PRECALCULUS 2021
Contents
Introduction
3
Chapter 1. Functions
1.1. General theory of functions
1.2. Types of functions
1.3. Solutions to questions
5
5
22
26
Chapter 2. Irrational and trigonometric inequalities
2.1. Irrational inequalities
29
29
Chapter 3. Exponential and logarithmic functions
3.1. Exponential functions
3.2. Logarithmic functions
3.3. Exponential equations
3.4. Logarithmic equations
3.5. Exponential inequalities
3.6. Logarithmic inequalities
31
31
35
38
41
44
48
Chapter 4. Limits of functions
4.1. Some topology
4.2. The definition of limit
4.3. Computing limits
4.4. Continuity
4.5. More computations of limits
51
51
57
72
87
94
Chapter 5.
Differentiation
101
Index of Terms
103
1
Introduction
The book is a first introduction to mathematical analysis, or infinitesimal
calculus, adjective which so explains the scholastic name ”precalculus” for
the related course. We can at least we can point at Newton and Leibnitz
as the inventors of infinitesimal calculus (17th century), and so as those
responsible for what you will study in this course.
The object of study in this book, so the main character, are functions.
What is a function? Putting aside for a moment the mathematical language, and looking for example at natural phenomena, we realize that we
can describe them by means of quantities whose measure depends on other
quantities: namely, temperature depends on where it is measured and it
may vary with time; the velocity of a particle or object depends on time,
as it may accelerate or decrease its speed during its movement; a force may
change from place to place and vary with time; and so on. This intuition,
as we will see, may be formalized in a rigorous way, that is, in a way we
can do some mathematics with it, into the notion of function. This will be
carried on using the language of sets: once we have sets, we may wonder if
we are able in some way to compare them: the concept of function provides
a way to meet this need. From this considerations and from the examples
above, it is quite clear why we are interested in functions, and in studying
their behaviour.
In some cases, functions can be visualized in a graph, that is, a picture,
encoding the main features, properties and behaviour of functions. Therefore, purpose of this course is to learn to draw graphs of functions, because
from them we can easily read off interesting information on functions.
To carry on the programme above, we need some sufficiently powerful
and reliable technology. As we will see, this is based on a simple nevertheless
deep idea: convergence. Since the idea is simple, it has been known since
long, however, a mathematical formulation of convergence into the elegant
and flexible concept of limit came only at the beginning of XIX century. with
mathematicians as Cauchy, Bolzano and Weierstrass (so note that Newton
and Leibnitz invented infinitesimal calculus without a crystalline definition
of limit). For a glimpse of it, let’s go back again to physical examples,
particularly, let’s think again to the velocity of an object: from the measure
of the space travelled by an object in a certain span of time, we can form
an idea of the average speed of that object, however, the true speed of the
object may change all the time and be different from the average speed;
3
4
INTRODUCTION
often it is this instant speed, intuitively measured along ”infinitesimally”
small intervals of time, which is relevant. The definition of limit will tame
this vague and ambiguous idea of ”infinitesimal”.
Down to earth, once unrolled, the definition of limit (and of other concepts in this course) boils down to inequalities, which thus will be our main
tool. In mathematics, inequalities are more important than equalities: indeed, inequalities provide bounds on quantities, and so control on them. On
the other hand, even quantities describing natural phenomena are measured
up to some error, and this is expressed in terms of some inequality. In other
words, inequalities are a must for this course.
Finally, a question: is precalculus easy or difficult? Difficult, but rewarding and eye-opening at the end.
CHAPTER 1
Functions
1.1. General theory of functions
1.1.1. First examples of functions.
(1) Imagine you are keeping your socks in a box or cupboard with
several compartments. Undoubtedly, each individual sock can only
go in one compartment - physical objects cannot be in two places
simultaneously. When you are tidying away your socks, putting
each into a compartment, you are defining a function from the set
of socks into the set of cupboard compartments.
(2) Upon registration at UNAM, a student is assigned a student ID.
Every student is assigned only one ID. This means that we have
again a function from the set of UNAM students into which other
set? Looking at UNAM student IDs, they are all strings of integers,
so we could say ”set of strings of integers”. Of course, sometimes
a more detailed description might be necessary.
Unlike the first example, where several socks may live in the same
compartment, the student ID uniquely identifies the student. Later
on, this property of a particular function will be called injective.
(3) The list
a b c d e
0 1 0 −1 2
defines a function from the set {a, b, c, d, e} into the set {−1, 0, 1, 2}.
We say ”a is mapped to 0, b is mapped to 1” etc..
The same function displayed in a different format:
e
2
d
1
c
0
b
−1
a
5
6
1. FUNCTIONS
The fact that the picture is depicting a function is reflected by the fact
that every element on the left is the starting point of exactly one arrow.
The function is not injective, since both a and c are mapped to b.
(4) The rule f (n) = 2n defines a function from the set N of natural
numbers into itself.
Since N is an infinite set, we cannot display this function in list format, but we can evaluate the function at any given natural number
by plugging in that number into the formula.
E.g. f (0) = 2 · 0 = 0, f (127) = 254.
(5) Plenty of examples of functions appeared in the course of Basic
Mathematics; like most of the functions discussed in this course,
those are functions from the (set of) real numbers into the real
numbers.
Prominent examples are the trigonometric functions sin x, cos x, . . .,
polynomial functions p(x) = an xn + . . . + a0 or rational functions
f (x) = p(x)
q(x) with polynomial functions p(x) and q(x).
1.1.2. The Cartesian product of two sets. From Basic Mathematics you will recall the notion of a set 1
Definition 1.1.1.
• The ordered pair (x, y) is a pair of objects x and y. The
order is significant a; x is called the first component of (x, y)
and y the second component.
• The Cartesian product of the sets A and B, denoted by A ×
B, is the set of all pairs (a, b), where a is in A and b in B.
A × B = {(a, b) | a ∈ A, b ∈ B}.
a(x, y) 6= (y, x) unless x = y
Example 1.1.2.
a) For A = {0} = B
A × B = {(0, 0)}.
b) For A = {1, 2, 3} and B = {1, 2},
A × B = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
and B × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}.
1Chapter 2
1.1. GENERAL THEORY OF FUNCTIONS
Questions 1.1.3.
a) For the sets A and B in example b) above, determine (A ×
B) \ (B × A).
b) Given nonempty sets A and B, show that A × B = B × A if
and only if A = B.
c) Given a set A with 5 elements and a set B with 2 elements,
show that |A × B| = |B × A| = 10.
d) Let n and m be natural numbers. Given a set A with m
elements and a set B with n elements, how many elements
does A × B have?
Example 1.1.4. More Examples:
• The set R × R = {(a, b) | a, b ∈ R} is the two-dimensional
plane (also called the Cartesian plane). The pair (a, b) of
real numbers identifies the point with coordinates (a, b).
y
The point (a, b)
b
•
0
a
x
• The set Z × Z = {(a, b) | a, b ∈ Z} corresponds to the set of
points in the Cartesian plane with integer coordinates.
y
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•1 •
•
•
•
•
•
•
•
•
•
0
1
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
x
Questions 1.1.5. Make a sketch indicating the following as sets of
points in the plane:
(1) {1} × R.
7
8
1. FUNCTIONS
(2) Z × {−1, 1}.
(3) Z × {y ∈ R | |y| < 1}.
1.1.3. The concept of function. Let A, B, C, D be sets.
Definition 1.1.6. A binary relation between elements of A and elements of B is any subset of A × B.
Functions are a special type of binary relation. Informally speaking,
a function from A to B is a rule that associates to each element of A
exactly one element of B.
Definition 1.1.7. A function f from A to B is a subset of A×B in
which every element of A appears as first component of exactly one
pair.
The set A is called the domain of the function, its elements the function’s arguments.
The set B is the codomain.
The fact that f is a function from A to B is succinctly denoted in the
form
f : A → B.
If (a, b) is an element of the function f , we say that f maps a to b,
or that f of a is equal to b,
written f (a) = b.
The set {f (a) | a ∈ A} is called image or range of f and its elements
are the values of the function. The image of f is denoted im f .
Two functions f : A → B and g : C → D are equal if and only if
A = C, B = D and f (a) = g(a) whenever a is in A.
In other words, f = g means that f and g have the same domain
and codomain and that the values of f and g on each argument are
equal.
Questions 1.1.8.
(1) Each of the below pictures displays a binary relation between
the sets A = {a, b, c, d} and B = {1, 2, 3, 4}.
Identify the functions.
1.1. GENERAL THEORY OF FUNCTIONS
9
d
4
d
4
c
3
c
3
b
2
b
2
a
1
a
1
d
4
d
4
c
3
c
3
b
2
b
2
a
1
a
1
(2) Which of the following functions are equal?
(a) f1 : R \ {0} → R, f1 (x) = 3x
x .
(b) f2 : R \ {0} → R, f2 (x) = 3.
(c) f3 : R → R, f3 (x) = 3.
(d) f4 : R → R, f4 (x) = sin2 (x).
(e) f5 : R → R, f5 (x) = 1 − cos2 (x).
2
2
(f) f6 : [0, 2π] \ { π2 , 3π
2 } → R, f6 (x) = (1 − tan x) · cos x.
π 3π
(g) f7 : [0, 2π] \ { 2 , 2 } → R, f7 (x) = cos 2x.
(h) f8 : [0, 2π] \ { π2 , 3π
2 } → [−1, 1], f8 (x) = cos 2x.
(i) f9 : [0, 2π]\ → R, f9 (x) = cos 2x.
1.1.4. The graph of a function.
Definition 1.1.9. Let A and B be sets and f : A → B a function.
The graph of f is the set
{(a, b) | a ∈ A, b ∈ B, (a, b) ∈ f }.
The graph of the function f is denoted Gr(f ).
The graph of a real-valued function f is what you see when you plot f in
the form of a curve in a coordinate system. The ”curve” is the set of points
(x, f (x)), which is precisely Gr(f ).
Below, a segment of the graph of the functions f (x) = − 21 x2 + 21 (red) and
g(x) = |x| − 1 ( green).
f
g
10
1. FUNCTIONS
The difference between a function and its graph lies in the fact that
two functions are only equal if they have the same domain and codomain;
however, the graph of a function is not dependent on the codomain.
Let f : A → B and g : C → D be functions. Then Gr(f ) = Gr(g) if
and only if A = C and f (a) = g(a) for a in A.
The functions f : R → [−1, 1] given by f (x) = sin x and g : R → R given
by g(x) = sin x are not equal, but their graphs are.
Whenever a reasonable picture of the Cartesian product of domain and
codomain of the function is possible, sketching the graph allows us to ”see”
a function.
Example 1.1.10. Let A = {1, 2, 3, 4, 5} and f : A → A be given by
f (1) = 1, f (2) = 3 = f (3), f (4) = 5, f (5) = 4. Below, find a sketch
of A × A and Gr(f ) indicated by red points.
5
4
3
2 (1,f(1))
1
1 2 3 4 5
Given an arbitrary curve in the plane or collection of points in a Cartesian product, we can perform the vertical line test to see if what we have
before us the graph of a function. Recall that a function assigns exactly
one value to every element of its domain; so if the domain of a function is
displayed on the horizontal axis, every vertical line intersects the graph of a
function in at most one point.
Example 1.1.11. The picture shows two graphs of functions R → R
and one set of points that is not the graph of any function R → R
1.1. GENERAL THEORY OF FUNCTIONS
11
1.1.5. Operations on real-valued functions, part I. In this section, we only consider functions whose codomain is a subset of R.
This implies that the values of the functions can be added, multiplied and so
on. This allow us to combine known functions using arithmetic operations.
Definition 1.1.12. Let A and B be sets. Let C and D be sets of real
numbers a. Let c ∈ R.
• f + g, the sum of the functions f and g, is the function with
domain A ∩ B and codomain R given by
(f + g)(x) = f (x) + g(x).
• The function cf has domain A, codomain R and is given by
c(x) = c · f (x).
• f · g, the product of the functions f and g, is the function
with domain A ∩ B and codomain R given by
(f · g)(x) = f (x) · g(x).
aThat is the same as saying they are subsets of R
Example 1.1.13.
• Let f : R → R and g : R → R be given by
f (x) = 2x, g(x) = sin2 x.
The domain of the functions f + g, cf , cg and f · g is R.
For x ∈ R, (f + g)(x) = f (x) + g(x) = 2x + sin2 x, e.g.
3
(f + g)(2) = 4 + sin2 2, (f + g)( π3 ) = 2π
3 + 4.
For x ∈ R, (2f )(x) = 2 · f (x) = 4x and (2g)(x) = 2 · g(x) =
2 sin2 x.
For x ∈ R, (f · g)(x) = f (x) · g(x) = (2x) · (sin2 x) =
2x(sin2 x), e.g. (f · g)( π4 ) = π2 · 12 = π4 .
• Let A = {1, 2, 3, 4, 5}, B = {0, 1, 2, 3, 4}, f : A → R be
given by f (a) = 12 a and g : B → R by g(b) = − 12 b.
The domain of f + g and f · g is A ∩ B = {1, 2, 3, 4};
for c ∈ {1, 2, 3, 4}, (f + g)(c) = f (c) + g(c) = 12 c − 12 c = 0
and (f · g)c = f (c) · g(c) = − 14 c.
12
1. FUNCTIONS
Questions 1.1.14. Let A = Z and Bp
= [−5, 5).
Let f : A → R be defined by f (z) = |z| and g : B → R by g(y) =
−|5y|.
Find the domain of f + g and f · g and determine (f + g)(x) and
(f · g)(x) for all x in this domain.
1.1.5.1. Images and pre-images. Let A and B be sets and f : A → B a
function.
We have already been calling the set {f (a) | a ∈ A} the image of f . This
definition can be generalised to arbitrary subsets of the domain.
Definition 1.1.15. Let C ⊆ A. The set {f (c) | c ∈ C} is called the
f- image of C a and denoted f (C).
aor simple the image of C if it is clear which function one is talking about
C
f (C)
Example 1.1.16.
a) Let f : R → R be given by f (x) = x2 .
Let C = {−1, 0, 1, 2, 3}. Then f (C) = {0, 1, 4 , 9}.
b) Let f : R → R be given by f (x) = −x − 1. Let C = [−1, 1].
Then f (C) = {−x − 1 | − 1 ≤ x ≤ 1} = [−2, 0].
The concept of pre-image reverses the viewing direction: Starting from
a subset of the codomain B, we determine the elements of the domain A
that are mapped by f to an element of D.
As illustration, picture yourself looking at the image of your face in a mirror;
while thus engaged, you see in the mirror the reflection of someone behind you and
you turn round to look at the person: You are seeking the pre-image of the person’s
reflection in the mirror.
1.1. GENERAL THEORY OF FUNCTIONS
Definition 1.1.17. Given any subset D of A, the set {a ∈ A | f (a) ∈
D} is called the f- pre-image of D a and denoted f − (D).
aor simple the pre-image of D if it is clear which function one is talking about
-
f − (D)
D
Example 1.1.18.
a) Let us consider the function assigning student IDs to students.
The image of the set of students who registered in 2017 is
the set of student IDs beginning with 2017.
They have since changed the ID-assigning algorithm; about
the pre-image of the set of student IDs that start with 2193
I cannot say more than that i it is the set of students whose
IDs start with 2193.
(1) Let A = {0, 1, 2, 3, 4}, B = Z, f : A → B be given by
f (a) = −a2 .
Let C = {1, 2}, D = [−5, 0), E = (0, +∞).
We have f (C) = {−1, −4}, f − (D) = {1, 2}, f − (E) = ∅.
c) Let f : R → [−1, 1] be given by f (x) = sin x. Let D = {0, 1}.
Determining the pre-image of D is tantamount to determining all solutions of the equations sin x = 0 and sin x = 1 and
collecting them in a set.
Recall that sin x = 0 means that x = πk for some integer k,
whereas sin x = 1 means that x = π2 + 2kπ for some integer
k.
Hence f −1 ({0, 1}) = {πk, π2 + 2kπ | k ∈ Z}.
Questions 1.1.19. Let A and B be sets and f : A → B be a function.
Let C ⊆ A and D ⊆ B.
For each the following values of A, B, C, D, and f , determine f (C)
and f − (D) :
a) A = Z = B, f (z) = −3z, C = [1, 100] ∩ Z, D = [−2, 24) ∩ Z.
13
14
1. FUNCTIONS
b) A = {a, b, c, d, e}, B = {−2, −1, 0, 1, 2, 3}, f (a) = 0 =
f (b), f (c) = −1, f (d) = f (e) = 3}, C = {b, c, e}, D =
{1, 2, 3}.
c) A = R, B = [−1, +∞), f (x) = x2 − 1, C = [0, 2], D =
[8, +∞).
1.1.6. Operations on real valued functions, part II. Let f and
g be functions whose codomains are subsets of R. We have defined the
(x)
;
functions f +g and f ·g. The function fg is defined by the rule ( fg )(x) = fg(x)
to find its domain, merely intersect the domains of f and g is not sufficient
since those values of x for which g(x) = 0, i.e. the elements of g − ({0}), need
to be excluded.
Definition 1.1.20. Let A and B be sets. Let C and D be sets of real
numbers, let f : A → C and g : B → D be functions.
The quotient of f by g, denoted as
(A ∩ B) \ g − ({0}) and defined by
(x)
( fg )(x) = fg(x)
.
f
g
is the function with domain
Example 1.1.21.
a) Let f, g : R → R be given by f (x) = 1 and g(x) = x2 ,
respectively.
We have g − ({0}) = {x ∈ R | x2 = 0} = {0}.
The domain of the function fg is thus R \ {0} and, for all x
in this set, ( fg )(x) = x12 .
Since f (x) 6= 0 for any real number x, the domain of
g
f
is R
x2
and ( fg )(x) = 1 = x2 .
b) Let A = {a, b, c, d, e}, B = {a, b, c, d, u}, C = R = D,
f (a) = −1, f (b) = −2, f (c) = f (d) = f (e) = 0}, g(a) = 0 =
g(b), g(c) = g(d) = g(u) = 3}.
The domain of fg is {a, b, c, d} \ g − ({0}) = {c, d}, ( fg )(c) =
0 = ( fg )(d).
The domain of fg is {a, b, c, d} \ g − ({0}) = {a, b}, ( fg )(a) =
0 = ( fg )(b).
c) Let A = B = C = D = R, f (x) = x2 − 3x + 1 and g(x) =
x2 − 7.
1.1. GENERAL THEORY OF FUNCTIONS
15
√
√
We have g − ({0}) = { 7, − 7}. Accordingly, the domain
√
√
2
of the function fg is R \ { 7, − 7} and ( fg )(x) = x x−3x+1
.
2 −7
The set f − ({0}) is the set of√roots √
of the polynomial x2 −
3x + 1, so that f − ({0}) = { 3+2 5 , 3−2 5 }. Accordingly, the
domain of the function
g
f
√
is R \ { 3+2 5 ,
√
3− 5
2 }
and ( fg )(x) =
x2 −7
.
x2 −3x+1
1.1.7. Composition of functions.
Definition 1.1.22. Let A, B, and C be sets. Let f : A → B and
g : B → C be functions.
The composite function g ◦ f is the function A → C defined by
(g ◦ f )(x) = g(f (x)) (x ∈ A).
g is applied to the result of applying f to x.
A
x
•
B
f
f (x)
•
C
g
g(f (x))
•
g◦f
Example 1.1.23. In each of the following examples, sets A, B, C, a
function f : A → B and a function g : B → C are given, and g ◦ f is
determined.
a) A = R \ {0}, B = R = C, f (x) = x1 , g(x) = −2x.
For x ∈ R \ {0}, (g ◦ f )(x) = g(f (x)) = g( x1 ) = −2 · x1 = − x2 .
b) A = R \ {0} = B, C = R, f (x) = −2x, g(x) = x1 .
1
1
For x ∈ R\{0}, (g◦f )(x) = g(f (x)) = g(−2x) = −2x
= − 2x
.
c) A = {x, y, z} = B = C, f (x) = y, f (y) = z = f (z), g(x) =
x = g(y), g(z) = y.
(g ◦ f )(x) = g(f (x)) = g(y) = x, (g ◦ f )(y) = g(z) = y,
(g ◦ f )(z) = g(z) = y.
16
1. FUNCTIONS
d) A = (Z \ {0}) × (Z \ {0}), B = Q \ {0} = C,
n
f ((n, m)) = m
, g(x) = x1 .
n
)=
For n, m ∈ Z \ {0}, (g ◦ f )((n, m)) = g( m
1
n
m
=
m
n.
Questions 1.1.24.
1
a) Let f : R \ {1} → R and g : R → R be given by f (x) = x−1
and g(x) = |x|.
Find the formula for g ◦ f and determine (g ◦ f )(0).
b) p
Let f, g : R → R be defined by f (x) = x2 − 7 and g(x) =
|x + 5|, respectively.
Find the formulas for f ◦ g, g ◦ f, f ◦ f, f ◦ (f + g), g ◦ (f · g).
c) Find functions f, g : R → R such that g ◦ f = −(f ◦ g).
1.1.7.1. ”Decomposing” functions. In Calculus, it is often helpful or indeed necessary to express a given function as composite of other, simpler
functions. This can generally be done in many ways and requires practice.
We provide some examples.
Example 1.1.25. All functions in this set of examples have R as their
domain and codomain.
• Let f (x) = x2 + 2x + 1.
The function f may be expressed as
f = h ◦ g, where g(x) = x + 1 and h(x) = x2 or
f = h1 ◦ g1 , where g1 (x) = x2 + 2x and h1 (x) = x + 1 or
f = h2 ◦ g2 ,(where
−2x
• Let f (x) =
0
(
x if
Let g(x) =
0 if
Then f = h ◦ g. a
2
1
g2 (x) = x3 , h2 (x) = x 3 + 2x 3 + 1.
if x ∈ [0, 1]
.
if x ∈
/ [0, 1]
x ∈ [0, 1]
and, finally, h(x) = −2x.
x∈
/ [0, 1]
aQuestion: Is h ◦ g equal to g ◦ h in this example? If not, can you describe g ◦ h?
Questions 1.1.26. Find three different ways (each) to write the
functions
f : R → R given by f (x) = x3 + 3x2 + 3x + 1 and
1.1. GENERAL THEORY OF FUNCTIONS
17
g : R → R given by g(x) = sin(x2 + 3x) as composition of two functions.
1.1.8. Maximal domains. We turn to a type of question common in
Precalculus and Calculus, finding the maximal possible domain of a function
into the real numbers.
Given the formula for a function f with codomain R whose domain is supposed to be a subset of R, you are asked to find the largest possible (or
maximal ) subset of R that can be the domain of a function defined by the
formula.
Some things to keep in mind when addressing this type of question are:
• The denominator of a fraction must be non-zero.
• The argument of a square (or,generally 2nth ) root must be nonnegative.
We illustrate this type of question by a series of examples, ranging from
the simple to the more complex:
Example 1.1.27. Find the possible domain of the real-valued function f defined by the following formulas:
a) f (x) = x1 .
Since division by a real number is permitted if and only if
the number is not equal to 0, the largest possible domain of
f is R \ {0}.
b) f (x) = 3x
x .
For the same reason as in a), the largest possible domain of
f is R \ {0}.
Note that
√ f (x) = 3 whenever x ∈ R \ {0}.
c) f (x) = x − 1.
Since a real number possesses a square root if and only if it
is non-negative, the maximal possible domain of f is the set
{x ∈ R |p
x − 1 ≥ 0} = [1, +∞).
d) f (x) = |x| − 1.
Since a real number possesses a square root if and only if it
is non-negative, the maximal possible domain of f is the set
{x ∈ R |q
|x| − 1 ≥ 0} = (−∞, −1] ∪ [1, +∞).
e) f (x) = x1 .
Let D be the domain of a function defined by this formula.
If x ∈ D, then x1 must exist, so x 6= 0. Moreover, x1 must
have a square root, so that x1 ≥ 0. So x > 0.
18
1. FUNCTIONS
The maximal possible domain of f is the set {x ∈ R | x 6= 0
and x1 ≥ 0} = (0, +∞).
√
x−1
.
f) f (x) = x2 −3x−2
Let D be the domain of a function defined by this formula.
If x ∈ D, then x − 1 has a square root, so that x ≥ 1.
Furthermore, x2 −3x−2 6= 0. As x2 −3x−2 = (x−1)(x−2),
this means that x is neither equal to 1 nor equal to 2.
The maximal domain of f is therefore [1, +∞) \ {1, 2} =
(1, 2) ∪ q
(2, +∞).
x−4
g) f (x) = x2 −3x−2
.
Let D be the domain of a function defined by this formula.
x−4
exists, so that x ∈
/ {1, 2}. Moreover,
If x ∈ D, then x2 −3x−2
x−4
x−4
should possess a square root, i.e. x2 −3x−2
≥ 0.
x2 −3x−2
This inequality may also be solved using a sign table. At
this point, you should be able to do this a, the procedure is
included for your convenience:
If x < 1, then all three terms x − 4, x − 2, x − 1 are negative and
x−4
x2 −3x−2 < 0.
x−4
If 2 > x > 1, then x−4 < 0 > x−2 and x−1 > 0, so x2 −3x−2
> 0.
x−4
If 4 > x > 2, then x−4 < 0 and x−1 > 0 < x−2, so x2 −3x−2 < 0.
x−4
If x > 4, then all three terms are positive, so x2 −3x−2
> 0.
The maximal domain of f is (1, 2) ∪ [4, +∞).
aInequalities are of vital importance for this.
Questions 1.1.28.
(1) Determine the maximal domain for the function f defined
by each of the following formulas:
a) f (x) = sin1 x .
q
b) f (x) = x21−1 .
p
c) f (x) = −|x|.
d) f (x) = tan x.
(2) Find a formula for a function R → R whose maximal domain
is the empty set.
1.1.8.1. The maximal domain of a composite function. Given functions
f and g defined on subsets of R and into R, we intend to determine the
maximal domain of the composite function g ◦ f .
Let x ∈ R. Since (g ◦ f )(x) = g(f (x)), the function g ◦ f is defined at x if
and only if
1.1. GENERAL THEORY OF FUNCTIONS
• The function f is defined at x and
• The function g is defined at f (x).
In other words,
Let Df , Dg be the largest possible domains of f and g, respectively
The largest possible domain of g ◦ f is
{x ∈ R | f (x) ∈ Dg } = f − (Dg ),
the f -preimage of Dg .
Note that f − (Dg ) ⊆ Df , so that the largest possible domain of g ◦ f
is always contained in the largest possible domain of f .
Example 1.1.29. Find the maximal possible domain of g ◦ f and of
f ◦ g, where
a) f (x) = x − 2, g(x) = x1 .
The maximal domain of f is R and that of g is R \ {0}. So
the domain of g ◦ f is {x ∈ R | f (x) 6= 0}.
Now f (x) = 0 means that x − 2 = 0, i.e. x = 2.
The maximal domain of g ◦ f is therefore equal to R \ {2}.
1
For R 3 x 6= 2, (g ◦ f )(x) = x−2
.
Every real number is a member of the domain of f , so
f ◦ g is defined wherever g is. The maximal domain of f ◦ g
is equal to that of g, i.e. R \ {0}.
For R 3 x 6= 0, (f ◦ g)(x) = x1 − 2.
√
b) f (x) = x1 , g(x) = x − 1.
The largest possible domain of f is R \ {0}, the largest possible domain of g is {x ∈ R | x − 1 ≥ 0} = [1, +∞).
The largest possible domain of g ◦ f is the set f − ([1, +∞),
i.e. the set of nonzero real numbers x such that x1 ≥ 1.
Since x1 ≥ 1 if and only if 1 ≥ x > 0, the domain of g ◦ f is
the interval (0, 1].
q
For x ∈ (0, 1], (g ◦ f )(x) = x1 − 1.
The largest possible domain of f ◦ g is the√ set {x ∈
[1, +∞) | g(x) 6= 0}. Now g(x) = 0 means that x − 1 = 0,
i.e. x = 1. It follows that the largest possible domain of f ◦g
is the set (1, +∞).
1
For x ∈ (1, +∞), (f ◦ g)(x) = √x−1
.
19
20
1. FUNCTIONS
(
1
if x > 0
, g(x) = x+1
x−1 .
if x ≤ 0
The maximal domain of f is R, that of g is R \ 1.
The domain of g ◦ f is R \ f − ({1}). Straight from the definition, the set f − ({1}) contains all positive real numbers;
for x ∈ R, x21+3 ≤ 13 , so that the maximal domain of f ◦ g is
(−∞, 0].
c) f (x) =
1
x2 +3
Questions 1.1.30. Given the following formulas for functions f and
g, determine the largest possible domains of f ◦ g and g ◦ f .
√
1
, g(x) = x.
a) f (x) = 1 − |x|
√
√
x.
b) f (x) = 3 x, g(x) = q
c) f (x) = cos x, g(x) = |x| − 12 .
√
d) f (x) = tan x, g(x) = x.
1.1.9. Computing the image/range. Let A and B be sets. Given a
function f : A → B, you will sometimes be required to determine the image
of f . This is not the type of question for which there is a standard recipe;
however, in the context of real-valued functions, there is an approach and
recurring arguments. Above all, this needs practice. We present several
examples.
In the context of s real-valued function, the name ”range” is more frequently
used in textbooks than ”image”, though they are the same thing. The range
of a function f is denoted ran f .
When computing the range of a real-valued function f , you want to find
out the possible solutions of y = f (x).We will develop stronger tools for this
during this course. For the beginning, some things to keep in mind when
addressing this type of question are the following:
•
•
•
•
•
Denominators must be non-zero.
Arguments of square roots must be non-negative.
Squares, square roots, and absolute values are non-negative.
Both f (x) = sin x and f (x) = cos x have range [−1, 1].
Try to find maximal and minimal values of the function.
Example 1.1.31. Determine the range of the function f , where
1.1. GENERAL THEORY OF FUNCTIONS
a) f : R \ {0} → R is given by f (x) = x1 .
/ ran(f ). For 0 6= y ∈ R,
If x ∈ R \ {0}, then x1 6= 0 a, so 0 ∈
1
y = 1 ∈ ran(f ). So ran(f ) = R \ {0}.
y
b) f : R → R, f (x) = x2 + 3.
Since x2 ≥ 0, x2 + 3 ≥ 3 for all real numbers x. Now
√ let
y ∈ [3, +∞). Then y − 3 ≥ 0 and y = f (x) with x = y − 3.
So ran(f ) = [3, +∞).
1
c) f : R → R, f (x) = |x|+3
.
1
Certainly, |x|+3 > 0 whenever x ∈ R; furthermore, |x| ≥ 0,
1
so |x| + 3 ≥ 3 and |x|+3
≤ 13 . So ran(f ) ⊆ (0, 13 ]. For
y ∈ (0, 13 ], we have 3 < y1 , so with x = y1 − 3, we get x = |x|,
1
so that |x| + 3 = y1 , i.e. y = |x|+3
. So ran(f ) = (0, 13 ].
d f : R → R, f (x) = sin2 x − 1.
Since ran(sin x) = [−1, 1], ran(sin2 x) = [0, 1] and ran(f ) =
[−1, 0].
aA fraction is equal to 0 if and only if the NUMERATOR equals 0.
Example 1.1.32. More examples:
Determine the range of the function f , where
e) f is given by the formula f (x) = 1+x
2−x and the domain of f
is as large as possible.
The largest possible domain of f is R \ {2}. Let y ∈ R. To
find out if y is in the range of f , we try to solve the equation
1+x
for x. This goes as follows: Given x 6= 2, y = 1+x
y = 2−x
2−x is
equivalent to y(2 − x) = 2y − xy = 1 + x. We take all terms
involving x to one side, getting 2y − 1 = x + xy = x(1 + y).
If y 6= −1, this yields y = f (x) with x = 2y−1
1+y . For y = −1,
we are getting −3 = 0, so there is no solution in this case.
Accordingly, ran(f ) = R \ {−1}. √
f) f is given by the formula f (x) = x − 5 and the domain of
f is as large as possible.
The maximal domain of f is [5, +∞). Square roots being
non-negative,
p
pran(f ) ⊆ [0, +∞). For y ∈ [0, +∞), we have
2
y = y = y 2 + 5 − 5, in other words y = f (y 2 + 5). It
follows that ran(f ) = [0, +∞).
21
22
1. FUNCTIONS
Questions 1.1.33. Determine ran(f ), where
1
a) The function f is given by f (x) = x−5
and the domain is as
large as possible.
b) The function f is given by f (x) = x21−5 and the domain is
as large as possible.
c) f : R → R and f (x) = sin2 x + cos2 x.
d) The function f is given by f (x) = sin1 x and the domain is as
large as possible.
1.2. Types of functions
In this section, we introduce properties that a function may or may not
possess. We start with the properties of injectivity and surjectivity and then
move on to properties of certain real valued functions.
1.2.1. Injective, surjective, and bijective functions.
1.2.1.1. Injective functions.
Definition 1.2.1. Let A and B be sets and let f : A → B be a
function. Then f is called injective (or one-to-one) if for all elements
a1 and a2 of A,
f (a1 ) = f (a2 ) if and only if a1 = a2 .
A
B
injective
A
B
not injective
When a function is visualised in this way, the property of injectivity
corresponds to each element of the codomain B being hit by at most one
arrow.
Remark 1.2.2.
1.2. TYPES OF FUNCTIONS
23
• The following is an equivalent formulation of the property
of injectivity:
A function is injective if the pre-image of each element of
its codomain has at most one element.
• Let A and B be finite sets and let f : A → B be an injective
function. It follows from the previous remark that A cannot
have more elements than B, i.e. |A| ≤ |B|.
• If f is a function R → R, then injectivity may be tested by
the horizontal line test:
The function f is injective if and only if every horizontal line
intersects the graph of f in at most one place.
y
2
1
−2π
− 3π
2
−π
− π2
0
0
π
2
π
3π
2
2π
x
−1
−2
The function f : [−2π, 2π] → R given by f (x) = cos x is not injective, as shown by the horizontal
line test.
Example 1.2.3. Find out if the following functions f are injective
a) f : N → N given by f (n) = 2n.
Let m, n ∈ N and suppose that f (n) = f (m). This means
2n = 2m. Dividing by 2 on both sides, this yields n = m.
So f is injective.
b) f : [1, 4] → R given by f (x) = x2 .
√
For a non-negative real number y, x = y is the only nonnegative real number satisfying x2 = y. So f is injective.
c) f : [−1, 4] → R given by f (x) = x2 .
The function f is not injective, seeing that f (1) = f (−1) =
1.
d) f : R → R given by f (x) = |x| − x.
The function f is not injective, seeing that f (x) = 0 whenever x ≥ 0.
e) f : (−∞, 0] → R given by f (x) = |x| − x.
Let 0 ≥ x ∈ R. Then |x| = −x and f (x) = −x − x = −2x.
If −2x1 = −2x2 , then x1 = x2 , so f is injective.
24
1. FUNCTIONS
f) f : [0, 2π) → [−1, 1] given by f (x) = sin x. You will recall
that sin(π − θ) = sin θ is true for all angles θ (why?).
In particular, f ( π4 ) = sin π4 = sin(π − π4 ) = f ( 3π
4 ). So f is
not injective.
g) f : R \ {1} → R given by f (x) = x+1
x−1 .
Let x, y ∈ R \ {1} with f (x) = f (y).
y+1
x+1
= y−1
.
This means that x−1
Cross-multiplying (i.e. multiplying by the nonzero real number
(y − 1)(x − 1)), this equation becomes
(x + 1)(y − 1) = xy − x + y − 1 = (x − 1)(y + 1) = xy + x − y − 1.
Cancelling like terms on both sides, this is equivalent to
−x + y = x − y = −(x − y).
The only real number equal to its negative is 0, so that x − y = 0
and x = y. Hence f is injective.
(
h) f : N → N given by f (n) =
n + 1 if n is even
.
n − 1 if n is odd
With a piecewise defined function f such as this one, it is typically necessary to distinguish several cases, since the definition of
f depends on the argument.
In the present case, we need to distinguish between the cases ”m
and n even”, ”m even and n odd” and ”m and n odd”.
If n and m are both even, then f (n) = n + 1 and f (m) = m + 1.
Subtracting 1 on either side of the equation n + 1 = m + 1 yields
that n = m.
If m is even and n is odd, then f (m) = m + 1 and f (n) = n − 1.
Since m is even, m + 1 is odd and since n is odd, n − 1 is even, so
that f (m) and f (n) cannot be equal.
Finally, assume that both n and m are odd numbers. Then
”f (n) = f (m)” means that n − 1 = m − 1. Adding one on both
sides, we get n = m.
So f is injective.
Questions 1.2.4. Determine which of the following functions are
injective. Provide a reason for your answer.
a) f : R → R, f (x) = x2 − x − 3.
b) The function f defined by the formula f (x) = x+1
x−5 on the
largest possible domain.√
c) f : Q → R, f (x) = x2 − 2x.
d) f : R → R, f (x) = |x| + x.
e) The function f given by f (x) = tan x on the largest possible
domain.
1.2. TYPES OF FUNCTIONS
25
1.2.1.2. Surjective and bijective functions.
Definition 1.2.5. Let A and B be sets. The function f : A → B is
called surjective (or onto) if im f = B.
A function is surjective if and only if its image is equal to the whole of
its codomain, or, equivalently, for each y in the codomain of f , there is an
x in the domain satisfying f (x) = y.
A
B
surjective
A
B
not surjective
Example 1.2.6. Examples: Find out if the following functions f are
surjective
a) f : N → N given by f (n) = 2n.
The function f is not surjective. If m is odd, then there is
no natural number satisfying m = 2n.
b) f : Q → Q given by f (q) = 2q.
Let p ∈ Q; then p = 2 · ( p2 ) = f ( p2 ), so f is surjective.
c) f : R → [−1, 1] given by f (x) = cos x.
is a surjective function, as you learned in Basic Mathematics.
Every real number in [−1, 1](is the cosine of some angle.
n
if n is even
d) f : N → N given by f (n) = 2
.
1 if n is odd
Let m ∈ N; then 2m is even and m = 2m
2 = f (2m), so f is
surjective.
1
f) f : R \ {1} → R \ {0} given by f (x) = x−1
.
Let 0 6= y ∈ R. Then y = f (x) with x = 1+y
y . So f is
surjective.
Definition 1.2.7. Let A and B be sets. The function f : A → B is
called bijective if it is both injective and surjective.
26
1. FUNCTIONS
Questions 1.2.8. Determine which of the following functions are
injective, surjective, or bijective.
Provide a reason for your answer.
√
a) f : [0, +∞) → [−1, +∞), f (x) = x − 1.
x+1
b) f : R \ {−5} → R, f (x) = x+5
.
x+1
c) f : Q \ {0} → R, f (x) = |x| .
d) f : R → R, f (x) = x3 − 1.
e) f : R → [−1, 1], f (x) = sin2 x − cos2 x.
1.2.2. Inverse functions.
Definition 1.2.9. Let A be a set. The identity function on A,
denoted idA , is the function from A into A defined by
idA (x) = x for all x ∈ A.
The identity function on any set is obviously a bijective function.
Lemma 1.2.10. Let A and B be non-empty sets and let f : A → B be
an injective function. Then there exists a function g : B → A such that
g ◦ f = idA .
Any function g : B → A with g ◦ f = idA is called a left inverse of f .
Proof. Let b ∈ B. Either b ∈ im (f ) or b ∈
/ im (f ). If ∈ im (f ),
then there is a unique element of a satisfying f (a) = b; this is because f is
injective.
We select an arbitrary element c of A and define the function g : B → A as
follows:
If b ∈ im (f ), then g(b) is the unique element a of A with f (a) = b.
If b ∈
/ im (f ), then g(b) = c.
Now we verify that g : f = idA . Indeed: For a ∈ A, we have f (a) ∈ im (f )
and a is the only element of A mapped by f to f (a). So f (g(a)) = (f ◦g)(a) =
a.
1.3. Solutions to questions
Question 1.1.3 a): (A × B) \ (B × A) = {(3, 1), (3, 2)}.
Question 1.1.3 b): If A = B, then A × B = A × A = B × A.
Suppose that A 6= B. This means some element of A is not in B or some
element of B is not in A.
If a ∈ A \ B and b0 is any element of B, the pair (a, b0 ) is in A × B but not
1.3. SOLUTIONS TO QUESTIONS
27
in B × A; if b ∈ B \ A and a0 is any element of A, the pair (a0 , b) is in B × A
but not in A × B.
Question 1.1.3 c): Each of the 5 elements of A appears as the first component
of a pair in A × B; given a fixed element of A, two elements of B appear as
second components of a pair (a, ). So there is a total number of 5 · 2 pairs
in A × B.
The argument for B × A is analogous.
Question 1.1.3 d): The set A × B has m · n elements. Each element of A is
the first component of n = |B| ordered pairs, and there are m elements in
A.
Question 1.1.14: The domain of f +g and f ·g is A∩B
p = {−5, −4, −3, −2 −
1, 0, 1, 2, 3, 4}. For c in this domain,
(f
+
g)(c)
=
|c| − |5c| (e.g. (f +
p
√
√
g)(−2)
√ = 2 − 10) and (f · g)(c) = |c| · (−|5c|) = −|5 cc (e.g. (f · g)(3) =
−|15 3|.
Question 1.1.19 a): f (C) = {−3z | z ∈ Z, 1 ≤ z ≤ 100} = [−300, −3] ∩
Z = {−300, −299, . . . , −4, −3}, f − (D) = {z ∈ Z | − 2 ≤ −3z ≤ 23}. If
−3z ≥ −2, then z ≤ 23 and if z is an integer, this means that z ≤ 0. If
−3z ≤ 23, then z ≥ − 23
3 and if z is an integer, this means that z ≥ −7. So
−
f (D) = {−7, −6, −5, −4, −3, −2, −1, 0}.
Question 1.1.19 b): f (C) = {0, −1, 3}, f − (D) = {d, e}.
Question 1.1.19 c): f (C) = {x2 − 1 | − 1 ≤ x} = [0, +∞). Next, x ∈ f − (D)
if and only if x2 − 1 ≥ 8, so x2 − 9 ≥ 0 which means that x ≤ −3 or x ≥ 3.
In other words, f − (D) = (−∞, −3] ∪ [3, +∞).
1
Question 1.1.24 a): For 1 6= x ∈ R, (g ◦ f )(x) = | x−1
| (g ◦ f )(0) = | − 1| = 1.
Question
1.1.24
b):
For
x
∈
R,
(f
◦
g)(x)
=
|x
+
5|
− 7 and (g ◦ f )(x) =
p
2
2
4
2
2
|x − 2|p
, (f ◦ f )(x) = (x − 7) − 7 = x − 14x
p + 42, 2(f + g)(x)4 =
2 − 7 +
◦
(f
+
g))(x)
=
(x
|x + 5|) p− 7 = x −
x2 − 7 + |x + 5|, (f
p
2
14x + 42 + 2(x −q
1) |x + 5| + |x + 5|, (f · g)(x) = (x2 − 7) · |x + 5|, and
p
(g ◦ (f · g))(x) = |(x2 − 7) · |x + 5| = 5|. Question 1.1.24 c): There are
many solutions to this, one is f (x) = |x| and g(x) = −x.
1
Question 1.1.33 a): The largest possible domain of f is R \ {5}. Let y = x−5
1+5y
with x 6= 5. If xy = 1 + 5y. Unless y = 0, we have y = f (x) with x = y
. For y = 0, the equation xy = 1 + 5y becomes 0 = 1, so that 0 is not in
ran(f ) = R \ {0}.
Question
b): The maximal
domain
of f is {x ∈ R | x2 − 5 6= 0} =
√ 1.1.33
√
√
√
R \ { 5, − 5}. Let x ∈ R \ { 5, − 5} and y = x21−5 . Then x2 y = 1 + 5y.
As in the previous question, it follows that y 6= 0. Taking y 6= 0, we obtain
x2 = 1+5y
y . Since we know that a real number is a square if and only if
it is non-negative, we obtain that y ∈ ran(f ) if and only if y 6= 0 and
1+5y
≥ 0. Finally, 1+5y
≥ 0 means that y < 0 and 1 + 5y ≤ 0, so y ≤ − 15 ,
y
y
or y > 0 and 1 + 5y ≥ 0, which means y > 0. Putting it all together,
ran(f ) = (−∞, − 15 ] ∪ (0, +∞].
28
1. FUNCTIONS
Question 1.1.33 c): For all real numbers x, sin2 x + cos2 x = 1, so ran(f ) =
{1}.
Question 1.1.33 d): The largest possible domain of the function f is the
set {x ∈ R | sin x 6= 0}. We have learned that the zeros of the sine are
the precisely the integer multiples of π, so the maximal domain is D =
R \ {kπ | k ∈ Z}.
Let’s note first that 0 ∈
/ ran(f ). Next, we know that the range of sin x
is equal to [−1, 1]. Hence y ∈ ran(f ) if and only if y = u1 where u ∈
[−1, 0) ∪ (0, 1]. Let 0 6= y ∈ R. Then 0 > y1 ≥ −1 if and only if y ≤ −1 and
0 < y1 ≤ 1 if and only if y ≥ 1. So ran(f ) = (−∞, −1] ∪ [1, +∞).
CHAPTER 2
Irrational and trigonometric inequalities
In this chapter we study inequalities which contains radicals (irrational
inequalities) and inequalities which contains trigonometric functions (trigonometric inequalities). Irrational and trigonometric equations will also follow
as a special case.
2.1. Irrational inequalities
We present how to solve irrational inequalities which can be reduced to
the form
p
n
f (x) R g(x),
where f (x) and g(x) are functions, and n is a positive integer.
• We consider first the case
p
n
f (x) ≥ g(x),
and with n even. It includes both the case where the order sign is
> and the equality =.
The radicand of roots of even degree must be positive or at
most zero, then, once defined, a root of even degree is a positive
quantity, hence, assuming f (x) ≥ 0, if g(x) < 0 the inequality is
always true, because a positive or at most zero quantity is always
greater than a negative one; if, instead, g(x) ≥ 0, then it is safe to
take the n-th power of both sides thus getting rid of the radical.
Therefore, the solution set is the union of
(
(
g(x) < 0 [
g(x) ≥ 0
f (x) ≥ 0
f (x) ≥ g(x)n .
Note that in the second system, the existence condition f (x) ≥ 0
follows from f (x) ≥ g(x)n because, since n is even, g(x) ≥ 0.
• We consider now
p
n
f (x) ≤ g(x),
and with n even.
Again, in order for the radical to be defined, its radicand must
be positive. If g(x) < 0, p
the inequality is inconsistent, because a
n
radical of even degree, as f (x), is positive or at most zero as soon
as defined. So, assuming g(x) ≥ 0, we can safely take n-th power
29
30
2. IRRATIONAL AND TRIGONOMETRIC INEQUALITIES
of both sides, getting an equivalent systems of inequalities without
radicals


f (x) ≥ 0
g(x) ≥ 0


f (x) ≤ g(x)n .
• Let’s consider finally
p
n
f (x) R g(x),
with n odd.
The radical is now defined whatever f (x), and we can take n-th
power of both sides obtaining an inequality with same solution set
f (x) R g(x)n .
CHAPTER 3
Exponential and logarithmic functions
We introduce a new class of real valued functions: exponential and logarithmic functions; of the former, defined on the whole R, we have already
studied its restrictions to Q and their properties are well-known, so in this
chapter we are left with extending the definition and their features to R; the
latter will be defined as the inverses of the former on their images.
Exponential and logarithmic functions (like the trigonometric functions)
are not polynomials and they can’t even be constructed ”easily” from them,
as for example rational functions by a quotient, or piecewise, as for example
the absolute value. Nevertheless, like trigonometric functions, exponential
and logarithmic functions can be ”approximated”, in a way which si not
part of this course, by means of polynomials.
3.1. Exponential functions
More than a rigorous definition, we present a series of steps, which we
should prove but that we won’t do, leading to a definition of the symbol bx ,
extending what we already know when x is rational.
Fix b > 1.
(1) Suppose x is a positive integer n: in this case we know how to
define exponentiation
bx = bn = b| · {z
... · }b .
n times
(2) Suppose now x is a negative integer −n: we define
1
bx = b−n = n .
b
with bn defined as in step 1.
(3) Suppose x = 1/n (that is, a rational number which is the inverse
of an integer): the exponentiation by x is the (positive, when n is
even) nth-root of b
√
n
bx = b1/n = b.
One should prove that such number
√ exists, that is, that there is a
real number y, which we denote n b, such that y n = b. This follows
from deep properties of the real numbers, that is, from the way
R is constructed from Q: indeed, note that at basic math we told
you roughly how to construct Z from N and how to obtain Q from
31
32
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Z, and some of you even know how to pass from R to C, the field
of complex numbers, however, we didn’t tell you exactly what real
numbers are.
The idea, to be a bit more precise, is to take the set of B = {t ∈
R | tn < b} and to prove that the smallest among the real numbers
larger than any element of E, which we call y, does exist and has
the property y n = b; we then define b1/n to be such y.
(4) Suppose now that x is a generic rational number: then x can be
written as a fraction in a non-unique way
x=
m
p
= ,
n
q
where m and n are integers; if x > 0, we can assume that n and q are
nonzero naturals (that is, strictly positive integers). For example,
the fractions 3/2 = 6/4 = 300/200 = ... all define the same rational
number, which in decimal notation is 1.5.
If we can prove (it’s possible but we don’t do it) that
(bm )1/n = (bp )1/q
(observe that all the exponentiations used above have been defined
in previous steps), then we can define
bx = (bm )1/n ,
whatever the representation of x as fraction is, as we proved (or
said that we can prove) that the computation doesn’t depend on
it.
If x < 0, then we set bx to be the inverse of b−x , which has been
defined just above as −x > 0. The case x = 0 has already been
considered.
So, bx is well-defined for every rational number x.
(5) Suppose, finally, x is any real number: the idea is again to take
the set B(x) = {bt | t is rational and t ≤ x}, to prove that the
smallest of all numbers larger than any elements of B(x) exists,
and which we call y, and to prove that such number y is bx when
x is rational; as a consequence, it then makes sense to define bx to
be such number y for any real number x.
If 0 < b < 1, then we can write b = a1 for some real number a > 1, and
we set bx = a1x , where ax is given in the way we explained above.
We have defined in this way, for every real number b > 0, b 6= 1, a
function
b− : R → R
x 7→ bx ,
3.1. EXPONENTIAL FUNCTIONS
33
which we call exponential function of base b (note that, if b = 1, the function
bx is just the constant function 1x = 1, which so too trivial to be called with
such a high-sounding name as exponential function).
One can prove that the function bx shares the properties of exponentiation which were studied for integer exponents.
Theorem 3.1.1.
bx+y = bx by
(bx )y = bxy
(ab)x = ax bx ,
for every real numbers x and y and a, b > 0.
As to the first of the properties above, note that it says that exponential
functions convert sums into products.
We summarize further properties of bx .
Proposition 3.1.2. The function bx
(1) is positive, namely, bx > 0, for every x ∈ R;
(2) has the positive real half-line R>0 = {x ∈ R | x > 0} as
image;
(3) is increasing for b > 1, that is, if x < y then bx < by , and
decreasing for 0 < b < 1, that is, if x < y then bx > by ;
(4) is injective;
(5) intersects the y-axis at (0, b0 ) = (0, 1).
If you check the construction we sketched of bx , it should be intuitively
quite clear why bx is positive: the procedure began with integer powers of
positive numbers, and the operations we did did not affect the sign. Positivity has the following consequence
• the graph bx lies in the upper half plane;
• , in particular, exponential functions have no intersection with the
x-axis.
The second property (which is not implied by the positivity) says that
the equation bx = y has, for every y > 0, a unique solution x; injectivity
implies then that such solution x is unique: for example,
bx = 1
whether b > 1 or 0 < b < 1.
⇐⇒
x = 0,
34
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
The increasing/decreasing property, according if b > 1 or 0 < b < 1,
implies injectivity and will be relevant, as we will see, when studying exponential inequalities.
Below, the graph of exponential function, when b > 1 and 0 < b < 1, for
example 2 and 12 respectively, subsumes the properties above.
y
2x
(0, 1)
x
x
1
2
y
(0, 1)
x
3.2. LOGARITHMIC FUNCTIONS
35
Remark 3.1.3. Observe that, if a 6= b, then ax = bx if and only if
x = 0 (this will be clearer after the cha. In terms of graphs, this
means that the graph of ax and the graph of bx meet only at (0, 1).
Moreover, if 1 < a < b or 0 < a < b < 1 or 0 < a < 1 < b, then
ax < bx for x > 0, and ax > bx for x < 0, that is, the graph of ax lies
below that of bx for x > 0, above for x < 0.
The most important number in matheamtics (or one of the most important) is
e
the EULER NUMBER: it is an irrational number, that is it is not a fraction,
and it is a transcendental number, that is, there is no polynomial equation
with integer coefficient having e as a root. It is approximately equal to
2.71828. The exact definiton will be given in 4.4.11.
The corresponding exponential function
ex
is, in turn, one of the most important function in math, therefore, we will
refer to ex as THE exponential function. Since e > 1, this tells us about the
graph and features of ex . Note also the notation exp(x) for ex .
3.2. Logarithmic functions
In the previous section, we saw that exponential functions bx , with 0 <
b < 1 or b > 1, are injective and that their image is Rx>0 , therefore, as
functions R → R>0 are bijective and so invertible. Note here that if we
allow b = 1, then bx , as already remarked, is just the contant function 1,
which is not injective.
Definition 3.2.1. Let 0 < b < 1 or b > 1. The logarithmic function
with base b is the function
logb − : Rx>0 → R
x 7→ logb x
which is the inverse of the expoenential function b− : R → Rx>0 .
As in general for any function, x is called the argument of the logarithm.
When the base is the Euler number e, we denote the corresponding
logarithm as log x, that is, dropping the index e, or ln x: in this book we
will prefer the former.
36
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Let’s unroll the definition of logarithmic function of base b to understand
”in practice” what the logarithm of a number is: since bx is the power of
b with exponent x, for a given y ∈ Rx>0 , the definition of inverse function
says that the logarithm logb y of y with base b is the unique exponent which
must be given to the base b to get y. Namely, from the definition of inverse
function we have the following identities
Corollary 3.2.2.
blogb x = x
logb bx = x.
(1)
Note that by setting x = 1 in the second formula above, we see that
logb b = 1,
while by setting x = 0 in the same formula, we have that
logb 1 = 0.
By means of these identities, the properties of exponentiation in theorem
3.1.1 translate into the following properties of logarithms
Corollary 3.2.3. Let b ∈ R>0 − {1}.
(1) For x, y ∈ R>0
logb (xy) = logb x + logb y
logb
x
= logb x − logb y.
y
(2) For x > 0 and r ∈ R
logb xr = r logb x.
Logarithms in different bases are related by the following formula
Proposition 3.2.4. Let a, b ∈ R>0 − {1}, then
logb x
loga x =
.
logb a
By setting a = b in the formula above, we obtain
loga b =
1
.
logb a
3.2. LOGARITHMIC FUNCTIONS
37
As an application of corollary 3.2.3 and proposition 3.2.4 observe that
for a, b ∈ R>0 − {1} and r ∈ R − {0}
logar b =
1
1
1
1
=
= loga b = loga b r ,
r
logb a
r logb a
r
(2)
and note that for b = 1 the terms above which are defined are trivially equal
to zero.
Example 3.2.5. Note that logb (x + 1)(x + 2) is defined when (x +
1)(x + 2) > 0, that is, for x < −2 or x > −1. The correct way to
apply formulas for log is
logb (x + 1)(x + 2) = logb |x + 1| + logb |x + 2|
Indeed, one might be tempted to write logb (x+1)(x+2) = logb (x+1)+
logb (x+2), but this holds only for x > −1, where both x+1 and x+2
are positive, however, for x < −2, both x + 1 and x + 2 are negative,
so, logb (x + 1) + logb (x + 2) has no meaning; the absolute value fixes
the problem. In fact, note that the first formula in corollary 3.2.3
holds for x, y ∈ R>0 .
We obtain the graph of logb x by reflecting the graph of bx about the
bisector of the first and second quadrant y = x in the Cartesian plane. In
the figures below, the bases are b = 2 and b = 21 , representing the two cases
b > 1 and 0 < b < 1 respectively.
y
log2 x
(1, 0)
x
38
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
y
(1, 0)
x
log 1 x
2
We summarize further features of logarithmic functions, which can read
off from the graphs above.
Proposition 3.2.6. The function logb x
(1) is injective and surjective;
(2) for b > 1 is positive for x > 1 and negative for 0 < x < 1,
for 0 < b < 1, is positive for 0 < x < 1 and negative for
x > 1;
(3) is increasing for b > 1, that is, if x < y then logb x < logb y,
and is decreasing for 0 < b < 1, that is, if x < y then
logb x > logb y;
(4) logb x = 0 if and only if x = 1, that is, the graph of logb
intersects the x-axis only at 1.
Parallel to remark 3.1.3, we observe what follows regarding the relative
position of the graphs of logarithmic functions relative to different bases.
Remark 3.2.7. If a 6= b then loga x = logb x if and only if x = 1,
that is, the graph of loga x and the graph of logb x meet only at (1, 0).
Moreover, if 1 < a < b or 0 < a < b < 1 or 0 < a < 1 < b, then
loga x < logb x for x > 1, and ax > bx for 0 < x < 1, that is, the
graph of ax lies below that of bx for x > 1, above for 0 < x < 1.
3.3. Exponential equations
Exponential equations are equations where the variable is involved in
an exponent in at least one place. To solve this kind of equations, often
3.3. EXPONENTIAL EQUATIONS
39
the strategy consists in recognizing a polynomial equation under the guise
of exponential equation, reducing in this way to solving an equation of the
form bf (x) = a, which we analyse just below.
Example 3.3.1. Consider the equation
bf (x) = a,
where f (x) is some expression containing x. Observe that if a ≤ 0,
the equation has no solutions, because expoenential functions are
positive (proposition 3.1.2); otherwise, if a > 0, by (3.2.2) we can
write a = blogb a ; substituting into the given equation we obtain
bf (x) = blogb a ,
and since exponential functions are injective (proposition 3.1.2), the
equation above is equivalent to
f (x) = logb a.
Since logb a does not contain x, the solvability of the given equation
depends on f (x): for example, if f (x) is a polynomial, we use the
techniques for polynomial equations.
Example 3.3.2. A variant of the example above is
bf (x) = cg(x) ,
where f (x) and g(x) are expressions containing x. Since c > 0, then
we can write c = blogb c , and substituting into the given equation, we
obtain
bf (x) = (blogb c )g(x)
= b(logb c)g(x) ,
which, using injectivity of the exponential function, yields
f (x) = (logb c)g(x).
If, for example, f (x) and g(x) are polynomials, then the given equation boils down to a polynomial equation.
Let’s see some concrete example
Exercise 3.3.3. Solve the equation
2 · 22x−1 − 6 · 2x−1 + 2 = 0.
40
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Applying properties of exponentiation, the equation becomes
2 · 22x · 2−1 − 6 · 2x 2−1 + 2 = 0,
and, after simplifying,
22x − 3 · 2x + 2 = 0.
Setting t = 2x into the equation above
t2 − 3t + 2 = 0,
whose solutions are t = 1 and t = 2. Recalling that t = 2x yields two
equations of the form studied in example 3.3.1
2x = 1
or
2x = 2.
or
2x = 2 = 21 ,
Since 1 = 20 and 2 = 21
2x = 1 = 20
the injectivity of exponential functions gives solutions
x=0
or
x = 1.
Exercise 3.3.4. Solve the equation
4
= 0.
e
Clearly this equation has no solution, because the sum of three strictly
positive terms can not be zero.
e2x−1 + 5ex−1 +
Exercise 3.3.5. Solve the equation
5 · 8x + 3 · 23x = 16.
Observing that 23x = (23 )x = 8x , we have
5 · 23x + 3 · 23x = 16
⇐⇒
(5 + 3) · 23x = 16
⇐⇒
8 · 23x = 16
⇐⇒
⇐⇒
23x = 2
3x = 1
1
x= .
3
⇐⇒
3.4. LOGARITHMIC EQUATIONS
41
Exercise 3.3.6. Solve the equation
2 · 5x−1 =
The equation is equivalent to
4
125 · 5x−1 = · 2x+1
2
4 x+1
2 .
125
⇐⇒
53 · 5x−1 = 2 · 2x+1
⇐⇒
5x+2 = 2x+2 ,
and by remark 3.1.3, the equation above is equivalent to x + 2 = 0,
that is, x = −2. Alternatively, and more generally, passing to logarithms (exponentials are always positive and logarithms are injective
functions)
5x+2 = 2x+2
⇐⇒
log2 5x+2 = log2 2x+2
⇐⇒
(x + 2) log2 5 = x + 2
⇐⇒
(x + 2)(log2 5 − 1) = 0,
and since log2 5 6= 1, we get again x + 2 = 0, that is, x = −2. In a
similar way, one could have used the equality 5 = 2log2 5
5x+2 = 2x+2
⇐⇒
2(x+2) log2 5 = 2x+2
⇐⇒
(x + 2) log2 5 = x + 2,
and proceed as above.
3.4. Logarithmic equations
Logarithmic equations are equations that involve variables in the arguments or in the bases of logarithmic functions. We will study a few cases
where we have a way to solve such equations. But before starting, let’s
not forget that logarithms are defined only for positive values of their argument, thus, determining when a logarithmic equation is defined, and so has
meaning, will be the first step.
Example 3.4.1. Consider the equation
loga f (x) = b.
The equation is defined for those x ∈ R for which f (x) > 0: so
we have to look for solutions among such x. Since b = loga ab , the
equation becomes
loga f (x) = loga ab ,
and since loga (−) is injective, then the above equation is equivalent
to
f (x) = ab .
42
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
If f is a polynomial, then f (x) = ab is a polynomial equation in x
(note that x is not involved in the expoenential ab ).
Exercise 3.4.2. Consider
log2 (x + 3) + log2 x = 2.
Let’s soon observed that the first logairthm is defined for x + 5 > 0,
namely, x > −5, and the second for x > 0, so the equation makese
sense wher both the logarithms are defined, that is, for x > 0 (the
intersection of the sets {x > −5} and {x > 0}: this is the set where
we must look for solutions of the given equation. Using the properties
of logarithms and writing 2 = log2 s2 , we have
log2 x(x + 3) = log2 4,
and so
x(x + 3) = 4
⇐⇒
⇐⇒
x2 + 3x − 4 = 0
x = −4 or x = 1.
Since we are assuming x > 0, only x = 1 is solution of the given
equation.
In the next exercise, under the guise of a logarithmic equation, we discern
a polynomial equation.
Exercise 3.4.3. Consider
(log(x − 2))3 − (log(x − 2))2 = 4((log(x − 2))2 − 1).
We soon see that the equation makes sense only when x − 2 > 0, that
is, for x > 2, because for such values the arguments of the logarithms
are positive.
Setting, for simplicity, t = log(x − 2), the equation becomes polynomial
t3 − t2 = 4(t2 − 1)
⇐⇒
t2 (t − 1) − 4(t + 1)(t − 1) = 0
⇐⇒
(t − 1)(t2 − 4t − 4) = 0
√
√
t = 1 or t = 2 − 2 2 or t = 2 + 2 2.
⇐⇒
Since t = log(x − 2), we obtain
√
√
log(x − 2) = 1 or log(x − 2) = 2 − 2 2 or log(x − 2) = 2 + 2 2
√
⇐⇒
x − 2 = e or x − 2 = e2−2
2
√
or x − 2 = e2+2
2
3.4. LOGARITHMIC EQUATIONS
√
⇐⇒
x = e + 2 or x = e2−2
2
43
√
+ 2 or x = e2+2
2
+ 2.
In the next example, the variable x occurs in the base of the logarithm:
we change base formula proposition 3.2.4.
Exercise 3.4.4. Consider
logx+2 3 = log7 5.
The base of logarithm must be a number 0 < x + 2 < 1 or x + 2 > 1,
that is, −2 < x < −1 or x > −1, namely, x > −2 and x 6= −1. Let’s
write the two logarithms in the same base (see proposition 3.2.4):
log 5
log 3
log7 5 =
.
logx+2 3 =
log(x + 2)
log 7
Substituting into the given equation we obtain
log 3
log 5
log 3 · log 7
=
⇐⇒
log(x + 2) =
log(x + 2)
log 7
log 5
Note that e
e
log 3·log 7
log 5
log 3·log 7
log 5
> 0, so e
⇐⇒
x+2=e
⇐⇒
x=e
log 3·log 7
log 5
log 3·log 7
log 5
log 3·log 7
log 5
− 2.
− 2 > −2, and, since
log 7
log 5
6= 1,
− 2 6= 1
Exercise 3.4.5. Consider
log2 (4x+1 − 2) − 2x
= 1.
2x + 1
Note the equation makes sense when the denominator is not zero and
the argument of the logarithm is positive:
2x + 1 6= 0
1
x 6= −
2
1
x 6= −
2
1
x 6= −
2
4x+1 − 2 > 0
22(x+1) > 2
2x + 2 > 1
x>−
1
2
44
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
that is, x > − 21 . Doing common denominator
log2 (4x+1 − 2) − 2x
=1
2x + 1
⇐⇒
log2 (4x+1 − 2) − 2x − 2x − 1
=0
2x + 1
log2 (4x+1 − 2) − 2x − 2x − 1 = 0
⇐⇒
log2 (4x+1 − 2) = 4x + 1
⇐⇒
log2 (4x+1 − 2) = log2 24x+1
⇐⇒
22(x+1) − 2 = 24x+1
⇐⇒
22 · 22x − 2 − 2 · 24x = 0
⇐⇒
24x − 2 · 22x + 1 = 0
⇐⇒
and, setting 22x = t,
⇐⇒
⇐⇒
t2 − 2t + 1 = 0
t = 1.
So, substituting,
t = 22x = 1
⇐⇒
⇐⇒
2x = 0
x = 0.
3.5. Exponential inequalities
An exponential inequality is an inequality in which the variable x is involved in an exponent in at least one place. Beside the same techniques
studied for exponential equations, what is essential is property (3) in proposition 3.1.2 which must be applied carefully.
Exercise 3.5.1. Consider
2
(7x+2 )x+3 49x −1
≤ 1.
49x2
We have, since the base 7 > 1,
(7x+2 )x+3 49x
49x2
2 −1
≤1
2 −1−x2 )
⇐⇒
7(x+2)(x+3) 72(x
⇐⇒
7x
⇐⇒
⇐⇒
x2 + 5x + 4 > 0
x < −4 or x > −1.
2 +5x+6−2
≤1
≤1
3.5. EXPONENTIAL INEQUALITIES
Exercise 3.5.2. Consider
√3 1
√3 4x
x
>
3
3
Note first that the equation is not defined for x = 0, as a division
by zero occurs at the exponent
√ on the left hand side. Then, since
the base of the exponential 33 < 1, according to property (3) in
proposition 3.1.2 we must reverse the inequality relation
1 − 4x2
<0
x
(1 + 2x)(1 − 2x)
⇐⇒
< 0.
x
Lets’ study the sign of each factor:
• 1 + 2x > 0 ⇔ x > − 21 ;
• 1 − 2x > 0 ⇔ x < 12 .
• x > 0.
Portraying the study in a sign table
1
< 4x
x
⇐⇒
− 12
1 + 2x
1 − 2x
x
(1+2x)(1−2x)
x
1
2
0
−
+
+
+
−
−
−
+
−
−
+
+
−
+
−
+
0
?
0
Therefore, the solution is x < − 12 or 0 < x < 12 .
Exercise 3.5.3. Consider
√
x
x
√
< ( x)x .
The inequality makes sense for x > 0, as both the radicand X in the
exponent on the left hand side must be positive, as well as the base
x at the left hand side, and noticing that 00 is also not defined.
45
46
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Since log is injective and the base e > 1
√
√
x
log x x < log( x)x = log x 2
⇐⇒
√
x
x log x < log x
2
√
x
x−
⇐⇒
log x < 0.
2
Let’s study the sign of each factor, under the condition x > 0:
√
√
• x − x2 > 0 ⇔ 2 x > x, and since x > 0, we can safely
square, getting 4x > x2 , that is, x2 − 4x < 0, which holds,
since x > 0, for 0 < x < 4;
• log x > 0 ⇔ x > 1.
Reporting the study in a sign table
0
√
x−
x
2
log x
√
x−
x
2
log x
1
4
+
+
−
−
+
+
−
+
−
0
0
Therefore, the solution is 0 < x < 1 or x > 4.
Exercise 3.5.4. Solve
4x+1 − 2x+2
≥ −8.
2x − 1
For the inequality to have meaning, the denominator 2x − 1 on the
left hand side must not vanish, that is, x 6= 0.
Using properties of exponentiation and making common denominator
4x+1 − 2x+2
4 · 22x − 4 · 2x
≥
−8
⇐⇒
+8≥0
2x − 1
2x − 1
4 · 22x − 4 · 2x + 8 · 2x − 8
⇐⇒
≥0
2x − 1
4(22x + 2x − 2)
⇐⇒
≥ 0.
2x − 1
Setting now t = 2x > 0
⇐⇒
t2 + t − 2
≥ 0,
t−1
3.5. EXPONENTIAL INEQUALITIES
and factoring the numerator
(t + 2)(t − 1)
⇐⇒
≥ 0,
t−1
that is, since t − 1 6= 0 for x 6= 0 as we are assuming,
⇐⇒ t + 2 ≥ 0
⇐⇒ 2x + 2 ≥ 0,
which holds for every x ∈ R, x 6= 0.
Exercise 3.5.5. Consider
r 2 x
2 2x
+3>3
− 1.
3
3
Note first that the equation makes sense for every x ∈ R, despite
a square root: in fact, the radicand, being the sum of an exponential function, which so is always positive, with a positive number, is
always positive as well. Setting now t = ( 23 )x , we have
p
t2 + 3 > 3t − 1.
If 3t − 1 < 0, that is, for t < 13 , the inequality is true, as the left hand
side is always positive. If t ≥ 31 , both sides are positive, so we can
safely square yielding
t2 + 3 > (3t − 1)2
4t2 − 3t − 1 < 0
1
⇐⇒
− < t < 1,
4
1
which, together with the condition t ≥ 3 , gives
1
< t < 1.
3
Taking the union of the solution sets for t < 13 and t ≥ 13 , we find
⇐⇒
t < 1,
and substituting
2 x
<1=
2 0
.
3
3
Since the base 23 of the exponential is less than 1, when applying property (3) in proposition 3.1.2, we must reverse the inequality relation,
yielding the solution x > 0.
47
48
3. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
3.6. Logarithmic inequalities
A logarithmic inequality is an inequality in which the variable x is involved in the argument or in the base in at least one place. The techniques
studied for logarithmic equations apply, care must be used when using property (3) in proposition 3.2.6, and, of course, we shouldn’t forget logarithms
are defined only when their arguments are positive.
Exercise 3.6.1. Consider
log 1 (x + 1) > log 1 (5 − x).
2
4
The logarithms in the inequality are defined for x + 1 > 0 and 5 − x >
0, so the inequality makes sense only in the interval −1 < x < 5,
where we will look for solutions.
By (2)
1
log 1 (5 − x) = log( 1 )2 (5 − x) = log 1 (5 − x),
4
2
2
2
so, substituting, the given inequality becomes
1
⇐⇒
2 log 1 (x + 1) > log 1 (5 − x)
log 1 (x + 1) > log 1 (5 − x)
2
2
2
2
2
2
⇐⇒
log 1 (x + 1) > log 1 (5 − x)
2
2
and by property (3) in proposition 3.2.6 for the case when the base
1
1
2 is such that 0 < 2 < 1,
⇐⇒
(x + 1)2 > 5 − x
⇐⇒
⇐⇒
x2 + 3x − 4 > 0
x < −4 or x > 1.
Since we are working under the condition −1 < x < 5, the solution
of the given equation is 1 < x < 5.
Exercise 3.6.2. Solve
log(x2 + 3)
≤2
log(x + 1)
Observe that the logarithm at numerator is always defined as x2 +3 >
0 for all x, while the logarithm at the denominator is defined for
x > −1.
Making common denominator
log(x2 + 3)
≤2
log(x + 1)
⇐⇒
log(x2 + 3) − 2 log(x + 1)
≤0
log(x + 1)
3.6. LOGARITHMIC INEQUALITIES
log(x2 + 3) − log(x + 1)2
≤ 0.
log(x + 1)
Let’s study the sign of numerator and denominator. As for the numerator
⇐⇒
log(x2 + 3) − log(x + 1)2 > 0
⇐⇒
log(x2 + 3) > log(x + 1)2
⇐⇒
⇐⇒
⇐⇒
x2 + 3 > x2 + 2x + 1
2x < 2
x < 1,
and recalling the existence condition x > −1, we have that the numerator is positive for −1 < x < 1. As for the denominator
log(x + 1) > 0
⇐⇒
⇐⇒
x+1>1
x>0
From the sign table
−1
Numerator
Denominator
Expression
0
1
+
+
−
−
+
+
−
+
−
?
we see the solution is −1 < x < 0 or x ≥ 1.
0
49
CHAPTER 4
Limits of functions
The notion of limit, that is, the idea of convergence, is one the central
concepts in analysis. As we will see, the derivative of a function is a limit,
and integration is also defined in terms of some limiting process.
The idea of limit is quite intuitive, in fact, ancient Greeks had some
notion of it, and Leibnitz and Newton, who invented ”infinitesimal calculus”, had for sure clear what a limit is; however, what makes it even more
interesting is that the definition of limit is so subtle that a precise formalization came only at the beginning of XIX century with Cauchy, Bolzano,
and later Weierstrass, and it involves some other deep ideas which we will
try to present at some extent (actually, mathematicians make definitions
after theorems, and a definition is a good definition if hypothesis of some
theorem).
4.1. Some topology
We will be concerned only with functions f : A ⊆ R → R: while limits
can be defined for functions between more exotic sets, it is nonetheless the
study of the case of real valued and real defined functions which will guide
with extending the concept to more general aituations.
The idea of limit is that of describing the behaviour of function as above
”near” a point x0 , that is, an element of R (where the domain A is contained)
which is ”close enough” to the domain; note that we do not pretend x0 is an
element of the domain, but we require x0 to have ”sufficiently many” points
of A close to it (for the function to behave in some way). By behaviour
of f ”near” x0 we mean that we are not interested in what f does at x0
(in fact, in case, we are already able to evaluate functions at points of its
domain), as f even need not to be defined at x0 . In even simpler terms, we
want to asnwer to the follwoing questions: what does the function do when
approaching to x0 ?
Since we are doing mathematics, we want to clarify in a mathematical
way all the vague statements in quotes above. The branch of mathematics
doing so is called topology, and in this way it provides the suitable formalism
to extend the notion of convergence to more general settings (for example,
one could be interested not only in expressing nearness for real numbers but
also in saying what we mean that a function is near to another function).
What might be suitable to express nearness? A notion of distance. Do
we have such a notion on R? Yes, we do have it: the absolute value! In fact,
51
52
4. LIMITS OF FUNCTIONS
the absolute value |x| of a real number x provides a reasonable notion of
distance of the point x from 0, and |x − y| will then be the distance between
the real number x and y. Since that of distance is a ”geometrical” idea, we
will from now often call elements of R as ”points” and we will think of R as
a ”line”.
Remark 4.1.1. We are not defining here what we mean by distance,
but it will be a precise definition of this notion which will allow the
extension of limit to different situations. However, the curious reader
will be wondering about which reasobale properties a distance should
fulfill: well, we pretend that the distance between two points is a
positive number and that the distance between two points vanishes
if and only if the two points coincide; then a distance from a point A
to a point B should not depend if measured from A to B or from B
to A; last, we would like that going ”straight” from A to B is shorter
than going from A to C first and then from C to B.
Recall from basic math the properties of the absolute value:
|A − B| ≥ 0
|A − B| = 0
⇐⇒
A=B
|A − B| = |B − A|
|A − B| ≤ |A − C| + |C − B|
are not they just what we said to expect from a distance?
The interested reader will find out that in applications even negative distances might make sense (in Relativity, the physical theory of
gravitation).
The next definition provides us with a tool to talk about nearness. Note
that the notion of distance is still hidden in it.
Definition 4.1.2. Let x0 ∈ R and > 0. We define the ball B(x0 , )
of center x0 and radius to be the interval (x0 − , x0 + ).
Why the set defined above deserves the name of ball becomes evident if
we observe, recalling how to expend inequalities with absolute value, that
B(x0 , ) = {x ∈ R | |x − x0 | < }.
Interpreting the absolute value as distance, the inequality characterizing the
elements of B(x0 , ) reads as ”the real number x whose distance from x0 is
less than , and a ball is the set of points whose distance from a chosen
point, called center, is less than the radius of the sphere which bounds the
4.1. SOME TOPOLOGY
53
ball; viewing R as a line, balls that we have just defined are ”1-dimensional”
balls. Note also that the sign < is a premonitory symptom of a overwhelming
entrance of inequalities into the story, actually, almost everything will boil
down to some inequality, so we advise the reader to review what studied
about inequalities in case any doubt about them is still hiding somewhere.
x0 − x0
x0 + R
Remark 4.1.3. We will also call neighborhood of x0 any subset of the
form B(x0 , ) and we will use the notation N (x0 ). More generally, a
neighborhood of x0 is defined to be any subset of R which contains
a ball centered at x0 . This has some advantage: for example, every
interval, even if not centered at x0 but containing x0 would be a
neighborhood of x0 according to this definition, and the green set
below
x0
R
would be as well a neighborhood of x0 , however, our definition has
the advantage of being conceptually easier, besides, the definition of
limit which we will give in this chapter will not depend on which of
these two definitions of neighborhood we will assume.
We stress again that what is essential in our definition of ball or neighborhood is to have a notion of distance: for example, any scaling the absolute
value may serve as well as a distance on R, whose set of neighborhoods
remains inchanged; if we were able to define a distance on a certain set of
functions, then we would be able to talk about ball centered at some function
and of a given radius.
Example 4.1.4. The curious reader may consider the following example. Let I = [a, b] ⊂ R be a closed interval and consider
`∞ (I) = {f : I → R | |f (x)| < M for some M ∈ R}
the set of what we reasonably call bounded functions on I, and define
for two such functions f and g as
|f − g| = max |f (x) − g(x)|,
x∈I
where the notation max means you have to take the largest of all
|f (x) − g(x)| for x ∈ I. Would it be a good notion of distance on the
set `∞ (I)? Can you figure out how a neighborhood of f looks like?
54
4. LIMITS OF FUNCTIONS
By means of neighborhood we can now express what we mean that an
elements of R has ”sufficiently many points of the domain” close it, where
sufficiently is relative to our purpose of studying a function near, that is, in
a neighborhood of a point.
Definition 4.1.5. Let A ⊆ R. We say x0 ∈ R is an accumulation
point of A (or limit point, or cluster point) if every neighborhood of
x0 contains an element of A − {x0 }.
Let’s use some symbol, namely, ∀, which means ”for every”, ∃, which
means ”exists”, and : or, more informally, s.t.., which means ”such that”.
to write the above definition in a more compact form:
x0 is an accumulation point of A if and only if
∀B(x0 , ) ∃a ∈ A − {x0 } such that a ∈ B(x0 , ),
or even,
∀ > 0 ∃a ∈ A − {x0 } s.t. a ∈ B(x0 , ).
We soon make a remark about the definition: if x0 itself happens to be
a point of A, then, for it to be an accumulation point, we pretend to find
a point of A distinct from x0 in any of its neighborhoods, otherwise, every
point of A would be accumulation point of A, without necessarily having
many points of A close to it. Indeed, we give a name to such points.
Definition 4.1.6. A point of A which is not an accumulation point
of A is called an isolated point.
Let’s see few examples to clarify visually the notion of accumulation
point (and of isolated point).
Example 4.1.7. Let A = { n1 | n ∈ N − {0}}. We want to prove that
0 is an accumulation point of A. Observe, letting n + 1, 2, 3, 4, 5, ...,
that the elements of A are
n 1 1 1 1 o
A = 1, , , , , ...
2 3 4 5
and note that 0 ∈
/ A. If we represent R as a line, then A looks like a
sequence of dots ”accumulating” near 0
4.1. SOME TOPOLOGY
0
11 1 1
76 5 4
1
3
1
2
55
1
R
If we can prove that every B(0, ) contains an element of A (since
0∈
/ A, such element will be distinct from 0), then we will have proved
that 0 is an accumulation point of A. Since the elements of A are
the numbers of the form 1/n, let’s see if ∀ > 0 we are able to find
n ∈ N − {0} such that
1
∈ (−, ),
n
that is
1
− < < .
n
Since n is positive, then − < 0 < 1/n holds ∀ > 0 and ∀n ∈ N−{0};
solving 1/n < , we find
1
n> ,
that is, choosing such an n, we are then sure that 1/n falls into B(0, ).
In symbols we proved that ∀ > 0 (that is, for every ball centered at
x0 ), ∃n (that is, we can find an element in A) such that − < n1 < and clearly n1 6= 0 (that is, n1 ∈ A − {0} and its distance from 0 is less
than ).
For example, if = 14 = 0.25, then if we pick any n > 1 = 4 we are
sure that the element n1 of A is close to 0, that is, its distance from 0
is less than the given = 14 . If = 1019 , then if we pick n > 1 = 109
we are then sure that the element n1 of A is close to 0 with distance
less than 1019 . In the picture below we see a neighborhood of 0 whose
radius is a number between 1/6 and 1/5
0
11 1 1
76 5 4
1
3
1
2
1
R
Note, instead, that 12 = 0.5, though it belongs to A, is not an accumulation point of A:
0
11 1 1
76 5 4
1
3
1
2
1
consider the neighborhood
1 1 1
1 1
1 4 6
B ,
=
− , +
=
,
= (0.4, 0.5),
2 10
2 10 2 10
10 10
R
56
4. LIMITS OF FUNCTIONS
that is, theball of radius =
1
that n1 ∈ B 12 , 10
; from
1
10
= 0.1, let’s study if there is n such
1
6
4
< <
10
n
10
we find
10
10
<n< ,
6
4
and since
10
10
<2<
< 3,
6
4
we can only choose n = 2, which yields n1 = 12 , but in the definition of
accumulation point we want an element of A−{1/2}, so the definition
of accumulation point for 12 is not fulfilled.
1<
Exercise 4.1.8. Verify whether:
(1) 1 is an accumulation point of
n
o n
o
1
1
A= 1+
|
n
∈
N
∪
1
−
|
n
∈
N
;
(n + 1)2
(n + 1)2
(2) 0 is an accumulation point of
B = [0, 1] ∩ (R − Q);
Example 4.1.9. Consider the segment I0 = [0, 1], from it remove the
interval ( 13 , 32 ), and call I1 what is left, that is,
h 1i h2 i
I1 = 0,
∪ ,1 .
3
3
Continue in this way by removing the middle third of each of the
intervals forming I1 , that is, ( 19 , 29 ) and ( 79 , 89 ), getting
h 1i h2 3i 6 7i h8 i
I2 = 0,
∪ ,
∪[ ,
∪ ,1 ,
9
9 9
9 9
9
and so on obtaining a sequence of sets I0 ⊃ I1 ⊃ I2 ⊃ I3 ⊃ ..., each
of which is a union of 2n intervals of lenght 3−n . Define P to be
P = I0 ∩ I1 ∩ I2 ∩ I3 ∩ ....
P is called Cantor set and is an example of fractal: it has neither dimesion 1 nor dimension 2, in fact, it has dimension log 2/ log 3 (whatever
dimension could mean).
4.2. THE DEFINITION OF LIMIT
57
Let’s try to figure out how P looks like by sketching the first terms
in the sequence I0 ⊃ I1 ⊃ I2 ⊃ ..., whose intersection is by definition
P:
I0
I1
I2
0
1
0
0
1
9
2
9
1
3
2
3
3
9
6
9
1
7
9
8
9
1
I3
I4 is then obtained by removing the middle third of all the segments
forming I3 , and so on.
Let’s prove that every point of P is an accumulation point of P , that
is, P has no isolated point. Let x ∈ P and consider a ball B(x, ) of
center x and radius > 0. Since P is the intersection of the sets In ,
then x ∈ In for every n, in particular, since In is a union of intervals
Ink (in the picture, I1 is union of two intervals, which we call I11 and
I12 ; I2 is union of four intervals, which we call I21 , I22 , I23 , I24 ; I3
is the union of eight intervals, and so on), then x ∈ Inj for some j.
Since Inj has length 3−n , then, choosing n large enough (3−n < ,
that is, n > log3 1 ), we have Inj ⊂ B(x, ). Then, by the way P
is constructed, the two endpoints of Inj are in P , and and by the
previous remark they are also in B(x, ); since at least one of them is
not x, it follows x is an accumulation point of P .
4.2. The definition of limit
Since the limit of a function f at some point x0 wants to describe the
behavior of the function near x0 and not at x0 , we will define it only when
x0 is an accumulation points of the domain dom(f ) of f , even if x0 is not in
the domain of the function; the idea is that the limit of f at x0 is the value
which the function approaches when x gets closer to x0 and which need not
to be f (x0 ) provided this is defined, that is, when x0 ∈ dom(f ) (intuitively,
when x0 is an accumulation point of dom(f ), according to our definition,
there are many x ∈ dom(f ) getting closer and closer to x0 and so it makes
sense to wonder about the beahvior of f near x0 ). If x0 is an isolated point
of dom(f ), we can only evaluate f at x0 , that is, we can only compute
f (x0 ), but the limit is not defined (intuitively, there is a neighborhood of
x0 containing no x ∈ dom(f ) but x0 , so f has no way to behave in any way
near x0 ).
58
4. LIMITS OF FUNCTIONS
In the following example, x0 ∈ dom(f ), however, if the limit of f at x0
is the beahvior of f near x0 then this can not agree with the value f (x0 ) of
f at x0 : let f : R → R
(
1 if x 6= 0
f (x) =
2 if x = 0
y
2
1
x
x0 = 0
then, since the behavior of f is to be constant equal to 1 in any neighborhood of 0, and so it approaches the value 1 when x is getting closer (or
tending) to 0 in dom(f ), it is reasonable to call 1 the limit of f for x tending
to 0, despite f (0) = 2. Similar considerations hold for f : R − {0} → R,
f (x) = 1: the function now is not even defined at 0, however, it is still
reasonable to say that 1 is the limit of f for x tending to 0, because f is
constant equal to 1 near 0, and so 1 is the value that f approaches when x
is tending to 0.
In some cases, we will see that the limit at x0 is indeed f (x0 ): we will
give a name (continuous functions) to such nice functions.
It may happen that near x0 the function does not approach any value,
as in the picture below
y
x
The picture shows the graph of the function sin x1 on the domain x > 0:
at 0, the function is not defined, but 0 is an accumulation point of the
4.2. THE DEFINITION OF LIMIT
59
domain {x > 0}; near it, the function f oscillates faster and faster and so it
does not get close to any real number.
The ideas we told about above are formalized in the following way.
Definition 4.2.1. Let f : A → R, x0 be an accumulation point of A
and L ∈ R. We say that the function f has limit L, or converges to
L, for x tending to x0
lim f = L
x→x0
if and only if for every neighborhood B(L, ) of L there exists a neighborhood B(x0 , δ) of x0 such that for every x ∈ B(x0 , δ) ∩ A − {x0 }
then f (x) ∈ B(L, ). We also write f (x) → L as x → x0 .
In a more compact form, we can rephrase the definition of limit as:
lim f = L ⇐⇒ ∀ > 0 ∃δ > 0 : ∀x ∈ B(x0 , δ) ∩ A − {x0 } ⇒ f (x) ∈ B(L, ),
x→x0
or, making explicit the inequalities defining neighborhoods,
lim f = L ⇐⇒ ∀ > 0 ∃δ > 0 : ∀x ∈ A, 0 < |x − x0 | < δ ⇒ |f (x) − L| < .
x→x0
y
L+
L
L−
x0
(x0 − δ, x0 + δ)
x
Example 4.2.2. Let’s check, using the definition of limit, that
1
lim
= 1.
x→0 x + 1
We pick > 0 and we have to find δ as in the definition above.
1
Observe that f (x) = x+1
∈ B(1, ) means
1
1
− 1 < ⇐⇒ − <
−1<
x+1
x+1
60
4. LIMITS OF FUNCTIONS
(
1
x+1
1
x+1
(
1−x−1−x−
x+1
1−x−1+x+
x+1
(
(1+)x+
x+1
(1−)x−
x+1
⇐⇒
⇐⇒
⇐⇒
−1<
− 1 > −
<0
>0
>0
< 0.
While the first inequality is easily solved after studying the sign of
the numerator and denominator, as for the second, because 1 − may
be negative, we have to distingush three cases

x < −1 or x > − 1+


1st inequality

−1 < x < 1−
if 0 < < 1
or
x
>
−1
if
>1
x
<
−

2nd inequality.

−1


x > −1
if = 1
Observe that for every > 0 we have 0 < < 1 + , so 0 < 1+
< 1,
and so −1 < − 1+ < 0; for 0 < < 1 we have 1 − > 0, hence
1− > 0; for > 1, then 0 < − 1 < , and so − −1 < −1. Therefore,
for x ∈ (− 1+
, 1−
) the system of inequalities above is fulfilled, thus,
if we choose δ as the minimum
n o
δ = min
,
,
1− 1+
, 1−
), we have f (x) ∈ (1 − , 1 + ).
then, for x ∈ (−δ, δ) ⊆ (− 1+
y
x = −1
0<<1
1+
1
1−
− 1+
x
1−
is 0
4.2. THE DEFINITION OF LIMIT
61
y
x = −1
1+
>1
1
− −1
x
− 1+
1−
is 0
The solution of the inequalities can be simplified after the following
remark. Since we have to find a δ > 0, we could assume from the
beginning to look for 0 < δ < 1, so that if x ∈ (−δ, δ) ⊆ (−1, 1), then
the denominator x + 1 > 0; in this way the system of inequalities is
equivalent to the simpler one
(
(1 + )x + > 0
(1 − )x − < 0
and so

 x > − 1+
−1 < x <

x > −1
1−
1st inequality
if 0 < < 1
2nd ineqaulity.
if ≥ 1
We conclude now with δ as found above.
Looking at the picture below, we can try visualize the definition of limit
limx→x0 f = L as follows: ”for > 0” tranlates into ”for every horizontal
strip we can choose in the Cartesian plane around the line y = L” (the pale
blue strip), ”there exists δ > 0” means ”we can find a vertical strip around
x = x0 ” (the thin yellow strip), the inequality ”0 < |x − x0 | < δ” implies
the inequality |f (x) − L| < ” reads as ”in the strip around x = x0 , but at
most for the line x = x0 itself, the graph of f all lie in the intersection with
the strip around y = L” (the green rectangle).
62
4. LIMITS OF FUNCTIONS
y
1+
L
1−
x0
x
Exercise 4.2.3. Prove that
lim x2 + 2 = 3.
x→−1
Let > 0 and let’s solve the inequality
|x2 + 2 − 3| < .
Recall that, according to the definition of limit, we have to find δ > 0
such that if −1 − δ < x < −1 + δ (but at most x = −1), then the
above inequality is true. Removing the absolute value, we obtain
1 − < x2 < 1 + .
(3)
√
√
(1) If < 1, then
< 1 + has solution − 1 + < x < √1 + ,
2 (note 1− > 0 since < 1) has solution x < − 1 − while 1−
<
x
√
√
or√x > 1 −√; therefore, √
equation (3) has solution −√ 1 + < x <
− 1 − or√ 1 − < x < 1 + , √
that is, −1 − (−1 + 1 + )
√< x <
−1 + (1 − 1 − ) or −1 − (−1 − 1 − ) < x < −1 + (1 + 1 + );
0
00
0
the latter
√ has not the form −1 − δ < x < −1 + δ , because δ =
−1 − 1 − is negative, and so it doesn’t contain any neighborhood
0
00
of −1; the√former can be written as
√ −1 − δ < x < −1 + δ with
0
00
δ = −1 + 1 + > 0 and δ = 1 − 1 − > 0 (since < 1); chosing
now δ = min{δ 0 , δ 00 }, for 0 < |x−(−1)| < δ, inequality (3) is satisfied.
(2) If ≥ 1, then 1 − ≤ 0, so 1 − √< x2 is verified
√ by every
x 6= 0, while x2√< 1 + has solution −√1 + < x < 1 + , that
is, −1 − (−1
√ 1 + ), therefore, choosing
√ + 1 + ) <√x < −1 + (1 +
δ = −1 + 1 + (note 1 + − 1 < 1 + + 1), we have that for
0 < |x − (−1)| < δ inequality (3) is fulfilled.
x2
4.2. THE DEFINITION OF LIMIT
63
Example 4.2.4. We prove that
1
@ lim sin .
x→0
x
We show that no L ∈ R can be the limit above, namely, we prove
that for every L ∈ R we are to find > 0 such that for every δ > 0
there exists x with 0 < |x − 0| < δ but with f (x) ∈
/ B(L, ).
If L 6= 0, then take = |L| (or less): note that is the distance
between L and the origin 0, so 0 ∈
/ B(L, ); this means that if f (x) =
sin x1 = 0 then f (x) ∈
/ B(L, ); but sin x1 = 0 exactly for x1 = πk,
1
, with k ∈ Z − {0}. Now, for
for every k ∈ Z, that is, for x = πk
1
1
1
every δ > 0, πk ∈ B(0, δ) if and only if | πk
| < δ, that is, |k| > πδ
; in
1
other words, for such k, we have found x = πk ∈ B(0, δ) − {0} with
f (x) ∈
/ B(L, ), as we wanted to show.
y
L
O
x
In the figure above, with L = 1/2 and = 0.3: we see that the
function leaves the strip around y = L in any neighborhood of of the
origin O.
It remains to consider the case L = 0. Take = 1 (or less), and
observe that in this way 1 ∈
/ B(L, ). So, if f (x) = sin x1 = 1 then
f (x) ∈
/ B(L, ); but sin x1 = 1 exactly for x1 = π2 + 2πk, for every
2
2
. For every δ > 0, π(1+4k)
∈ B(0, δ) if
k ∈ Z, that is, x = π(1+4k)
2
2−δπ
and only if | π(1+4k)
| < δ, that is, for k < − 2+δπ
4πδ or k > 4πδ ; in
1
other words, for such k, we have found x = πk
∈ B(0, δ) − {0} with
f (x) ∈
/ B(L, ), as we wanted to show.
64
4. LIMITS OF FUNCTIONS
y
O
x
In the figure above, with L = 0 and = 0.8: we see that the function
leaves the strip around the x-axis y = 0 in any neighborhood of of
the origin O.
Therefore, a limit might not exists, however, if it exists then it is unique.
Theorem 4.2.5. If f : A ⊆ R → R has limit for x → x0 , where x0 is
an accumulation point of A, then the limit is unique.
Proof. Suppose both L and L0 are limits of f as x → x0 , and L 6= L0 .
0|
0 0
Let = |L−L
2 . Note that B(L, ) ∩ B(L , ) = ∅. Then, by definition of
0
limit, there are δ > 0 and δ > 0 such that if x ∈ B(x0 , δ) ∩ A − {x0 } then
f (x) ∈ B(L, ), and if x ∈ B(x0 , δ 0 ) ∩ A − {x0 } then f (x) ∈ B(L0 , 0 ). Let
δ 00 = min{δ, δ 0 }, then for every x ∈ B(x0 , δ 00 ) ∩ A − {x0 } =
6 ∅ (it is not empty
because x0 is an accumulation point of A) we have f (x) ∈ B(L, )∩B(L, 0 ) =
∅, which is a contradiction because the empty set has no elements.
The argument in the proof above is portrayed in the figure below: in
this case δ 00 = δ, and for every x ∈ B(x0 , δ 00 ) the point (x, f (x)) of the graph
of the function should lie both in the light blue and orange strips, which
is not possible since they are disjoint because of the ch, this leading to a
contradiction:
4.2. THE DEFINITION OF LIMIT
L0 + 65
y
L0
L0 − = L + L
L−
x0 − δ
x0 −
δ0
x0 + δ
x0
x0 + δ 0
x
A variant of the definition of limit is that of right/left limit, that is, we
study the behaviour of a function in a right/left neighborhood of an accumulation point x0 , that is, (x0 , x0 + δ) and (x0 − δ, x0 ) respectively.
Definition 4.2.6. Let f : A → R, let x0 be an accumulation point of
A, let L ∈ R. We say that the function f has limit L, or converges
to L, for x tending to x0 from the right (or from above)
lim f = L
x→x+
0
if
∀ > 0 ∃δ > 0 : ∀x ∈ A, x0 < x < x0 + δ ⇒ |f (x) − L| < .
and similarly:
Definition 4.2.7. Let f : A → R, let x0 be an accumulation point of
A, let L ∈ R. We say that the function f has limit L, or converges
to L, for x tending to x0 from the left (or from below)
lim f = L
x→x−
0
if
∀ > 0 ∃δ > 0 : ∀x ∈ A, x0 − δ < x < x0 ⇒ |f (x) − L| < .
Example 4.2.8. As an example, consider the function whose graph
is as in the figure below
66
4. LIMITS OF FUNCTIONS
y
3
2
1
x
x0
then we see that the function has no limit for x → x0 , but it has right
and left limits
f (x0 ) = 2
@ lim f
x→x0
lim f = 1
x→x−
0
lim f = 3
x→x+
0
Note that if f has limit for x → x0 , then its right and left limits for
x → x0 exists and are equal, and conversely (as we have already remarked
many times, the value of the function at x0 , provided the function is defined
at x0 , is unrelated to the limit at x0 or to left or right limits).
We want to extend now the definition of limit in such a way to capture
other situations: for example, we want a notion of limit allowing us to
describe the behaviour of a function which is defined for larger and larger
positive (or negative) x; to be able to generalize the definition we have given
for limit for x → x0 , we need to generalize the concept of neighborhood.
Let’s add to R two new symbols +∞ and −∞
Definition 4.2.9.
e = R ∪ {+∞, −∞},
R
and let’s extend the natural order < on R to R̃ by setting
−∞ < x < +∞
for every real number x ∈ R.
We make the following conventions: if x is real
x
x
x + ∞ = +∞, x − ∞ = −∞,
=
= 0,
+∞
−∞
if x > 0 then x · (+∞) = +∞,
x · (−∞) = −∞,
if x < 0 then x · (+∞) = −∞,
x · (−∞) = +∞.
4.2. THE DEFINITION OF LIMIT
67
Definition 4.2.10. A neighborhood of +∞ is any set of the form
(M, +∞), for M ∈ R, similarly, a neighborhood of −∞ is any set of
the form (−∞, N ), for N ∈ R.
Translating into inequalities, x ∈ (M, +∞) if and only if x > M , similarly, x ∈ (−∞, N ) if and only if x < N .
−∞
N
M
+∞
Definition 4.2.11. A subset A ⊆ R is bounded if ∃x0 ∈ R and > 0
such that A ⊆ B(x0 , ).
That is, A is bounded if ∃M, N ∈ R such that N < x < M for every
x ∈ A. In an obvious way we can define the notions of bounded below and
bounded above.
We say A is unbounded if it is not bounded. We say that A is unbounded
from above if there is no M such that x < M ∀x ∈ A (equivalently, A
has nonempty intersection with any neighborhood of +∞), similarly, A is
unbounded from below if there is no N such that N < x ∀x ∈ A (equivalently,
A has nonempty intersection with any neighborhood of −∞).
For example, Z as a subset of R is unbounded, any interval of the form
(a, +∞), where a ∈ R, is unbounded from above but bounded from below,
clearly, any ball is bounded.
Therefore, it makes sense to talk about limit at +∞, respectively at
−∞, for functions whose domain is unbounded from above, respectively
unbounded from below.
With the notion of neighborhood of +∞ and −∞ we can now extend
the notion of limit in a natural way.
Definition 4.2.12. Let f : A → R, where A is unbounded from above,
and L ∈ R. We say that the function f has limit L, or converges to
L, for x tending to +∞
lim f = L
x→+∞
if and only if for every neighborhood B(L, ) of L there exists a neighborhood (M, +∞) of +∞ such that for every x ∈ (M, +∞) ∩ A then
f (x) ∈ B(L, ). We also write f (x) → L as x → +∞.
68
4. LIMITS OF FUNCTIONS
In a more compact form, we can rephrase the definition of limit as:
lim f = L ⇐⇒ ∀ > 0 ∃M : ∀x ∈ (M, +∞) ∩ A ⇒ f (x) ∈ B(L, ),
x→+∞
or, making explicit the inequalities defining neighborhoods,
lim f = L ⇐⇒ ∀ > 0 ∃δ > 0 : ∀x ∈ A, x > M, ⇒ |f (x) − L| < .
x→+∞
Definition 4.2.13. Let f : A → R, where A is unbounded from below,
and L ∈ R. We say that the function f has limit L, or converges to
L, for x tending to −∞
lim f = L
x→−∞
if and only if for every neighborhood B(L, ) of L there exists a neighborhood (−∞, N ) of −∞ such that for every x ∈ (−∞, N ) ∩ A then
f (x) ∈ B(L, ). We also write f (x) → L as x → −∞.
In a more compact form, we can rephrase the definition of limit as:
lim f = L ⇐⇒ ∀ > 0 ∃N : ∀x ∈ (−∞, N ) ∩ A ⇒ f (x) ∈ B(L, ),
x→−∞
or, making explicit the inequalities defining neighborhoods,
lim f = L ⇐⇒ ∀ > 0 ∃δ > 0 : ∀x ∈ A, x < M, ⇒ |f (x) − L| < .
x→−∞
y
L+
L
L−
y=L
N
M
L0 + L0
L0 − y = L0
Definition 4.2.14. If
lim f = L,
x→+∞
lim f = L0
x→−∞
x
4.2. THE DEFINITION OF LIMIT
69
we say that f has a horizontal asymptote at +∞, respectively at −∞,
and that the line y = L, respectively y = L0 is a horizontal asymptote
for f (see the figure above).
By end-behavior of a function f we refer to the behaviour of a function
as x approaches +∞ or −∞, that is,
lim f and
x→+∞
lim f.
x→−∞
We formalize now the idea of a function being unbounded near an accumulation point of its domain in the following way.
Definition 4.2.15. Let f : A → R, let x0 be an accumulation point
of A. We say that the function f has limit +∞ for x tending to x0
lim f = +∞
x→x0
if and only if for every neighborhood (M, +∞) of +∞ there exists a
neighborhood B(x0 , δ) of x0 such that for every x ∈ B(x0 , δ)∩A−{x0 }
then f (x) ∈ (M, +∞). We also write f (x) → +∞ as x → x0 .
In a more compact form
lim f = +∞ ⇐⇒
x→x0
∀M ∃δ > 0 : ∀x ∈ B(x0 , δ) ∩ A − {x0 } ⇒ f (x) ∈ (M, +∞),
or
lim f = +∞ ⇐⇒
x→x0
∀M ∃δ > 0 : ∀x ∈ A, 0 < |x − x0 | < δ ⇒ f (x) > M.
70
4. LIMITS OF FUNCTIONS
y
x = −1
M
−1 − δ
x
−1 + δ
−1
In a similar way we define f → −∞ for x → x0 :
lim f = −∞ ⇐⇒
x→x0
∀N ∃δ > 0 : ∀x ∈ B(x0 , δ) ∩ A − {x0 } ⇒ f (x) ∈ (−∞, N ),
or
lim f = +∞ ⇐⇒
x→x0
∀M ∃δ > 0 : ∀x ∈ A, 0 < |x − x0 | < δ ⇒ f (x) < N.
The definition of right/left limits when these are infinite is straightforward.
Definition 4.2.16. If any of the following limits exists at x0 and are
as follow
lim f = +∞
x→x+
0
lim f = −∞
x→x+
0
lim f = +∞
x→x−
0
lim f = −∞ :
x→x−
0
then we say that the f has a vertical asymptote at x0 and that the
line x = x0 is a vertical asymptote for f .
For the function f shown in the previous figure, the line x = −1 is a
vertical asymptote for f .
Exercise 4.2.17. Verify
1
= +∞.
x→0 x2
lim
4.2. THE DEFINITION OF LIMIT
71
By definition of limit, given M we have to solve x12 > M and show
there is δ > 0, such that −δ < x < δ (but at most x = 0) is solution
of the previous inequality.
(1) If M ≤ 0, every x fulfils x12 > 0 > M , so any δ > 0 works.
2
(2) If M > 0, then,
q since x >
q 0 for all x but 0, we have to solve
x2 <
1
M,
finding −
1
M
<x<
1
M;
thus we can take δ =
1
M.
Finally, we define infinite limits on unbounded domains.
Definition 4.2.18. Let f : A → R, A unbounded from above or from
below depending on whether the limit is for x → +∞ or x → −∞:
lim f = +∞ ⇐⇒ ∀K ∃M : ∀x ∈ A, x > M ⇒ f (x) > K
x→+∞
lim f = −∞ ⇐⇒ ∀K ∃M : ∀x ∈ A, x > M ⇒ f (x) < K
x→+∞
lim f = +∞ ⇐⇒ ∀K ∃M : ∀x ∈ A, x < M ⇒ f (x) > K
x→−∞
lim f = −∞ ⇐⇒ ∀K ∃M : ∀x ∈ A, x < M ⇒ f (x) < K.
x→−∞
y
K
M0
M
x
K0
lim f = +∞,
x→+∞
lim f = −∞
x→−∞
Example 4.2.19. The function ax has the x-axis as horizontal asymptote for x → −∞ when a > 1, and for x → +∞ when 0 < a < 1.
The function loga x has the y-axis as vertical asymptote.
The function tan x has the lines x = π2 + kπ, k ∈ Z, has vertical
asymptotes.
The function cot x has the lines x = kπ, k ∈ Z, has vertical asymptotes.
72
4. LIMITS OF FUNCTIONS
The functions sin x and cos x have no limit for both x → +∞ and
x → −∞ (compare with the proof that sin x1 has no limit for x → 0),
therefore they have no horizontal asymptotes.
The function csc x has the lines x = π2 + kπ, k ∈ Z, has vertical
asymptotes.
The function sec x has the lines x = kπ, k ∈ Z, has vertical asymptotes.
We will see more examples once we will have learnt how to compute
limits.
4.3. Computing limits
To compute limits, we will begin with the following strategy based on
the two ingredients:
• to compute limits of some special and simple functions, verifying
them by means of the definition of limit,
• and to give criteria to compute limits of functions constructed from
these simple functions.
For example, suppose we know how to compute the limits of the functions
f (x) = x and g(x) = sin x, and suppose also we have a rule to compute the
limits of additions of functions, then we will be able to compute the limit of
x + sin x.
As for the second ingredient, the next theorem tells us how to compute
limits of sum, products and quotient of functions knowing the limits of the
e but for some ”special cases”.
addenda or factors, using the algebra on R,
Theorem 4.3.1. Let f, g : A ⊆ R → R and x0 be an accumulation
point of A, or +∞, −∞ if A is unbounded from above, respectively
from below. If
e
lim f = L,
lim g = L0 ,
L, L0 ∈ R,
x→x0
x→x0
then
lim (f + g) = L + L0 ,
x→x0
lim (f · g) = L · L0 ,
x→x0
lim
x→x0
f
L
= 0,
g
L
g 6= 0,
e
where the addition and multiplication of limits are computed in R
according to the convention we made, and provided the forms below
4.3. COMPUTING LIMITS
do not occur
∞−∞
0·∞
0
0
73
∞
.
∞
Proof. We check just the addition formula in the simplest case x0 ∈ R
and L, L0 ∈ R. Take > 0: by definition of limit there are δ and δ 0 such
that for every x ∈ A with 0 < |x − x0 | < δ we have |f (x) − L| < 2 , and for
every x ∈ A with 0 < |x − x0 | < δ 0 we have |g(x) − L0 | < 2 . Then, setting
δ 00 = min{δ, δ 0 }, for x ∈ A with 0 < |x − x0 | < δ 00 , we have
|(f + g)(x) − (L + L0 )| = |(f (x) − L) + (g(x) − L0 )|
≤ |g(x) − L| + |g(x) − L0 |
< +
2 2
= ,
but this means
lim (f + g) = L + L0 .
x→x0
Corollary 4.3.2. The theorem above holds, mutatis mutandis, for
right and left limits.
Definition 4.3.3. The expressions
∞−∞
0·∞
0
0
∞
∞
are called indeterminate forms.
When indeterminate forms occur, the theorem above doesn’t tell us anything (there are three further indeterminate form that we will meet later).
Intuitively, what goes wrong with them is that two functions may converge with different speeds: x2 grows much faster than x, when x becomes
larger and larger, that is, they both tend to +∞ when x → +∞ but x2 does
so faster than what x does, thus, when studying the limit of the difference
x2 − x for x → +∞, the indeterminate form ∞ − ∞ occurs, however, as x2
wins over x, then x2 − x → +∞ when x → +∞.
Note that, instead, the functions x + 1 and x both tend to +∞ when
x → +∞, so if computing the limit of their difference (x + 1) − x using
the theorem above, we would get the indeterminate form ∞ − ∞, but the
difference (x + 1) − x is just the constant function 1, so it converges to 1.
74
4. LIMITS OF FUNCTIONS
All this can be formalized, and when an indeterminate form appears,
the problem will be to establish which of two functions is converging faster.
The next theorem, known as the ”two policemen theorem” or ”prisoner
function theorem” or ”squeeze theorem” or ”sandwich theorem”, is another
example where bounding, and so controlling, a given function by means of
two other functions whose behavior is known, can provide information on
the given function.
Theorem 4.3.4. Let f, g, h : A ⊆ R → R, and x0 an accumulation
point of A. Suppose that
lim f = lim h = L ∈ R,
x→x0
x→x0
and in some neighborhood of x0 , we have
f (x) ≤ g(x) ≤ h(x).
Then also
lim g = L.
x→x0
y
h
x
g
f
Proof. Let > 0, then there are δ, δ 0 > 0 such that for x ∈ A such that
0 < |x − x0 | < δ ⇒ L − < f (x) < L + 0 < |x − x0 | < δ 0 ⇒ L − < h(x) < L + .
4.3. COMPUTING LIMITS
75
Then if δ 00 is the radius of a neighborhood of x0 where f (x) ≤ g(x) ≤ h(x), by
setting δ 000 = min{δ, δ 0 , δ 00 }, we have that for x ∈ A such that 0 < |x − x0 | <
δ 000
L − < f (x) ≤ g(x) ≤ h(x) < L + ,
that is,
lim g = L.
x→x0
Variants of this theorem hold when x0 is +∞ or −∞, for right and left
limits, when limits are infinite, and for combinations of these. We state one
and we draw another one of such cases.
Theorem 4.3.5. Let f, g, : A ⊆ R → R, and x0 an accumulation
point of A. Suppose that in some neighborhood of x0
f (x) ≤ g(x).
Then
lim f = +∞ ⇒ lim g = +∞,
x→x0
x→x0
lim g = −∞ ⇒ lim f = −∞,
x→x0
x→x0
g
y
f
x
We now begin verifying some ”elementary” limits using the definition,
then to compute others using the theorems above.
Example 4.3.6. Let x0 ∈ R, then the limit of a constant function
with value c ∈ R is c itself:
lim c = c.
x→x0
76
4. LIMITS OF FUNCTIONS
In fact, given > 0, the inequality |c − c| = 0 < holds for every x
and so any δ > 0 solves the problem.
This mean, for example, that
lim 4 = 4.
x→x0
Example 4.3.7. Let x0 ∈ R, then
lim x = x0 .
x→x0
Indeed, let > 0 and let’s solve |x − x0 | < : this means to look for
δ > 0 such that 0 < |x − x0 | < δ implies 0 < |x − x0 | < ; clearly,
taking δ = solves the problem.
So, for example, we have
lim x = 3.
x→3
Example 4.3.8. As consequence of example 4.3.6 and the product
formula 4.3.1, and under the same hypothesis of 4.3.1, for every constant c ∈ R,
lim cf = c lim f.
x→x0
x→x0
In practice, with f (x) = x and c = −1, we have
lim −x = − lim x = −x0 .
x→x0
x→x0
Example 4.3.9. If n ∈ N, x0 ∈ R, then
lim xn = xn0
x→x0
The formula follows from the product formula for limits 4.3.1 and the
example 4 above:
n
lim xn = lim x = xn0 .
x→x0
x→x0
For a specific example
lim x3 =
x→2
lim x
x→2
3
= 23 = 8.
4.3. COMPUTING LIMITS
77
Example 4.3.10.
lim |x| = |x0 |.
x→x0
In fact, if x0 6= 0, then, in any neighborhood of x0 of radius δ < |x0 |,
we have |x| = x if x0 > 0 and |x| = −x if x0 < 0, and so we can apply
what we know about the function x from example 4 or from example
4.3.8 respectively.
If x0 = 0, then, for > 0 we have
|x| − |0| = |x| = |x|,
so, choosing δ = , if |x| < δ = , we have
|x| − |0| < .
Thus, for example,
lim |x| = | − 4| = 4.
x→−4
Example 4.3.11. Let P (x) be a polynomial, then
lim P = P (x0 ).
x→x0
Indeed, if P (x) = an xn +an−1 xn−1 +...+a1 x+a0 , then by the addition
formula for limits 4.3.1 and by combining examples 4.3.8 and 5
lim P = an lim xn + an−1 lim xn−1 + ... + a1 lim x + lim a0
x→x0
=
x→x0
x→x0
n−1
n
an x0 + an−1 x0 + ... +
x→x0
x→x0
a1 x0 + a0
= P (x0 ).
As a specific example
lim (2x4 + 3x3 − 2x − 1) = 2(−2)4 + 3(−2)3 − 2(−2) − 1
x→−2
= 32 − 24 + 4 − 1
= 11.
Example 4.3.12. Let x0 ∈ R, x0 =
6 0, then
1
1
lim
=
.
x→x0 x
x0
Use the quotient formula for limits:
1
1
1
lim
=
=
.
x→x0 x
limx→x0 x
x0
78
4. LIMITS OF FUNCTIONS
And so, for example,
lim
x→3
1
1
= .
x
3
Example 4.3.13.
1
= +∞.
x
Given M we have to find δ > 0 such that if 0 < x − 0 = x < δ then
1
x > M : this inequality is satisfied for every x when M < 0, and so
δ can be anything; instead, when M > 0, the inequality has solution
1
1
x< M
, so it is enough to choose δ = M
.
lim
x→0+
Example 4.3.14. Similarly
lim
x→0−
1
= −∞.
x
Note that we have proved that the function
at 0.
1
x
has a vertical asymptote
y
x
Example 4.3.15. By the theorem on quotient of limits 4.3.1 and the
e we quickly find out that
conventions for R,
1
1
lim
= 0 = lim
,
x→+∞ x
x→−∞ x
That is, 1/x has the x-axis as horizontal asymptote, both at +∞ and at
−∞.
4.3. COMPUTING LIMITS
Example 4.3.16. More generally, suppose two functions f and g are
defined in a neighborhood of 0, g ≥ 0, 0 is an accumulation point of
the domain of fg , and
lim f = L ∈ R>0 ,
x→0
lim g = 0,
x→0
then
f
= +∞.
g
To verify it, given M > 0 we have to find δ > 0 such that, for
x ∈ dom fg with 0 < |x − 0| = |x| < δ, then fg > M . So, let’s apply
the definition of limit to f and g: there are δf , δg > 0 such that for
x ∈ dom(f ), 0 < |x| < δf then |f (x) − L| < L2 (we have taken = L2 ),
and for for x ∈ dom(g), 0 < |x| < δg then |g(x)−0| = |g(x)| = g(x) <
L
2M . Then, setting δ = min{δf , δg }, for 0 < |x − 0| < δ, observing
1
that f (x) > L − L2 = L2 and g(x)
> 2M
L ,
lim
x→0
f (x)
L 2M
>
= M.
g(x)
2 L
As a concrete example
x+1
= +∞.
x2
Note that g(x) = x2 ≥ 0; if this condition is not fulfilled, for example,
with g(x) = x, then the limits does not exsist, though for g(x) = x
both right and left limits exist.
lim
x→0
Remark 4.3.17. If also f → 0 when x → 0, then the indeterminate
form 00 appears, and in general we can’t say anything about the limit
of fg , actually, everything can happen: if g(x) = x3 and f1 (x) = x4 ,
f2 (x) = Lx3 , where L ∈ R − {0}, f3 (x) = x2 , f4 (x) = x then the
quotients fg1 , fg2 , fg3 , fg4 are defined on R − {0}, they all rpesent the
indeterminate form 00 , but
f1
x→0 g
f2
lim
x→0 g
f3
lim
x→0 g
f4
lim
x→0 g
lim
x4
= lim x = 0
x→0 x3
x→0
3
Lx
= lim 3 = lim L = L
x→0 x
x→0
2
x
1
= lim 3 = lim = @
x→0 x
x→0 x
x
1
= lim 3 = lim 2 = +∞,
x→0 x
x→0 x
= lim
79
80
4. LIMITS OF FUNCTIONS
so, when the indeterminate form 00 occurs, everything can then happen for the limit of the quotient fg .
Example 4.3.18.
lim x = +∞
x→+∞
lim x = −∞.
x→−∞
To verify the limits, given M > 0 we have to find N > 0 such that if
x > N then x > M : it is trivial, and it suffices to take N = M
Example 4.3.19. Using the product formula for limits 4.3.1 together
with the example 7, we deduce, for every n ∈ N, n 6= 0,
1
lim xn = +∞,
lim
= 0.
x→+∞
x→+∞ xn
Example 4.3.20. Let P (x) = an xn + an−1 xn−1 + ... + a1 x + a0 be a
polynomial (an 6= 0), then
(
+∞ if an > 0
lim P =
x→+∞
−∞ if an < 0
(
+∞ if an > 0 and n even, or an < 0 and n odd
lim P =
x→−∞
−∞ if an < 0 and n even, or an > 0 and n odd.
Indeed, writing
1
1
1 P (x) = xn an + an−1 + ... + a1 n−1 + a0 n
x
x
x
first we note that xn → +∞ and that the second factor converges
to an 6= 0 for x → +∞, then the claim follows from the theorem on
product of limits 4.3.1.
So, for example, we have
lim x2 − x + 7 = +∞
x→+∞
lim −x2 − x + 7 = −∞
x→+∞
lim x2 − x + 7 = +∞
x→−∞
(∞ − ∞)
4.3. COMPUTING LIMITS
81
lim −x2 − x + 7 = −∞
(∞ − ∞)
x2 − x + 7 = +∞
(∞ − ∞)
x→−∞
lim x3 −
x→+∞
lim x3 −
x→−∞
lim −x3 −
x→+∞
lim −x3 −
x→−∞
x2 − x + 7 = −∞
x2 − x + 7 = −∞
x2 − x + 7 = +∞
(∞ − ∞).
Note that in some of these examples, as remarked, the indeterminate
form ∞ − ∞ occurs.
Remark 4.3.21. Example 4.3.20 says that the limit of a polynomial
at +∞ or at −∞ is controlled by its monomial of highest degree, and
is determined by degree and coefficient of such monomial; intuitively,
the monomial of highest degree is the one diverging faster to infinite,
and so it ”wins” over the monomials of lower degree.
The next two examples show how to deal with limits of rational functions.
Example 4.3.22. Consider the limit
x3 − x
lim 2
x→+∞ x + 2x + 5
if we compute the limits of the numerator and of the numerator,
according to what said in example 4.3.20, we get in both case +∞,
so the theorem on quotient of limits can’t be applied because of the
∞
. To overpass it, we can factor the monomial
indeterminate form ∞
of highest degree out at both numerator and denominator:
x 1 − x12
x3 1 − x12
+∞(1 − 0)
= lim
lim
=
= +∞
5
2
x→+∞ 1 +
x→+∞ 2
1+0+0
x 1+ 2 + 5
x + x2
x
x2
in fact, the indeterminate form has disappeared, so we can apply the
theorem 4.3.1 and conclude the limit is +∞. Note also that the division x does not make any problem as we can work in a neighborhood
of +∞ not containing 0.
82
4. LIMITS OF FUNCTIONS
Example 4.3.23. Consider now the limit
x−1
lim 2
x→1 x + 2x − 3
trying to apply directly theorem 4.3.1 leads to the indeterminate form
0 , however, by factoring the denominator (there must be a factor x−1
for the polynomial to vanish at 1),
x−1
x−1
lim
= lim
x→1 (x − 1)(x + 3)
x→1 x2 + 2x − 3
1
= lim
x→1 x + 3
1
=
4
In this way, the indeterminate form has disappeared and we could
apply theorem 4.3.1. As in the previous example, we can divide by
x−1 because the limit has nothing the value of the function at x0 = 1.
A generalization of the above to non integer exponents is left below as
an exercise.
Exercise 4.3.24. Consider the function f defined by f (x) = (x−a)α ,
with a, α ∈ R, defined on {x ∈ R | x > a}. Then


+∞ if α > 0
α
lim (x − a) = 1
if α = 0
x→+∞


0
if α < 0
and


if α > 0
0
α
lim (x − a) = 1
if α = 0

x→a+

+∞ if α < 0
Example 4.3.25. As an application of exercise 4.3.24, we have that
√
1
lim x = lim x 2 = 0
x→0+
lim
x→+∞
√
x→0+
1
x = lim x 2 = +∞
x→+∞
4.3. COMPUTING LIMITS
83
where, with the notation of exercise 4.3.24, a = 0 and α =
and
1
1
lim √ = lim x− 2 = +∞
+
+
x x→0
x→0
1
1
lim √ = lim x− 2 = 0
x→+∞
x x→+∞
1
2
> 0;
where a = 0 and α = − 12 < 0
Example 4.3.26.
lim sin x = 0,
lim cos x = 1
x→0
x→0
Suppose x > 0. Observe that 0 ≤ sin x < x: in fact, from the picture
below, showing the circle with center the origin and radius 1, since
d
sin x = P Q and x is the length of the arc P
R, then in the right
triangle 4P QR, the side P Q is shorter than the hypothenuse P R,
d
which is shorter than the corresponding arc P
Q. Since x and the
constant function 0 both converge to 0, as soon as x → 0, by the
prisoner function theorem 4.3.4, we deduce
lim sin x = 0.
x→0+
Since sin(−x) = − sin x, we deduce that also
lim sin x = 0,
x→0−
and so the conclusion.
x
T
P
x
O
Q R
x
84
4. LIMITS OF FUNCTIONS
As for cos x, observe first that since for 0 < x < π/2 we have 0 <
cos x < 1, thus cos2 x < cosx (and similarly, sin2 x < sin x), so
0 < 1 − cos x < 1 − cos2 x = sin2 x < sin x,
and now the prisoner function theorem 4.3.4 implies 1 − cos x → 0
for x → 0, that is, cos x → 1 for x → 0.
Example 4.3.27.
sin x
= 1.
x
If we try to apply the theorem 4.3.1, we find the indeterminate form
0
0 , and so we can’t conclude anything with that. Rather, let’s look
again to the figure in example 4.3.26 (we can assume x positive in
the right neighborhood [0, π2 ) of 0 and verify the right limit is 1, then,
since the function is even, the left limit must also be 1): observe that
T R = tan x, the area of the triangle 4ORP is OR · P Q/2 = 21 sin x,
the area of the triangle 4ORT is OR · T R/2 = 21 tan x, and the area
d
of the slice of circle determined by the arc P
R is 12 x, so we have
lim
x→0
1
1
1 sin x
1
sin x < x < tan x =
,
2
2
2
2 cos x
from which
1<
x
1
<
,
sin x
cos x
and so
sin x
< 1.
x
Since cos x (as proved in example 4.3.26) and 1 both converge to 1
when x → 0, by the two policemen theorem 4.3.4 sinx x also converges
to 1 when x → 0.
cos x <
Example 4.3.28.
lim sin x = sin x0
x→x0
Let’s use the following trigonometric formula
a+b
a−b
sin a − sin b = 2 cos
sin
.
2
2
Then
x − x0
x + x0
| sin x − sin x0 | = 2 cos
sin
2
2
x + x0
x − x0
= 2 cos
sin
2
2
4.3. COMPUTING LIMITS
85
x − x0
2
x − x0
≤2
2
= |x − x0 |,
≤ 2 sin
so, given > 0, choosing δ = , then |x − x0 | < δ = implies
| sin x − sin x0 | < .
The behavior of the function sin x as explained in example 4.3.28 is
shared also by the other trigonometric functions: namely, if the function is
defined at x0 , then the limit for x → x0 is computed by evaluation at x0 .
Note the vertical asymptote of tan x and sec x at π2 + kπ, k ∈ Z, and of cot x
and csc x at kπ for k ∈ Z: indeed, by example 4.3.16, these functions have
right and left limits equal to either +∞ or −∞ at such points.
x=
π
2
x=
3π
2
x
y = tan x
x = − π2
Example 4.3.29. We summarize the situation, as for limits, of exponetial and logarithmic functions:
lim ax = ax0
x→x0
lim loga x = loga x0 .
x→x0
where x0 ∈ R in the case of the exponential function, and x0 ∈ R>0
in the case of the logarithm; intuitively, from their graph we see also
(
+∞ if a > 1
lim ax =
x→+∞
0
if 0 < a < 1
(
0
if a > 1
lim ax =
x→−∞
+∞ if 0 < a < 1
86
4. LIMITS OF FUNCTIONS
(
+∞
lim loga x =
x→+∞
−∞
(
+∞
lim loga x =
x→0+
−∞
if a > 1
if 0 < a < 1
if a > 1
if 0 < a < 1.
Therefore, expoential functions ax have the x-axis y = 0 as horizontal
asymptote, more precisely, at +∞ for a > 0 and at −∞ for a < 0,
while logarithms loga x have the y-axis x = 0 as vertical asymptote,
with limit +∞ or −∞ according if a > 0 or a < 0.
Intuitively, if a function has a line as horizontal or vertical asymptote
means the the graph of the function in the Cartesian plane is approaching
that line and touching it eventually ”at infinity”. This intuition can be made
rigorous, however, it will not be done in these notes. Also lines other than
horizontal and vertical ones may be asymptotes to graphs of functions (of
course, functions, in general, have no asymptotes).
Definition 4.3.30. Let f : (a, +∞) → R (or f : (−∞, a) → R). We
say that the line y = mx + q is a slant asymptote to the graph of f
for x → +∞ (or for x → −∞) if
lim (f (x) − mx − q) = 0
x→+∞
(or
lim (f (x) − mx − q) = 0)
x→−∞
Example 4.3.31. The function f : R − {0} → R
1
f (x) = x −
x
has the line y = x as slant asymptote for both x → +∞ and x → −∞:
in fact
1
lim x − − x = 0
x→±∞
x
4.4. CONTINUITY
y
87
y=x
x
Proposition 4.3.32. Let f : (a, +∞) → R. The line y = mx + q is
a slant asymptote to the graph of f for x → +∞ if and only if
f (x)
(1) ∃ lim
= m ∈ R,
x→+∞ x
(2) ∃ lim (f (x) − mx) = q ∈ R.
x→+∞
A similar result holds for slant asymptotes for x → −∞.
In the previous example with f (x) = x − 1/x, we have
x−
x→±∞
x
m = lim
1
x
= lim 1 −
x→±∞
1
=1
x2
and
1
− x) = 0,
x
so, substituting into y = mx + q, we find f has slant asymptote of equation
y = x at both +∞ and at −∞.
q = − lim (x −
x→±∞
4.4. Continuity
When x0 is both an element and an accumulation point of the domain of
a function f , having as limit for x → x0 the value f (x0 ) of the function f at
x0 is what characterizes functions which are continuous at x0 . Intuitively, a
88
4. LIMITS OF FUNCTIONS
function which is continuous has a graph which may have ”kinks” but not
”jumps”
Definition 4.4.1. A function f : A ⊆ R → R is continuous at
x0 ∈ A if and only if for every neighborhood B(f (x0 ), ) of f (x0 )
there is a neighborhood B(x0 , δ) of x0 such that for x ∈ B(x0 , δ) ∩ A
then f (x) ∈ B(f (x0 ), ).
We say that f is continuous if it is continuous at every point x0 ∈ A.
Equivalently, f is continuous at x0 if and only if
∀ > 0 ∃δ > 0 : x ∈ A, |x − x0 | < δ ⇒ |f (x) − f (x0 )| < .
Observe that, unlike the definition of limit, x0 is required to be a point
of A while it is not required to be an accumulation point of A. Note also
that the inequalty |x − x0 | < δ in the definition of continuity replaces 0 <
|x − x0 | < δ in the definition of limit: that is, unlike limits, continuity ”has
to do with what the function does at x0 ”. In fact, while the limit of f is not
defined at isolated points, a function is trivially continuous at any isolated
point.
As anticipated at the beginning of this ection, computing limits at points
where a function is continuous is easy:
Theorem 4.4.2. If x0 ∈ A is an accumulation point of the domain
of f , then f is continuous at x0 if and only if
lim f = f (x0 ).
x→x0
So, if a function is continuous at an accumulation point x0 of the domain
of f , then the limit of f for x → x0 is computed by evaluating the function
at x0 .
Example 4.4.3. The computations of limits done in section 4.3 prove
that polynomials, rational functions, trigonometric functions, exponential and logarithmic functions are continuous.
4.4. CONTINUITY
89
Remark 4.4.4. The computations of limits done in section 4.3 also
prove that constructions done with continuous functions, like addition, multiplication, division when defined, of continuous functions,
yield continuous functions.
Another interesting property of continuous functions is that it is easy to
compute the limits of compositions.
Theorem 4.4.5. Let f : A ⊆ R → R and g : B ⊆ R → R be such
that g ◦ f exists,
∃ lim f = y ∈ B,
x→x0
and g is continuous at y and y is an accumulation point of B, then
lim g(f (x)) = g lim f .
x→x0
x→x0
We deduce that composition of continuous functions is continuous:
Corollary 4.4.6. Consider f : A ⊆ R → R, g : B ⊆ R → R
and x0 ∈ A such that f is continuous at x0 and g is continuous at
f (x0 ) ∈ B, then g ◦ f is continuous at x0 .
Let’s see an application of the theorem about limits of compositions of
functions:
Example 4.4.7.
(1)
(2)
1
1 lim log 1 −
= log
lim 1 −
= 0,
x→+∞
x→+∞
x
x
1
lim e1− x = e
limx→+∞
1− x1
x→+∞
= e.
However, let sgn : R → R be the sign function
(
|x|
, if x 6= 0
sgn(x) = x
0,
if x = 0,
then
1
lim sgn x2 sin
x→0
x
90
4. LIMITS OF FUNCTIONS
does not exist; indeed, note that the function sgn is not continuous
at 0
y
1
x
−1
Intuitively, the function x2 sin x1 , whose graph is shown after theorem
4.3.4, oscillates about the x-axis near 0, and so its sign changes in
any neighborhood of 0.
Even when f and g are not continuous, we can say something about
the limit of the composition, under some hypotheses as stated below, particularly, that f ”does not oscillate too much about its limit” as in the
pathological case in example 4.4.7. We will use frequently the theorem below, which will allow ”to substitute”, in a suitable sense, when computing
limits.
Theorem 4.4.8. Let f : A ⊆ R → R and g : B ⊆ R → R be such
that g ◦ f exists,
∃ lim f = y ∈ B and
∃ lim g,
x→x0
t→y
and either there is a neighborhood N (y) of y in B where g(t) 6= y for
every t ∈ N (y) − {y}, then
lim g(f (x)) = lim g.
x→x0
t→y
Example 4.4.9. Here is a typical application of theorem 4.4.8: to
compute
sin x1
lim
1 .
x→+∞
x
4.4. CONTINUITY
91
We may set x1 = t, observe that t → 0 as x → +∞, note that the
hypothesis of theorem 4.4.8 are satisfied, and so compute
lim
sin x1
x→+∞
1
x
sin t
=1
x→0 t
= lim
by means of 4.3.27.
For a counterexample, see again the pathological case in example
4.4.7.
When computing limits involving exponential functions we may meet
the other indeterminate forms (compare with definition 4.3.3).
Definition 4.4.10. The expressions
1∞ ,
00 ,
∞0
together with those in definition 4.3.3, are called indeterminate forms.
The first of these occurs in the following limit: one can prove that,
nonetheless, the limit exists and is finite; such limit is the definition of the
number e, which we didn’t give when we defined exponential functions.
Definition 4.4.11.
e = lim
x→+∞
1+
1 x
.
x
Another reason which makes continous functions very interesting is that
they admit maxima and minima on reasonable sets.
Definition 4.4.12. Let f : A ⊆ R → R. We say that f is bounded
(bounded from above, bounded from below) if its image is bounded
(bounded from above, bounded from below).
Definition 4.4.13. We say that p ∈ A is a global (or absolute) maximum point for f if f (x) ≤ f (p) ∀x ∈ A, it is a global (or absolute)
minimum point if f (x) ≥ f (p) ∀x ∈ A. In this case, we say that f (p)
is a global (or absolute) maximum, respectively minimum, for f .
92
4. LIMITS OF FUNCTIONS
Definition 4.4.14. We say that p ∈ A is a local (or relative) maximum point for f if there is a neighborhood B(p, δ) of p such that
f (x) ≤ f (p) ∀x ∈ B(p, δ) ∩ A, it is a local (or relative) minimum
point if there is a neighborhood B(p, δ) of p such that f (x) ≥ f (p)
∀x ∈ B(p, δ) ∩ A. In this case, we say that f (p) is a local (or relative)
maximum, respectively minimum, for f .
In other words, f has a global maximum, respectively a global minimum,
when restricted to B(p, δ) ∩ A.
y
x
The function on the left is bounded from above but not from below, has
a global maximum but no global or local minima, while the function on the
right is bounded from below but not from above, has a global minumum but
no global or local maxima.
y
x
4.4. CONTINUITY
93
This function is not bounded, both from above or below, has a local
maximum and a local minimum, no global maximum and no global minimum.
Note that if a function has a global maximum (global minimum) then
it is bounded from above (bounded from below), the converse is not true:
exponential function ax are all bounded from below, but have no global or
relative minima (in the picture the case a > 1).
y
x
From the previous examples, we realize that existence of local/global
maxima/monima depends also on the domain. We would like to have some
condition on the domain which ensures that continuous functions defined on
it admit maxima and minima.
Definition 4.4.15. A subset A of R is called compact if it is bounded
and contains all of its accumulation points.
Example 4.4.16.
(1) The canonical example of a compact subset of R is a closed interval [a, b], with a, b ∈ R.
(2) Finite unions of closed intervals or any finite set are other
examples of compact subsets.
(3) As R is not bounded, that is, it is not contained in any ball
B(x, ), then R is not compact.
(4) The Cantor set (example 4.1.9) is compact.
Theorem 4.4.17 (Weierstrass). If f : A ⊆ R → R is continuous and
A is compact, then f has a global maximum and a global minimum.
94
4. LIMITS OF FUNCTIONS
Though it may look obvious and easy to state, the Weierstrass theorem
is a deep result and its proof is not easy.
y
x
The conditions in the theorem are all necessary, that is, without them
the conclusion may fail.
(1) If A is not bounded, for example A = R, and so it is not compact,
a continuous function might not have have a global maximum and a global
minimum: for example, any exponential functions ax .
(2) The interval (−π/2, π/2) does not contain its accumulation points
π/2 and π/2, so it is not compact, and indeed, a continuous function like
tan x has no global maximum and no global minimum on it.
(3) Extending tan x to a function f on the compact set [−π/2, π/2] by
setting
(
tan x, if x ∈ − π2 , π2
f (x) =
0,
if x = − π2 , π2 ,
yet f has no absolute maximum or absolute minimum, indeed f is not
continuous.
4.5. More computations of limits
Limits are hard to compute when indeterminate forms occur. In this
section we provide other results usefull to overpass indeterminate forms.
We will often apply theorems 4.4.5 and 4.4.8 in the following computations.
Example 4.5.1. For every a, b, c > 0
eax
lim
= +∞
x→+∞ xb
xb
lim
= +∞
x→+∞ (log x)c
4.5. MORE COMPUTATIONS OF LIMITS
95
We are not proving now these limits. Just note the indeterminate
form ∞/∞.
Intuitively, the exponential function diverges to infinite (that is, the
limit is infinite) faster than any power of x, in particular, any polynomials, and these in their turn diverges to infinite faster than the
logarithm.
Example 4.5.2. For a ∈ R
lim
x→+∞
1+
a x
= ea .
x
(4)
x
Indeed, recall 1 + x1 → e as x → +∞, and note that
a x h
1 xa ia
1+
= 1+ x
→ ea .
x
a
Example 4.5.3. From (4) we deduce
1
x
lim 1 + ax
= ea
x→0
(5)
Example 4.5.4. For a ∈ R
log(1 + ax)
= a.
x
This limit follows by passing to logarithms in the formula (5).
lim
x→0
(6)
Example 4.5.5. In particular, for a = 1 in formula (6), we have
Example 4.5.6.
lim
log(1 + x)
=1
x→0
x
(7)
ax − 1
= log a,
x→0
x
(8)
lim
96
4. LIMITS OF FUNCTIONS
in particular, for a = e
ex − 1
= 1,
(9)
x→0
x
in fact, by setting, y = ax − 1, so that x = loga (1 + y), we have
ax − 1
y
y
=
=
log a,
x
loga (1 + y)
log(1 + y)
lim
which, using the formula (7), converges to log a for y → 0.
Example 4.5.7.
(1 + x)a − 1
= a.
(10)
x→0
x
If a = 0, the function is the zero function, so the limit is obvious;
otherwise, setting y = a log(1 + x), we have, using formula (9) twice,
lim
(1 + x)a − 1
(1 + x)a − 1 a log(1 + x)
=
·
x
a log(1 + x)
x
y
e − 1 y/a
→ a.
=a
y ey/a − 1
We now apply the results above to specific examples.
Exercise 4.5.8.
lim
x→+∞
cos
1
1 sin2 (1/x)
.
x
We have, using formula (5),
1
1
r
1 sin2 (1/x)
1 sin2 (1/x)
cos
=
1 − sin2
x
x
1
1 2 sin21(1/x)
= 1 − 2 sin2
→ e−1/2 .
2
x
Exercise 4.5.9.
2
lim x log 1 + sin
.
x→+∞
x
We have, by formula (7) and the formula sinx/x → 1 for x → 0,
2
log 1 + sin x sin 2
2
x
x log 1 + sin
=2
→2
2
x
sin x2
x
4.5. MORE COMPUTATIONS OF LIMITS
Exercise 4.5.10.
√
x
lim x( 2 − 1).
x→+∞
We have, by formula (8),
√
21/x − 1
x
x( 2 − 1) =
→ log 2.
1
x
Exercise 4.5.11.
lim
x→+∞
p
x2 + x − x.
We have, by formula (10),
r
p
1
2
x +x−x=x
1+ −1
x
1
2
1 + x1
−1
1
=
→ .
1
2
x
Or,
p
√x2 + x + x
p
x2 + x − x =
x2 + x − x √
x2 + x + x
2
2
x +x−x
=√
x2 + x + x
x
= q
x
1 + x1 + 1
= q
1
1+
1
x
1
→ .
2
+1
Exercise 4.5.12.
e2x − 1
.
x→0 sin(4x) cos(x)
By applying 9 and 4.3.27,we have
lim
e2x − 1
1 e2x − 1 4x
1
1
=
→ .
sin(4x) cos(x)
2 2x sin(4x) cos(x)
2
97
98
4. LIMITS OF FUNCTIONS
Exercise 4.5.13.
log(x + 2)
.
x→+∞ log(x + 1)
lim
By 7 we have
2
log
x
1
+
x
log(x + 2)
=
log(x + 1)
log x 1 + x1
log x + log 1 + x2
=
log x + log 1 + x1
log 1+ x2 log x 1 +
log x
=
log 1+ x1 log x 1 +
log x
1+ x2
log
1+
2
x log x
2
x
=
log
1+
1
x
Exercise 4.5.14.
lim
x→±∞
1+ x1
−→
1+1·0
=1
1+1·0
1
x log x
x2
xe x2 +1 − xex .
Let’s consider first the limit for x → −∞, and note that
x2
e x2 +1 = e
1
1+ 12
x
−→ e,
and that by 4.5.1 and theorem 4.4.5
x
−x
xex = −x = − −x −→ 0,
e
e
therefore
x2
xe x2 +1 − xex −→ −∞ · e − 0 = −∞.
For x → +∞
x2
x2
−x
xe x2 +1 − xex = xex e x2 +1 − 1
4.5. MORE COMPUTATIONS OF LIMITS
= xex
e
−x3 +x2 −x
x2 +1
− 1 −x3 + x2 − x
−x3 +x2 −x
x2 +1
x2 + 1
99
−→= −∞.
CHAPTER 5
Differentiation
101
Index of Terms
limit, 3
103