) as n → ∞.
g(xn ) − g(c)
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137
Derivatives
Since the sequence xn
∞
was chosen arbitrarily, this proves that
f g(x) − f g(c)
→ f ′ g(c) as x → c.
g(x) − g(c)
n=1
Hence, according to the Product Rule for limits of functions, the limit
(f ◦ g)(x) − (f ◦ g)(c)
x→c
x−c
lim
exists and the function f ◦ g is differentiable at c. Furthermore, the derivative of f ◦ g at c
is equal to f ′ g(c) · g ′ (c).
EXERCISE 11
Find functions f and g such that h = f ◦ g, where h is the function defined
by
(a) h(x) = 3 sin(x + 2x2 )
(c) h(x) = e1−x
EXAMPLE 12
2
(b) h(x) = 2 + sin2 x
√
(d) h(x) = sin x.
Let h be the function on IR \ {0}, defined by
h(x) = sin
1
x
and let c > 0. Then, h = f ◦ g, where g is the function on J = (0, ∞), defined by g(x) =
1
x
and f is the function on I = IR defined by f (x) = sin x.
Since the functions f and g are differentiable, the Chain Rule implies that the function h is
differentiable at c and that
1
1
1
1
h (c) = cos · − 2 = − 2 cos .
c
c
c
c
′
The same result can be obtained for c < 0, so that the function h is differentiable and the
1
1
derivative h′ is the function on IR \ {0}, defined by h′ (x) = − 2 cos .
x
x
EXERCISE 12
Find the derivative of the functions in the foregoing exercise.
EXERCISE 13
Find f ′ in terms of g ′ if f is the function defined by
(a) f (x) = g x + g(1)
(b) f (x) = g x · g(1)
(c) f (x) = g x + g(x)
(d) f (x − 1) = g(x2 ).
4
THE DERIVATIVE OF THE INVERSE FUNCTION
In Section 8 of Chapter 1 we introduced the inverse function and in Chapter 6 we discussed
the continuity of inverse functions. In this section we state and prove the Inverse Function
Theorem. This theorem gives an easy way to find the derivative of the inverse function. We
start with a first impression of the result.
Suppose that a function f is invertible and that both f and f −1 are differentiable. Then
the derivative of f −1 can easily be related to the derivative of f . Note that
(f ◦ f −1 )(y) = y,
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138
Derivatives
for all y in the domain Df −1 of the inverse function (remind that this domain is equal to the
range Rf of the function f ). Using the Chain Rule differentiating both sides of the equation
leads to
′
′
f ′ f −1 (y) · f −1 (y) = 1 ⇐⇒ f −1 (y) =
1
f′
.
f −1 (y)
Observe that the validity of this calculation hinges upon the assumption that f and f −1 are
both differentiable. The next theorem discusses the differentiability of f −1 Note also that
troubles occur if f ′ vanishes. If, for instance, f ′ (x0 ) = 0 and y0 = f (x0 ), then f −1 is not
differentiable at y = y0 , since if it were, we would have
′
0 = f ′ f −1 (y0 ) · f −1 (y0 ) = 1,
which is impossible. Geometrically, this corresponds to the fact that if a curve is reflected
in the line y = x, then a horizontal tangent is mapped onto a vertical line (reconsider Figure
10 of Chapter 1).
THEOREM 7
INVERSE FUNCTION THEOREM
Let f be a differentiable and invertible function on an interval I containing
c. Further let d = f (c) and assume that f ′ (c) 6= 0. Then the inverse function
f −1 is differentiable at d and
′
f −1 (d) =
PROOF
1
f′
.
f −1 (d)
In order to prove that the limit
f −1 (y) − f −1 (d)
y→d
y−d
lim
exists, let yn
∞
n=1
be a sequence in f (I) converging to d such that yn 6= d for all n ∈ IN.
For every n ∈ IN there is a unique xn ∈ I such that yn = f (xn ). Since the function f −1 is
continuous,
lim xn = lim f −1 (yn ) = f −1 (d) = c.
n→∞
n→∞
Now for any n ∈ IN
xn − c
1
f −1 (yn ) − f −1 (d)
,
=
=
f (xn ) − f (c)
yn − d
f (xn ) − f (c)
xn − c
which implies that
1
f −1 (yn ) − f −1 (d)
1
1
= ′
= lim
= ′ −1 .
n→∞
n→∞ f (xn ) − f (c)
yn − d
f (c)
f f (d)
xn − c
lim
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139
Derivatives
∞
was arbitrarily chosen, this implies that the function f −1 is diffe′
1
rentiable at y = d and that f −1 (d) = ′ −1 .
f f (d)
As the sequence yn
n=1
Note that in the foregoing theorem the condition that f ′ (c) 6= 0 is essential. Consider for
instance the function f : x → x3 . The function f is invertible. However f ′ (0) = 0 and the
√
inverse function f −1 : y → 3 y is not differentiable at d = f (0) = 0.
EXAMPLE 13
Let f be the function defined by
f (x) = x3 + x.
If x′ < x′′ , then
f (x′ ) = (x′ )3 + x′ < (x′′ )3 + x′′ = f (x′′ ).
Hence, the function f is invertible. Although we cannot find a nice formula for f −1 , the
inverse function is differentiable.
Since f ′ (x) = 3x2 + 1, for any d ∈ IR
′
f −1 (d) =
1
2
3 [f −1 (d)]
+1
.
So, because f (1) = 2, f −1 (2) = 1 and
′
f −1 (2) =
1
2
3 [f −1 (2)] + 1
= 14 .
Note that f −1 (1) is the unique solution of the equation
x3 + x = 1.
Hence, it isn’t possible to determine f −1 (1) exactly. However, the solution f −1 (1) can be
approximated with any degree of accuracy. Obviously, the same holds for the derivative
′
1
f −1 (1) =
.
−1
3 [f (1)]2 + 1
EXAMPLE 14
Let n ∈ IN and let f be the power function on I = (0, ∞) defined by
f (x) = xn .
According to Exercise 2.16, this function is strictly increasing. So the function f is invertible.
The inverse function f −1 of f on (0, ∞) is given by
1
f −1 (y) = y n .
For the function f it holds that f ′ (x) = nxn−1 > 0 for every x > 0. According to the Inverse
Function Theorem, f −1 is a differentiable function on (0, ∞), and for every y > 0
′
f −1 (y) =
1
1
=
n−1
−1
f ′ f −1 (y)
n f (y)
1
1 −1
1
1
= yn .
=
= 1
n−1
n−1
n
n yn
ny n
8
140
EXERCISE 14
Derivatives
Let m, n ∈ IN and let h be the function on (0, ∞), defined by
m
h(x) = x n .
Prove that the function h is differentiable and that
m
−1
m
h (x) = x n
n
′
for all x > 0.
EXERCISE 15
Use the Arithmetic Rules for differentiable functions to determine the derivative of the function f on (0, ∞), defined by
√ √
f (x) = 3 3 x( x + x3 ).
EXERCISE 16
Prove that the function h on (−1, ∞), defined by
p
h(x) = 3 1 + x|x|,
is differentiable and determine its derivative.
EXAMPLE 15
(THE FUNCTIONS ARCSIN AND ARCTAN)
The function sin restricted to the interval [− π2 , π2 ] is strictly increasing and therefore invertible. The inverse of this function which we denote as arcsin is defined on the interval
[−1, 1].
y
π
2
y
π
2
arcsin
−1
1
arctan
x
x
−
−
FIGURE 5
π
2
π
2
The graph of the function arcsin and the function arctan
According to the Inverse Function Theorem, the function arcsin is differentiable on (−1, 1)
and
arcsin′ (y) =
1
1
1
1
,
=q
=
= p
cos(arcsin(y))
sin′ (arcsin(y))
2
1
− y2
1 − sin (arcsin(y))
for all y ∈ (−1, 1).
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141
Derivatives
The function tan restricted to the interval (− π2 , π2 ) is strictly increasing and therefore invertible. The inverse of this function which is denoted as arctan is defined on IR.
The function arctan is differentiable on IR and in Exercise 17 you are asked to prove that
arctan′ (y) =
EXERCISE 17
1
.
1 + y2
(a) Prove that for − 12 π < t < 12 π
1
= 1 + tan2 t.
cos2 t
(b) Prove that for all y ∈ IR,
arctan′ (y) =
EXERCISE 18
1
.
1 + y2
Explain why the function cos is invertible on the interval [0, π]. The corresponding inverse function is denoted as arccos.
Prove that for every y ∈ (−1, 1),
−1
arccos′ (y) = p
.
1 − y2
EXAMPLE 16
As you know from secondary school, the natural logarithm ln is a strictly
increasing function on the interval (0, ∞) and
ln′ (x) =
1
.
x
So the function ln is invertible. According to the Inverse Function Theorem, its inverse
function, denoted by exp, is a differentiable function (on IR) and
exp′ (y) =
1
1
=
= exp(y).
1/ exp(y)
ln′ exp(y)
One can show that exp(x) = ex for all x.
5
SOME ECONOMICAL CONCEPTS BASED ON THE DERIVATIVE
Next we discuss two applications of the derivative in economics.
EXAMPLE 17
MARGINALITY
Assume that total costs K depend on the weekly production q, so
K = k(q),
where k is a differentiable (cost) function on (0, ∞).
Corresponding to a weekly production of q = 700 we have
k ′ (700) = lim
q→700
k(q) − k(700)
.
q − 700
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142
Derivatives
So for a small change ∆q of the production level (with respect to the production level 700)
the change ∆K in the costs satisfies
=∆K
}|
{
z
k(q) − k(700)
≈ k ′ (700),
q − 700
| {z }
=∆q
or in other words:
∆K
|{z}
the change in
the costs
= k(700 + ∆q) − k(700) ≈ k ′ (700) · ∆q .
|{z}
the change in
the level of production
If the production level q rises by one, ∆q = 1 and
∆K = k(q + 1) − k(q) ≈ k ′ (q) .
| {z }
marginal cost function
In words: the derivative of a cost function, which is called the marginal cost function, can
be seen as an approximation of the change in cost if the production level is raised by one
unit.
EXAMPLE 18
Suppose q(x) bushels of wheat are harvested per acre of land when x kilos
of fertilizer per acre are used. If pw is the Euro price per bushel of wheat and pf is the Euro
price per kilo of fertilizer, then the corresponding profits in Euros per acre are
π(x) = pw q(x) − pf x.
Assume that q is a differentiable function and that profits are maximal for x∗ > 0. Then π
is a differentiable function and π ′ (x∗ ) = 0. So
π ′ (x∗ ) = 0 =⇒ pw q ′ (x∗ ) − pf = 0 =⇒ pw q ′ (x∗ ) = pf .
Let us give an economic interpretation of this condition.
Suppose x∗ units of fertilizer are used and we contemplate increasing x∗ by one unit. What
do we gain? If x∗ increases by one unit, then q(x∗ + 1) − q(x∗ ) more bushels are produced.
Now, by the foregoing example,
q(x∗ + 1) − q(x∗ ) ≈ q ′ (x∗ ).
For each of these bushels, we get pw Euros. So by increasing x∗ by one unit, we gain
≈ pw q ′ (x∗ ) Euros. On the other hand, by increasing x∗ by one unit, we lose pf Euros
because this is the cost of one unit of fertilizer.
Hence, we can interpret the condition pw q ′ (x∗ ) = pf as follows: in order to maximize profits,
you should increase the amount of fertilizer to the level x∗ at which an additional kilo of
fertilizer equates your gains and losses.
8
143
Derivatives
EXAMPLE 19
ELASTICITY
Suppose that the equation
q = 1000(1 − 12 p +
1 2
16 p )
(0 ≤ p ≤ 4)
describes the relationship between the number q of car drivers that uses everyday a certain
ferry-boat and the ferry-rate of p Euros. So q = d(p), where
d(p) = 1000(1 − 12 p +
1 2
16 p )
(0 ≤ p ≤ 4).
Today the ferry-rate is p = 2.50 and d(2.50) ≈ 141.
If the price increases with 25 cents, the relative change of the price equals
2.75 − 2.50
= 0.1,
2.50
corresponding with a change of 10%. This increase of the price causes a decrease of
d(2.50) − d(2.75) ≈ 43
car drivers daily using the ferry-boat, which corresponds with a change of the demand of
d(2.75) − d(2.50)
× 100% = −30.6%.
d(2.50)
So (for this increase of the price) the decrease of the demand (in percents) is about three
times the increase of the price (in percents).
The factor −3, which is equal to the quotient
d(2.75) − d(2.50) . 2.75 − 2.50
d(2.75) − d(2.50)
2.50
=
·
,
d(2.50)
2.50
0.25
d(2.50)
is called the price elasticity of d(emand) if the price is raised from 2.50 to 2.75.
More generally,
d(p + ∆p) − d(p)
p
d(p + ∆p) − d(p) . ∆p
=
·
d(p)
p
∆p
d(p)
is the price elasticity of demand if the price is raised from p to p+∆p. Because this elasticity
depends on ∆p (and ∆p should be small), we often prefer to consider the limit
Ed (p) = lim
∆p→0
p
p
d(p + ∆p) − d(p)
·
= d′ (p) ·
,
∆p
d(p)
d(p)
which is called the elasticity of the demand (function) d at a price p.
Note that
Ed (p) = −500 + 125p
and that Ed (2.50) = −3 31 . Since
p
1000(1 − 12 p +
1 2
16 p )
relative change of the demand ≈ Ed (2.50) × relative change of the price,
8
144
Derivatives
for a price increase of 2% (with respect to the original price of 2.50 Euros) demand decreases
approximately by
3 13 × 2 = 6 32 %.
More generally, elasticity of ’smooth’ functions can be defined as follows.
elasticity
DEFINITION
For a positive, differentiable function f on an open interval I containing c
Ef (c) = f ′ (c) ·
c
f (c)
is called the elasticity of f at c.
EXERCISE 19
Prove that the function h on (0, ∞) defined by
3
h(x) = 4x 2 ,
has constant elasticity.
EXERCISE 20
Let f and g be positive, differentiable functions on an open interval I containing c. Prove that
Ef g (c) = Ef (c) + Eg (c).
EXERCISE 21
Assume that a demand function d is a differentiable function of the price p.
The revenue function r for the demand function d is given by r(p) = p · d(p).
(a) Show that r′ (p) = (1 + Ed (p))d(p), if Ed (p) is the elasticity of demand
d at a price p.
(b) Let Er (p) be the elasticity of the revenue r at the price p. Show that
Er (p) = 1 + Ed (p).
6
STANDARD DERIVATIVES
In this section we will give the derivatives of the basic functions and their motivation as far
as possible and useful.
The power function
The derivative of the power function f : x → xa is given by f ′ (x) = a xa−1 .
If we take the domain of the function to be the positive real numbers, then this formula
holds for all real numbers a. If a is an integer number then the domain of f can be taken to
be all real numbers (except 0 if a < 0).
Motivation.
8
145
Derivatives
In Example 10 we have proved this result for natural numbers and in Exercise 14 for rational
exponents. We will not motivate why the formula can be extended to all real values of a
(and for positive values of x), as we do not offer the necessary mathematical background in
this course.
The sine function
The derivative of the sine function f : x → sin x is given by f ′ (x) = cos x.
Motivation.
Let c be a real number. Then, according to the formula
sin a − sin b = 2 sin 12 (a − b) cos 12 (a + b),
for x 6= c
2 sin 12 (c − x) cos 12 (c + x)
f (c) − f (x)
sin c − sin x
=
=
c−x
x−c
c−x
=
sin 21 (c − x)
cos 21 (c + x).
1
(c
−
x)
2
sin 1 (c − x)
sin h
= 1. So we may conclude
= 1, so that lim 1 2
x→c
h→0 h
2 (c − x)
According to Exercise 5.16, lim
that
sin 21 (c − x)
cos 12 (c + x) = cos c.
1
x→c
2 (c − x)
sin′ c = lim
The cosine function
The derivative of the cosine function f : x → cos x is given by f ′ (x) = − sin x.
Motivation.
As cos x = sin
1
2π
− x , the Chain Rule and the formula cos 21 π − x = sin x, imply that
cos′ x = cos 21 π − x · −1 = − cos 21 π − x = − sin x.
The tangens function
1
.
cos2 (x)
The formula, which can be proved by using the Quotient Rule, holds for all real values
The derivative of the tangens function f : x → tan x is given by f ′ (x) =
except . . . , − 25 π, − 32 π, − 12 π, 12 π, 32 π, 52 π, . . .
The inverse trigonometric functions
1
The derivative of the function f : x → arcsin x on (−1, 1) is given by f ′ (x) = √
.
1 − x2
−1
The derivative of the function f : x → arccos x on (−1, 1) is given by f ′ (x) = √
.
1 − x2
1
The derivative of the function f : x → arctan x is given by f ′ (x) =
.
1 + x2
In Section 4 of this chapter we have motivated these formulas by means of the Inverse
Function Theorem.
The natural logarithm function
The derivative of the function f : x → ln x on (0, ∞) is given by f ′ (x) =
1
.
x
8
146
Derivatives
This result will be discussed in Section 6 of Chapter 10.
The logarithm function
The derivative of the function f : x → a log x on (0, ∞) is given by f ′ (x) =
1
.
x ln a
The exponential function
The derivative of the exponential function f : x → ex is given by f ′ (x) = ex .
The derivative of the exponential function f : x → ax (with a > 0) is given by f ′ (x) = ax ln a.
EXERCISE 22
Find the derivative of the following functions.
√
(a) h: u → (10A)2u 3 1 − u2 with u ∈ [−1, 1].
p
(b k)2 + 3c2
(b) G: k →
with k 6= 2.
2k − 4
1
(c) f : u → arctan
1 + u2
(d) C: h → ln(1 + tan2 h).
EXERCISE 23
Suppose that f −1 is the inverse of the function f defined by f (x) = x4 + 1.
According to the Inverse Function Theorem we have that
′
f −1 (y) =
1
1
.
=
f ′ (f −1 (y))
4(f −1 (y))3
′
As f (1) = 2 and f (−2) = 17, we conclude that f −1 (2) =
′
1
f −1 (17) = − 32
.
What’s wrong with this reasoning?
1
4
and
8
147
Derivatives
Mixed exercises
EXERCISE 24
Let f be a function on IR which is continuous at 0 and for which lim
x→0
f (x)
x
exists.
(a) Prove that f (0) = 0.
(b) Prove that f is differentiable at 0.
EXERCISE 25
Show by means of the definition that the function f on (0, ∞), defined by
f (x) = x−1 , is differentiable and determine its derivative.
EXERCISE 26
Show by means of the definition that the function f on IR, defined by
3
f (x) = 3x + 2|x − 1| 2 ,
is differentiable at 1.
EXERCISE 27
Let f be a function on an open interval I containing c. Assume that f is
differentiable at c and that f ′ (c) > 0.
Prove that a δ > 0 exists such that f (x) > f (c) for every c < x < c + δ.
[Clue: Exercise 5.27.]
EXERCISE 28
Let g be the function defined by g(x) = x2 and let f be a differentiable
function on IR with the property that for every x ∈ IR
(f ◦ g)′ (x) = (g ◦ f )′ (x).
Prove that f (1) = 1 or f ′ (1) = 0.
EXERCISE 29
Consider the function f defined by
f (x) =
4x
if x ≥ 1
2
2x + 2 if x < 1.
Prove that f is differentiable and determine its derivative.
EXERCISE 30
Consider the function g defined by
g(x) =
4x
2
2x
if x ≥ 1
if x < 1.
Prove that g is not differentiable at 1.
EXERCISE 31
Find the derivative (including its domain) of the function f defined by
(a) f (x) = ln(x2 ) ex
(c) f (x) =
2
ex − e−x
ex + e−x
x2 − 1
x2 + 1
sin x2 2
(b) f (x) = ln
(d) f (x) =
cos x2
.
148
8
Derivatives
9
9
149
Significance of the derivative
SIGNIFICANCE OF THE DERIVATIVE
In this chapter we shall see that a lot of information about a function can be extracted from
the first and second derivative of that function. As you know from secondary school, the sign
of the first derivative of a function provides useful information about the monotone behavior
of that function. Furthermore extreme values of a function are related to the sign switch
of the first derivative. Finally it will appear that the ’curvature’ of the graph of a function
depends on the sign of the second derivative of that function. In this context, we introduce
the concepts of convex and concave functions that play an important role in economics.
A basic theoretical result is the Mean Value Theorem. It is the main result of this chapter
and the heart of the proofs about the monotonicity of functions. It is also used to prove
Taylor’s Theorem about polynomial approximations of functions. In the last section we
apply this theorem to prove an other widely used tool for evaluating limits of functions: de
l’Hôpital’s Rule.
1
EXTREME VALUES
In economics one wants to know when profits are maximal and when costs are minimal. So
points where a function has a maximal or a minimal value are relevant in applications.
We will start with a number of definitions. The following figure shows the graph of a
continuous function f defined on a compact interval [a, b].
y
E
C
f
G
H
A
D
F
a
b
B
FIGURE 1
A continuous function on a compact interval
x
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150
Significance of the derivative
Note that the function f takes on its greatest value at E. We say that the function has a
global maximum at this point. The function has a global minimum at B. In addition to the
global maximum occurring at E, the function also has peak values at A, C and H. We say
that the function has local maxima at these points. The function has a local minimum at D
and F . We give the formal definitions now.
DEFINITION
Let f be a function on an interval I containing c.
The function f has a global maximum at c if
global maximum
f (c) ≥ f (x)
for all x ∈ I. The function has a local maximum at c if there exists some
local maximum
ε > 0 such that
f (c) ≥ f (x)
for all x ∈ I ∩ (c − ε, c + ε).
minimum
Definitions of global minimum and local minimum can be obtained by reversing the inequa-
extremum
lities in the above definition. We shall use the term extremum to designate either a maximum
or a minimum.
Note that the Theorem of Weierstrass states that each continuous function defined on a
compact interval has a global minimum and a global maximum
Now suppose we have to determine all the (local) extrema of some function. Observe that
the graph shown in Figure 1 has a horizontal tangent at the points B, C, F and G. This
means that the derivative of the function must be zero at these points. This fact will be
proven in the following theorem.
THEOREM 1
Let f be a function on an open interval I. If f assumes a maximum or a
minimum at c ∈ I and f is differentiable at c, then f ′ (c) = 0.
PROOF
Suppose that f takes on a maximum at c. Then an ε > 0 exists such that
f (c) ≥ f (x),
whenever c − ε < x < c + ε. Note that
f ′ (c) = lim
x→c
So for a sequence xn
∞
n=1
f (x) − f (c)
.
x−c
in (c − ε, c) converging to c,
≤0
z
}|
{
f (xn ) − f (c)
→ f ′ (c) as n → ∞.
xn − c
| {z }
<0
According to Theorem 3.1, this implies that f ′ (c) ≥ 0.
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Significance of the derivative
By considering a sequence at the right-hand side of c, one can prove in a similar way that
f ′ (c) ≤ 0. So we may conclude that f ′ (c) = 0.
The converse of Theorem 1 is definitely not true: it is possible that the derivative of a
function f is zero at c even if f doesn’t have a minimum or maximum at c (consider point
G in Figure 1). A simple example of this situation is provided by the function f defined by
f (x) = x3 .
For this function f ′ (0) = 0, but f has no maximum or minimum anywhere. Compare this
with the situation at point G in Figure 1.
EXERCISE 1
Prove that the function f defined by f (x) = x3 has no extremum at zero.
Figure 1 suggests that a continuous function f defined on a closed interval [a, b] can have
an extreme value at c only if c is one of the following three types of points:
stationary point
singular point
1
c ∈ (a, b) is a stationary point of f , that is: f ′ (c) = 0
2
c is one of the endpoints a and b
3
c is a singular point of f , that is: f is not differentiable at c.
In Figure 1, B, C, F and G correspond to the stationary points of the function, whereas D
and E correspond to the singular points of the function.
EXERCISE 2
Determine all stationary points of the following functions:
(a) f : x → x3 − 3x2 − 9x + 12
(c) F : v → ln(1 + (v − 2)2 )
2
(b) g: u → Au3 + Bu2 + Cu + D
(d) h: y → e−y (y 2 − 2y − 7).
THE MEAN VALUE THEOREM
In this section we will discuss the main result of this chapter. Its proof will be based on the
following result (which is in fact a simple version of the Mean Value Theorem).
THEOREM 2
ROLLE’S THEOREM
Let f be a continuous function on an interval [a, b] that is differentiable on
(a, b).
If f (a) = f (b), then there exists a number τ in (a, b) such that f ′ (τ ) = 0.
PROOF
According to the Theorem of Weierstrass, the function f takes on a global max-
imum and a global minimum on the interval [a, b].
Suppose first that the maximum value or the minimum value occurs at τ ∈ (a, b). Then
f ′ (τ ) = 0 by Theorem 1, and the proof is complete.
Next suppose that the function has both its maximal value and its minimal value at the
endpoints of the interval. Since f (a) = f (b), in that situation, the maximum and minimum
values of f we can choose any τ ∈ (a, b).
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Significance of the derivative
The geometric interpretation of Rolle’s Theorem (see Figure 2) is that if the graph of a
differentiable function touches a horizontal line at a and b, then for some number τ between
a and b there is a horizontal tangent.
y
f
f (a) = f (b)
a
τ
b
x
FIGURE 2
The graph of f has a horizontal tangent
EXERCISE 3
Let g be a continuous function on an interval [a, b], which is differentiable on
(a, b). Suppose that g ′ (x) 6= 0 for every x ∈ (a, b).
Prove that g(x) 6= g(x′ ) for all x, x′ ∈ [a, b] with x 6= x′ .
[Clue: a proof by contradiction could be helpful.]
In Example 12 of Chapter 5, we showed how the Intermediate Value Theorem can be used
to prove that an equation has at least one solution. In the next example we will indicate
how Rolle’s Theorem can be applied to prove that an equation has at most one solution.
EXAMPLE 1
We will prove that the equation
x3 + x − 3 = 0
has a unique solution.
If f is the function defined by
f (x) = x3 + x − 3,
then f (0) = −3 and f (2) = 7 and the (polynomial) function f restricted to the interval [0, 2]
is continuous.
Since f (0) < 0 < f (2), according to the Intermediate Value Theorem there exists a number
c ∈ (0, 2) such that f (c) = 0. So the foregoing equation has at least one solution.
In order to prove that there is only one solution, assume that
f (c) = f (d) = 0,
where c 6= d. Say c < d. The function f restricted to the interval [c, d] is continuous and
differentiable on the interval (c, d). Since f (c) = f (d), Rolle’s Theorem implies the existence
of a τ ∈ (c, d) such that f ′ (τ ) = 0. However, f ′ (τ ) = 3τ 2 + 1 6= 0.
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Significance of the derivative
EXERCISE 4
Consider the equation
x5 + 2x3 + x − 5 = 0.
(a) Prove that this equation has a solution.
(b) Prove that this equation has a unique solution.
If we allow a function f on [a, b] to have different values at the endpoints, then there will
be a number τ between a and b such that the tangent to the graph at the point τ, f (τ )
will be parallel to the chord between the endpoints of the graph. This is represented in the
following figure.
y
f
f (a)
f (b)
a
FIGURE 3
τ
b
x
The Mean Value Theorem
This is the essence of the Mean Value Theorem.
THEOREM 3
MEAN VALUE THEOREM
Let f be a continuous function on an interval [a, b] that is differentiable on
(a, b). Then there exists a number τ in (a, b) such that
f ′ (τ ) =
f (b) − f (a)
.
b−a
As discussed in part (c) of Exercise 1.30, the chord joining the points a, f (a)
and b, f (b) is the graph of the linear function ℓ on [a, b] defined by
PROOF
f (b) − f (a)
ℓ(x) =
(x − a) + f (a).
b−a
We will apply Rolle’s Theorem to the function h on [a, b] which is the difference of the
functions f and ℓ. So
h(x) = f (x) − ℓ(x).
One can easily verify that h(a) = h(b) = 0.
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Significance of the derivative
Because the functions f and ℓ are continuous on [a, b], the function h is continuous on [a, b].
Similarly, the function h is differentiable on the interval (a, b). Hence, the function h satisfies
the conditions of Rolle’s Theorem. According to that theorem, there exists a τ ∈ (a, b) such
that
h′ (τ ) = 0 ⇐⇒ f ′ (τ ) − ℓ′ (τ ) = 0 ⇐⇒ f ′ (τ ) =
f (b) − f (a)
.
b−a
Consider the function f on an interval [a, b], defined by f (x) = x2 .
f (b) − f (a)
Determine a number τ ∈ (a, b) such that f ′ (τ ) =
.
b−a
EXERCISE 5
THEOREM 4
Let f be a continuous function on an interval [a, b] that is differentiable on
(a, b). If f ′ (x) = 0 for all x ∈ (a, b), then f is constant on [a, b].
PROOF
Let c ∈ (a, b]. We will prove that f (c) = f (a).
The function f restricted to the interval [a, c] is continuous and differentiable on the interval
(a, c). So according to the Mean Value Theorem a number τ ∈ (a, c) exist such that
f (c) − f (a)
= f ′ (τ ).
c−a
Since τ belongs to (a, b), it follows that f ′ (τ ) = 0. Thus, f (c) − f (a) = 0 or f (c) = f (a).
EXERCISE 6
Let f and g be continuous functions on an interval [a, b], which are differentiable on (a, b). Suppose that f ′ = g ′ on (a, b).
Prove that a constant C exists such that f (x) = g(x) + C for every x ∈ [a, b].
EXERCISE 7
(a) Let f and g be two differentiable functions such that f (0) = g(0) and
f ′ (x) ≤ g ′ (x) for all x > 0.
Prove that f (x) ≤ g(x) for all x ≥ 0.
(b) After introducing the proper functions f and g, use part (a) to prove that,
for all x ≥ 0,
sin x ≤ x.
(c) As in part (b), prove that cos x ≥ 1 − 21 x2 for all x ≥ 0 and use this
inequality to evaluate the limit
lim
x↓0
EXAMPLE 2
cos x − 1
.
x
Let f be a continuous function on the interval [0, 1] that is differentiable
on (0, 1). Furthermore, f (0) = 0, f (1) = 1 and f ′ (x) ≤ 1 for all x ∈ (0, 1).
We will prove that f (x) = x for all x ∈ [0, 1].
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Significance of the derivative
Let c ∈ (0, 1). Then the function f restricted to the interval [0, c] is continuous and differentiable on (0, c). According to the Mean Value Theorem, there exists a τ ∈ (0, c) such
that
f ′ (τ ) =
f (c)
f (c) − f (0)
=
.
c−0
c
Since f ′ (τ ) ≤ 1 and c > 0, this implies that f (c) ≤ c.
Furthermore, the function f restricted to the interval [c, 1] is continuous and differentiable
on (c, 1). According to the Mean Value Theorem, there exists a σ ∈ (c, 1) such that
f ′ (σ) =
f (1) − f (c)
1 − f (c)
=
.
1−c
1−c
Since f ′ (σ) ≤ 1 and 1 − c > 0, this implies that
1 − f (c) ≤ 1 − c =⇒ f (c) ≥ c.
Hence, f (c) = c. Since c was arbitrarily chosen, the proof is complete.
EXAMPLE 3
We will prove that
lim x2 ln x = 0.
x↓0
We consider the differentiable function f on the interval (0, ∞) defined by
f (x) = ln x.
Let x ∈ (0, 1). The function f restricted to the interval [x, 1] is continuous and it is differentiable on the interval (x, 1). According to the Mean Value Theorem, there exists a τ ∈ (x, 1)
such that
As x < τ < 1,
1<
− ln x
1
1
f (1) − f (x)
= f ′ (τ ) =⇒
= =⇒ ln x = − (1 − x).
1−x
1−x
τ
τ
1
1
1
1
x−1
< =⇒ −(1 − x) > − (1 − x) > − (1 − x) =⇒
< ln x < x − 1.
τ
x
τ
x
x
This inequality can be used as follows. By multiplying by x2 we obtain
x(x − 1) < x2 ln x < x2 (x − 1).
As lim x(x − 1) = lim x2 (x − 1) = 0, the Sandwich Lemma for functions implies that
x↓0
x↓0
lim x2 ln x = 0.
x↓0
EXERCISE 8
Let f be a differentiable function on IR such that f ′ (x) > 0 for every x ∈ IR
and let c ∈ IR. Suppose that f (c) = 0.
Prove that f (x) < 0 for every x < c and f (x) > 0 for every x > c.
3
MONOTONE FUNCTIONS
Intervals on which the derivative of a function is positive or negative provide useful information about the behavior of that function. This behavior is described in the following
definition.
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156
DEFINITION
Significance of the derivative
Let f be a function on an interval I.
The function f is called increasing if f (x) ≤ f (x′ ) for all x, x′ ∈ I with
(strictly) increasing
x < x′ .
If f (x) < f (x′ ) for all x, x′ ∈ I with x < x′ , then f is called strictly increasing.
The function f is called decreasing if f (x) ≥ f (x′ ) for all x, x′ ∈ I with
(strictly) decreasing
x < x′ .
If f (x) > f (x′ ) for all x, x′ ∈ I with x < x′ , then f is called strictly decreasing.
A function that is (strictly) increasing or (strictly) decreasing is called (strict-
(strictly) monotone
ly) monotone.
THEOREM 5
Let f be a continuous function defined on an interval [a, b]. If this function
is differentiable on (a, b), then
(a) f is increasing if and only if f ′ (x) ≥ 0 for all x ∈ (a, b)
(b) f is strictly increasing if f ′ (x) > 0 for all x ∈ (a, b)
(c) f is decreasing if and only if f ′ (x) ≤ 0 for all x ∈ (a, b)
(d) f is strictly decreasing if f ′ (x) < 0 for all x ∈ (a, b).
PROOF
We only give a proof of part (a). The proof of the other parts is similar.
(a) (⇒) Assume that f ′ (x) ≥ 0 for all x ∈ (a, b).
Let x, x′ ∈ [a, b] with x < x′ . Since the function f restricted to the interval [x, x′ ] is
continuous and differentiable on (x, x′ ), the Mean Value Theorem implies that there exists
a number τ ∈ (x, x′ ) such that
f (x′ ) − f (x) = f ′ (τ )(x′ − x).
Since f ′ (τ ) ≥ 0 and x′ − x > 0, we must have f (x′ ) − f (x) ≥ 0, so that f (x′ ) ≥ f (x). Thus
the function f is increasing.
(⇐) Assume that f is increasing and let c ∈ (a, b).
Then for any x 6= c,
f (x) − f (c)
≥ 0.
x−c
According to Exercise 7.13, this implies that
f (x) − f (c)
≥ 0.
x→c
x−c
f ′ (c) = lim
Note that the converse of part (b) as well as the converse of part (d) of the foregoing theorem
is false: the function g: x → x3 is strictly increasing and g ′ (0) = 0.
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Significance of the derivative
We will prove, by using Theorem 5, that the function
EXAMPLE 4
g: x → x3
is strictly increasing.
Let x < x′ ≤ 0. Since g ′ (x) = 3x2 > 0, Theorem 5 (b), applied to the function g restricted
to the interval [x, 0], implies that g(x) < g(x′ ).
Similarly, g(x) < g(x′ ) if 0 ≤ x < x′ .
Finally, if x < 0 < x′ , then g(x) < 0 < g(x′ ).
EXERCISE 9
(a) Let f be a differentiable function on an open interval I such that f ′ ≥ 0
on I. Prove that f is strictly increasing unless f ′ is zero on some open
interval.
(b) Prove that the function g, defined by
g(x) = x + sin x,
is strictly increasing.
EXERCISE 10
Let f be the function on IR \ {0}, defined by
f (x) = x +
1
.
x
Determine the intervals where f is strictly increasing and where it is strictly
decreasing.
Let f be a differentiable function on an open interval (a, b) containing c.
We proved in Theorem 1 that f ′ (c) = 0 is a necessary but not sufficient condition for an
extreme value at c. A simple way to prove that the function f has an extreme value at c is
to show that the derivative of f switches sign at c. For instance, if
• f ′ (c) = 0,
• f ′ < 0 on the interval (a, c), and
• f ′ > 0 on the interval (c, b),
then the function f has a minimum at c.
Instead of proving this sufficient condition, we will formulate a more general result.
THEOREM 6
THE SWITCHING SIGN TEST
Let f be a continuous function on an open interval (a, b) containing c.
Suppose that
(a) f is a differentiable function on the intervals (a, c) and (c, b),
(b) f ′ < 0 on the interval (a, c), and
(c) f ′ > 0 on the interval (c, b).
Then the function f has a minimum at c.
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Significance of the derivative
EXERCISE 11
Prove Theorem 6.
EXERCISE 12
Let f be a continuous function on [0, 1], which is differentiable on (0, 1).
Assume that f (0) = 0 and that f ′ is increasing on (0, 1).
f (x)
Prove that the function g on (0, 1), defined by g(x) =
, is increasing on
x
(0, 1).
EXAMPLE 5
The function f , defined by
f (x) = 1 + 3x2/3 ,
is continuous everywhere but differentiable except at zero. Furthermore, for all x 6= 0,
f ′ (x) =
2
3
2
.
· 3x−1/3 = √
3
x
Clearly, f ′ (x) < 0 if x < 0 and f ′ (x) > 0 if x > 0. This is presented in the following sign
diagram.
−
−
−
×
+
+
0
ց
ր
0
+
f′
f
So, according to the Switching Sign Test, the function f has a minimum at zero.
It is also possible to formulate sufficient conditions for an extreme value of a function in
terms of the second derivative of that function.
THEOREM 7
SECOND-DERIVATIVE TEST
Let f be a twice-differentiable function on an open interval I containing c
and assume that f ′ (c) = 0.
(a) If f ′′ (c) > 0, then the function attains a (local) minimum at c.
(b) If f ′′ (c) < 0, then the function attains a (local) maximum at c.
PROOF
We give a proof of part (a).
Assume that f ′′ (c) > 0. By definition
f ′ (x) − f ′ (c)
f ′ (x)
= lim
.
x→c
x→c x − c
x−c
0 < f ′′ (c) = lim
According to Exercise 5.27, a δ > 0 exists such that
f ′ (x)
> 0,
x−c
whenever x 6= c is in the interval (c − δ, c + δ). So if c < x < c + δ, then f ′ (x) > 0. And
similarly, if c − δ < x < c, then f ′ (x) < 0.
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Significance of the derivative
Since f is continuous at c (because it is differentiable at c), the function f restricted to the
interval (c − δ, c + δ) satisfies the conditions of the Switching Sign Test.
Hence, the function f attains a (local) minimum at c.
EXERCISE 13
Let f be a twice-differentiable function on an open interval I containing c.
Prove that f ′′ (c) ≥ 0 if the function attains a (local) minimum at c.
[Clue: a proof by contradiction might be helpful.]
EXERCISE 14
Find the extrema of the function f on IR \ {0}, defined by
f (x) = x +
4
1
.
x
CONVEX AND CONCAVE FUNCTIONS
In this section, we shall be concerned with the significance of the sign of the second derivative.
It will appear that if the second derivative of a function on an open interval is positive, then
the chord joining any two points of the graph lies above the graph.
If a function f on an open interval I has a positive second derivative, then the first derivative
is a strictly increasing function. By consequence, if we move along the graph of the function
(from left to right), the tangent to the graph turns in a counterclockwise direction. Intuitively
this means that the graph of the function bends upward.
y
y
f
f
a
b
x
a
b
x
If on the other hand, f has a negative second derivative, then the tangent will rotate in a
clockwise fashion. In that case we say that the graph bends downward.
We will make these ideas precise by introducing the concepts of convexity and concavity.
convex function
DEFINITION
Let f be a function on an open interval I.
The function f is said to be convex if any chord joining two points of the
graph lies above the graph of f .
concave function
The function is called concave if the function −f is convex.
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Significance of the derivative
In the next theorem we will describe how convexity can be checked by means of the (sign of
the) second derivative.
THEOREM 8
Let f be a twice-differentiable function on an open interval I. If f ′′ > 0 on
the interval I, then the function f is convex.
Let a, b ∈ I with a < b. We will prove that the chord between the points a, f (a)
and b, f (b) lies above the graph of f . This chord is the graph of a linear function, say g,
PROOF
on the interval [a, b]. Note that the derivative of this linear function is
f (b) − f (a)
.
b−a
We will prove that the difference d = g − f is strictly positive on the interval (a, b).
According to the Mean Value Theorem, applied to the function f restricted to the interval
[a, b], there exists a point c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
= g ′ (c).
b−a
It will appear that the difference between f and g is maximal at c.
y
f
g
a
FIGURE 4
c
b
x
The difference between f and g is maximal at c
Since f ′′ > 0 on the interval I, the function f ′ is strictly increasing on the interval I. Hence,
for any x ∈ (a, c), f ′ (x) < f ′ (c) = g ′ (c). So for such x,
d′ (x) = g ′ (x) − f ′ (x) > g ′ (c) − g ′ (c) = 0,
which implies that the function d is strictly increasing on the interval [a, c]. Hence,
d(x) > d(a) = 0 if a < x ≤ c.
Similarly, d(x) > 0 if c ≤ x < b, which proves that d ≥ 0 on the interval [a, b].
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Significance of the derivative
EXAMPLE 6
Let f be the function we introduced in Example 5, that is
√
3
f (x) = 1 + 3 x2 .
As for all x 6= 0, f ′ (x) = 2x−1/3 ,
f ′′ (x) = − 23 x−4/3 = −
2
√ .
3x 3 x
Clearly, f ′′ < 0 on the intervals (−∞, 0) and (0, ∞). This is presented in the following sign
diagram.
−
−
−
×
−
−
−
0
∩
∩
f ′′
f
0
According to the foregoing theorem, the function is concave (indicated by ∩) on these
intervals.
The graph of the function is represented in the next figure.
y
f
1
−2
FIGURE 5
EXERCISE 15
2
x
The graph of the function f
Determine the intervals where the function g, defined by
g(x) =
1
,
1 + x2
is convex and find those intervals where the function is concave.
EXERCISE 16
Determine the intervals where the function f on IR \ {0}, defined by
f (x) = x +
1
,
x
is convex and find those intervals where the function is concave.
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162
5
Significance of the derivative
TAYLOR’S THEOREM
For your pocket calculator and even for the most advanced computer it is impossible to
√
1
calculate for expressions like x, 2x and the precise value at x = 13 . The reason is quite
x
simple: your pocket calculator can only add and multiply numbers. If you use your pocket
q
calculator to determine 13 , then your calculator evaluates for a function ’close’ to the root
function the value at 31 . Approximating functions by polynomial ones is the subject of this
section.
Let f be a differentiable function on an open interval I, containing c.
In the foregoing chapter, the line
y = f ′ (c)(x − c) + f (c)
was called the tangent to the graph of f at the point c, f (c) . This tangent is the graph of
the linear function
ℓc : x → f ′ (c)(x − c) + f (c),
which we call the linear approximation for f at c.
y
f
ℓc
c
FIGURE 6
x
The linear approximation
That we call ℓc an approximation for f at c is, from an intuitive point of view, based on the
fact that the tangent fits the graph in an acceptable way: for x close to c, f (x) is close to
ℓc (x). We will give a formal definition of ’approximation’ now.
approximation
remainder
DEFINITION
Let f be a function on an open interval I containing c. A function h on I
is called an approximation for the function f at c if the remainder r, defined
by
r(x) = f (x) − h(x)
satisfies
lim r(x) = 0.
x→c
(x ∈ I),
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163
Significance of the derivative
Note that for the linear approximation
r(x) = f (x) − ℓc (x) = f (x) − [f ′ (c)(x − c) + f (c)] = f (x) − f (c) − f ′ (c)(x − c).
Hence, the Arithmetic Rules for limits of functions and the continuity of f at c imply that
lim r(x) = lim f (x) − f (c) − f ′ (c)(x − c) = 0.
x→c
x→c
√
Let f be the function on [0, ∞), defined by f (x) = x.
1
Since, for x > 0, f ′ (x) = √ , the linear approximation for f at 25 is given by
2 x
EXAMPLE 7
ℓ25 (x) = f (25) + f ′ (25)(x − 25) = 5 +
1
10 (x
− 25).
Putting x = 26, we get
√
26 = f (26) ≈ ℓ25 (26) = 5 +
1
10 (26
− 25) = 5.1.
If we use the square root function on a calculator, we obtain the value
√
which is by the way also an approximation of the true value of 26.
√
26 = 5.0990195 . . .,
If we don’t know the value of a function at some point and we use an approximation instead,
we would like to know how good the approximation will be. So we would like to have an
expression for the remainder whose size is easy to estimate. For the linear approximation,
one such an expression – there exist several – will be derived in the next theorem.
THEOREM 9
TAYLOR’S THEOREM
Let f be a twice-differentiable function on an open interval I containing c.
Then for each x ∈ I \ {c}, there exists a point τ between x and c such that
f (x) = lc (x) +
PROOF
f ′′ (τ )
f ′′ (τ )
(x − c)2 = f (c) + f ′ (c)(x − c) +
(x − c)2 .
2
2
Let x ∈ I \ {c} be fixed throughout this proof, and let
C=
f (x) − [f (c) + f ′ (c)(x − c)]
.
(x − c)2
f ′′ (τ )
for some τ between x and c.
2
To this end we introduce the function h on I, defined by
We will show that C =
h(t) = f (t) + f ′ (t)(x − t) + C(x − t)2 .
In view of the properties of the function f , the function h is differentiable on the interval I.
Further, h(x) = f (x) and by the definition of C,
h(c) = f (c) + f ′ (c)(x − c) + C(x − c)2 = f (x).
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164
Significance of the derivative
The function h restricted to the closed interval with endpoints x and c is continuous on this
interval and differentiable on the corresponding open interval.
Then, according to Rolle’s Theorem, there exists a point τ between x and c such that
h′ (τ ) = 0. However,
h′ (t) = f ′ (t) + [f ′′ (t)(x − t) − f ′ (t)] − 2C(x − t) = (x − t) [f ′′ (t) − 2C] ,
so that
h′ (τ ) = 0 ⇐⇒ (x − τ ) [f ′′ (τ ) − 2C] = 0 ⇐⇒ f ′′ (τ ) − 2C = 0 ⇐⇒ C =
f ′′ (τ )
.
2
We will estimate the size of the error in the approximation
EXAMPLE 8
√
26 ≈ 5.1
obtained in Example 7.
For x > 0, f ′′ (x) = −
f (x) = 5 +
1
√ . So for x 6= 25,
4x x
1
10 (x
− 25) +
f ′′ (τ )
(x − 25)2 = 5 +
2
1
10 (x
− 25) −
1
√ (x − 25)2 ,
8τ τ
where τ is between 25 and x.
For 25 < τ < 26, we have
|f ′′ (τ )| <
So
1
√ =
4 · 25 25
f ′′ (τ )
(26 − 25)2 <
2
Therefore 5.1 >
1
500 .
1
1000 .
√
26 > 5.1 − 0.001 = 5.099.
Let f be a twice-differentiable function on an open interval I with f ′′ > 0 on the interval I.
If c ∈ I, then according to Taylor’s Theorem, there exists, for each x ∈ I \ {c}, a point τ be
tween x and c such that
f (x) = f (c) + f ′ (c)(x − c) +
f ′′ (τ )
f ′′ (τ )
(x − c)2 = ℓc (x) +
(x − c)2 > ℓc (x).
2
2
|
{z
}
>0
Since f (c) = ℓc (c), this implies that f ≥ ℓc . In geometrical terms this means that the graph
of the function f lies above any tangent to this graph.
EXERCISE 17
Find the linear approximation at 1 for the function f on (0, 2), defined by
f (x) = x +
1
.
x
Also give a formula for the remainder.
Note that for the linear approximation ℓc for a function f at c it holds that
ℓc (c) = f (c)
and
ℓ′c (c) = f ′ (c).
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Significance of the derivative
One could try to find a ’better’ approximation for f by using quadratic or higher degree
polynomials and matching more derivatives at c. For example if f is twice-differentiable,
then the quadratic polynomial p2 defined by
f ′′ (c)
(x − c)2
2
p2 (x) = f (c) + f ′ (c)(x − c) +
satisfies p2 (c) = f (c), p′2 (c) = f ′ (c) and p′′2 (c) = f ′′ (c).
Taylor polynomial
This polynomial is called the Taylor polynomial of degree 2 for f at c.
EXAMPLE 9
Consider the function f on (0, 1) defined by
f (x) =
1
.
9x
1
3
The linear approximation ℓc for the function f at c =
(or the first-degree Taylor polynomial
p1 for the function f at 31 ), is given by
p1 (x) = f
1
3
1
3
+ f′
x−
The second-degree Taylor polynomial p2 for f at
p2 (x) = f
1
3
+ f′
1
3
= p1 (x) + 12 f ′′
= 3x2 − 3x + 1.
x−
1
3
1
3
1
3
x−
1
3
2
3
=
− x.
is given by
+ 21 f ′′
1
3
1 2
3
x−
1 2
3
Both approximations are represented in the figure below.
y
1
p2
2
3
1
3
p1
1
3
FIGURE 7
f
2
3
1
x
A linear and a quadratic approximation
If we approximate f (0.3) using p1 we get f (0.3) ≈ p1 (0.3) = 0.366 67 and using p2 we get
f (0.3) ≈ p2 (0.3) = 0.37, whereas f (0.3) =
EXERCISE 18
10
27
= 0.37037 . . ..
Find the Taylor polynomial of degree 2 at 1 for the function f , defined by
√
(a) f (x) = x (x ≥ 0)
3
(b) f (x) =
(x 6= − 12 ).
1 + 2x
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Significance of the derivative
Also for the second degree polynomial approximation an expression for the remainder can
be derived, but we will not do that here. And of course the process can be continued to
third and higher degree Taylor polynomials.
As an example, let f be a function defined on an open interval containing c such that f (3)
exists. Then the third degree Taylor polynomial p3 for f at c is the unique polynomial of
(3)
degree 3 that satisfies p3 (c) = f (c), p′3 (c) = f ′ (c), p′′3 (c) = f ′′ (c) and p3 (c) = f (3) (c).
EXERCISE 19
Prove that the coefficient of x3 in p3 is equal to 61 f (3) (c).
EXERCISE 20
Derive a formula for the Taylor polynomial p3 of degree 3 for the function f
at 1, where
(a) f (x) = x +
6
1
x
(b) f (x) =
√
x.
DE L’HÔPITAL’S RULE
In many situations one wants to evaluate the limit of a quotient of functions like
lim
x→c
f (x)
.
g(x)
If lim f (x) = ℓ, lim g(x) = m 6= 0 and g(x) 6= 0 in a neighborhood of c, then according to
x→c
x→c
the Arithmetic Rules for limits of functions, this limit is equal to ℓ/m.
Observe that it is also possible that the limit exists if m = 0. As an example of this situation
consider the limit
lim
x→0
sin x
=1
x
we discussed in Exercise 16 of Chapter 5. However, if m = 0 and the limit lim
x→c
then
lim f (x) = lim
x→c
x→c
f (x)
exists,
g(x)
f (x)
f (x)
· g(x) = lim
· lim g(x) = 0.
x→c g(x) x→c
g(x)
So it makes sense to consider the limit
lim
x→c
f (x)
g(x)
for the case that lim f (x) = lim g(x) = 0. In this section we will discuss two methods to
x→c
x→c
deal with such limits.
First of all, note that sometimes it is possible to evaluate such a limit by cancelling a common
factor in the quotient as in
(x + 1)(x − 2)
x2 − x − 2
= lim
= lim (x − 2) = −3.
x→−1
x→−1
x→−1
x+1
x+1
lim
In this section we will derive another technique, the so-called Rule of de l’Hôpital, that has
wider application. First we will consider a weak form, then it will appear that sometimes a
stronger version is needed.
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167
Significance of the derivative
EXERCISE 21
THEOREM 10
Evaluate the limits
x2
(a) lim
2
x→0 2x + x
(b) lim
x→2
2x2 − x − 6
.
3x2 − 7x + 2
THE RULE OF DE L’HÔPITAL (WEAK FORM)
Let f and g be differentiable functions on an open interval I containing c.
If
• g ′ (x) 6= 0 for every x ∈ I
• f (c) = g(c) = 0,
then
f ′ (c)
f (x)
= ′ .
x→c g(x)
g (c)
lim
PROOF
(a) First we will show that g(x) 6= 0 for every x ∈ I\{c}.
Assume that a d ∈ I exists with d 6= c and g(d) = 0. According to Rolle’s Theorem, applied
to the function g restricted to the interval with endpoints c and d, a point τ between c and
d exists such that g ′ (τ ) = 0. This contradicts the fact that g ′ (x) 6= 0 for every x ∈ I.
(b) For every x ∈ I with x 6= c
f (x) − f (c)
f (x)
f (x) − f (c)
x−c
.
=
=
g(x) − g(c)
g(x)
g(x) − g(c)
x−c
Because the limits of numerator and denominator at c exist and g ′ (c) 6= 0, according to the
Arithmetic Rules for limits of functions,
f (x) − f (c)
lim
f ′ (c)
f (x)
x→c
x−c
= ′ .
=
lim
x→c g(x)
g(x) − g(c)
g (c)
lim
x→c
x−c
EXAMPLE 10
In order to show that
lim
x→0
1 − cos x
= 0,
x
we introduce the differentiable functions f and g defined by f (x) = 1 − cos x and g(x) = x.
Then, for all x,
g ′ (x) = 1 6= 0
and f (0) = g(0) = 0. So the weak form of de l’Hôpitals Rule implies that
lim
x→0
EXERCISE 22
sin 0
1 − cos x
=
= 0.
x
1
Evaluate the limit
lim
x→0
1 √
x + 1 − 1 + 21 x .
x
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168
Significance of the derivative
Note that we cannot use the technique applied in the foregoing example to evaluate the limit
lim
x→0
1 − cos x
.
x2
So we need a stronger result. Before we discuss a stronger form of de l’Hôpital’s Rule, we
first derive the following generalization of the Mean Value Theorem.
THEOREM 11
CAUCHY’S MEAN VALUE THEOREM
Let f and g be functions that are continuous on an interval [a, b] and differentiable on (a, b). If g ′ (x) 6= 0 for every x ∈ (a, b), then there exists a
number τ in (a, b) such that
f ′ (τ )
f (b) − f (a)
= ′
.
g(b) − g(a)
g (τ )
PROOF
From the fact that g ′ (x) 6= 0 for every x ∈ (a, b) it follows with the Mean Value
Theorem that g(a) 6= g(b). As
f (b) − f (a)
f ′ (τ )
= ′
⇐⇒ [f (b) − f (a)]g ′ (τ ) − [g(b) − g(a)]f ′ (τ ) = 0,
g(b) − g(a)
g (τ )
we introduce the function h on [a, b], defined by
h(x) = [f (b) − f (a)]g(x) − [g(b) − g(a)]f (x).
We must prove that there exists a number τ in (a, b) such that h′ (τ ) = 0.
Note that the function h is continuous on [a, b] and differentiable on (a, b). Furthermore,
h(a) = f (b)g(a) − g(b)f (a) = h(b).
So, according to Rolle’s Theorem, there exists a number τ in (a, b) such that h′ (τ ) = 0.
To see that this result is a generalization of the Mean Value Theorem, we choose g(x) = x
for all x ∈ [a, b]. Obviously, g satisfies the conditions of Cauchy’s Mean Value Theorem and
the formula in the theorem simplifies to
f (b) − f (a)
= f ′ (τ ).
b−a
Using the Cauchy’s Mean Value Theorem, we are able to derive a stronger version of de
l’Hôpital’s rule. In this version the functions f and g are not necessarily differentiable at c.
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Significance of the derivative
THEOREM 12
THE RULE OF DE L’HÔPITAL (STRONG FORM)
Let f and g be continuous functions on an open interval I containing c.
Assume that
• f and g are differentiable on I \ {c}
• g ′ (x) 6= 0 for every x ∈ I \ {c}
• f (c) = g(c) = 0.
If
f ′ (x)
= ℓ,
x→c g ′ (x)
lim
then
lim
x→c
PROOF
Let xn
∞
f (x)
= ℓ.
g(x)
As in part (a) of Theorem 10 one can show that g(x) 6= 0 for every x ∈ I \ {c}.
n=1
be a sequence in I converging to c such that xn > c for all n. According to
Cauchy’s Mean Value Theorem, applied to the functions f and g restricted to the interval
[c, xn ], a number τn in (c, xn ) exists such that
f ′ (τn )
f (xn )
f ′ (τn )
f (xn ) − f (c)
= ′
⇐⇒
= ′
.
g(xn ) − g(c)
g (τn )
g(xn )
g (τn )
As τn ∈ (c, xn ) for any n, the Sandwich Lemma implies that τn → c as n → ∞.
f ′ (x)
= ℓ,
Since lim ′
x→c g (x)
f ′ (τn )
f (xn )
= lim ′
= ℓ.
lim
n→∞ g (τn )
n→∞ g(xn )
∞
As the sequence xn n=1 was arbitrarily chosen (at the right-hand side of c), this proves
f (x)
f (x)
that lim
= ℓ. Similarly, lim
= ℓ.
x↓c g(x)
x↑c g(x)
EXAMPLE 11
In order to evaluate the limit
lim
x→0
1 − cos x
,
x2
we introduce the continuous functions f and g defined by f (x) = 1 − cos x and g(x) = x2 .
The functions f and g are differentiable. Furthermore, for all x 6= 0,
g ′ (x) = 2x 6= 0
and f (0) = g(0) = 0. Finally, since
sin x
f ′ (x)
= lim
= 21 ,
x→0 2x
x→0 g ′ (x)
lim
the strong form of de l’Hôpitals Rule implies that
lim
x→0
1 − cos x
= 12 .
x2
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170
EXAMPLE 12
Significance of the derivative
In order to evaluate the limit
x3 − 1
lim p
,
x→1 3 (x − 1)2
we introduce the continuous functions f and g, defined by f (x) = x3 − 1 and g(x) =
p
3
(x − 1)2 .
The functions f and g are differentiable on IR \ {1}. Furthermore, for all x 6= 1,
2
g ′ (x) = √
6= 0
3
3 x−1
and f (1) = g(1) = 0.
Finally, since
√
f ′ (x)
= lim 92 x2 3 x − 1 = 0,
′
x→1
x→1 g (x)
lim
the strong form of de l’Hôpitals Rule implies that
x3 − 1
= 0.
lim p
3
x→1
(x − 1)2
EXERCISE 23
Evaluate the limits
ln2 x
x→1 (x − 1)2
ln x
x→1 x − 1
(b) lim
(a) lim
EXERCISE 24
(c) lim
x→1
| ln x|
.
|x2 − 1|
We ’prove’ that
lim
x→0
x
=0
x2 + sin x
as follows:
As numerator and denominator are zero at x = 0, we may apply de l’Hôpital’s
Rule leading to
lim
x→0 x2
1
x
= lim
.
+ sin x x→0 2x + cos x
If we apply this rule once again we get
0
1
= lim
= 0.
x→0 2 − sin x
x→0 2x + cos x
lim
Find the mistake we made.
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Significance of the derivative
Mixed exercises
EXERCISE 25
Consider the function f on [0, 2], defined by f (x) = x5 − 16x.
f (2) − f (0)
.
Determine a point τ ∈ (0, 2) such that f ′ (τ ) =
2
EXERCISE 26
Let f be a differentiable function such that lim f ′ (x) = 0. Assume that a
x→∞
positive constant c exists such that |f (x)| ≥ c, for every x ∈ IR.
Prove that
lim
x→∞
1
1
= 0.
−
f (x + 1) f (x)
[Clue: apply the Mean Value Theorem.]
EXERCISE 27
Prove that the polynomial p, defined by
p(x) = x3 + ax + b,
where a > 0 and b < 0, has precisely one zero. vskip8pt
EXERCISE 28
Determine the intervals, where the function h on [0, 2], defined by
h(x) = 1 −
p
3
(x − 1)2
is (strictly) increasing, and the intervals on which the function is (strictly)
decreasing.
EXERCISE 29
Let f be a continuous function on [a, b] that is differentiable on (a, b). Assume
that lim f ′ (x) = ℓ.
x→a
Prove that the function f has a right hand derivative at a and that f ′ (a) = ℓ.
EXERCISE 30
Let I be an open interval containing c and let f and g be functions defined
on I \ {c} such that g(x) 6= 0 for every x ∈ I \ {c}.
Assume that lim f (x) = ℓ 6= 0 and that lim g(x) = 0.
x→c
x→c
Prove that
f (x)
→∞
g(x)
as x → c.
EXERCISE 31 (A MODEL FOR DEMAND AND SUPPLY)
Assume that the demand D for and the supply S depend on the price p of
the good: D = d(p) and S = s(p).
Assume that the demand function d is a continuously differentiable function
with d′ < 0 (decreasing demand for increasing price). Similarly assume that
the supply function s is a continuously differentiable function with s′ (p) > 0
(increasing supply for an increase in price).
A price p∗ is called an equilibrium price if d(p∗ ) = s(p∗ ).
Prove that at most one equilibrium price exists.
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172
EXERCISE 32
Significance of the derivative
Let f be a continuous function on [a, ∞), which is differentiable on (a, ∞).
Assume that f ′ ≥ 0 on the interval (a, ∞).
Prove that f (x) ≥ f (a) for every x ≥ a.
EXERCISE 33
Let f be a differentiable function such that f ′ (x) =
Prove that lim x2 (f (x + 2) − f (x)) = 2.
1
.
1 + x2
x→∞
EXERCISE 34
Evaluate the limit
x3 − 7x2 + 16x − 12
.
x→2
x2 − 4x + 4
lim
EXERCISE 35
EXERCISE 36
Evaluate the following limits.
u
ex − 1
(b) lim
(a) lim
u→0 1 − eu
x→0
x
ex − 1 − x
arctan x
(e) lim
(d) lim
x→0
x→0
x
x2
tan x
x→0
x
tan x − x
(f) lim
.
x→0
x3
(c) lim
Let h be a twice-differentiable function such that 0 ≤ h′′ ≤ 13 . Consider the
function g, defined by
g(x) = h(x) + 3h′ (x)
and assume that g(0) = g(2).
Prove that h′ (2) ≥ −1.
EXERCISE 37
Let f be a twice-differentiable function with f ′ = f and f (0) = 1.
Find the linear approximation for f at 0. Also give a formula for the remainder.
EXERCISE 38
Let f be the function on − 21 , 21 , defined by f (x) = x2 .
Find the linear approximation for f at 0.
Prove that the remainder r satisfies |r| < 41 .
EXERCISE 39
Consider the function f defined by
f (x) =
x
.
(x − 1)2
(a) Find the domain of this function.
(b) Find the first and second derivative of this function.
(c) Find the extreme values of this function.
(d) Determine the intervals where the function is convex and determine the
intervals where the function is concave.
(e) Evaluate the limits lim f (x) and lim f (x).
x→∞
x→−∞
(f) Give a sketch of the graph of this function.