CALCULUS
Contents
1
2
3
4
Real numbers
7
1.1
Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2
Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.3
The Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.4
The Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
Sequences
19
2.1
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
2.2
Limit of a sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
2.3
Properties of limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
Functions
37
3.1
Monotone and Bounded Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
3.2
Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
3.3
Combining Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
Limits of functions
4.1
Limit of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
4
calculus
5
Continuity
61
6
Differentiation
73
6.1
Differentiation formulae and techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
6.2
The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
6.3
L’Hôpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
Outline
1. This module will run from 01 September - 19 September.
2. The coursework is made up of 1 main test and 2 “small” tests.
3. There will be no hand-in assignments.
4. The test dates are as follows;
Date of the Test
08/09/25
15/09/25
18/09/25
Type
Duration
Scope of the test
Small Test 1
Small Test 2
Main Test
10 minutes
10 minutes
2 hours
Content covered up to Wednesday 06/09/25
Content covered up to Wednesday 13/09/25
Content covered up to Friday 15/09/25
5. These tests will constitute your coursework mark which is 50% of your grade.
Chapter
Content
Week
Number of
Lectures
Amount of
study hours
Numbers systems
Number systems, axioms of real numbers, mathematical induction
Types of sequences, definition of limits, applications of the definition, proofs of limit properties, applying Squeeze Theorem for
sequences
Definition of limits, applications, proofs of properties, one-sided
limits, application of Squeeze Theorem for functions
Definition of limits, application of this definition., proofs of
properties of limits, intermediate value theorem.
Definition of a derivative, application of the definition. proofs of
properties of a derivative, Mean value theorem. L’Hôpital’s rule
Riemann sums, Definition of a definite integral, Application of
the definition. Proofs of properties of a definite integral, indefinite integrals
Partial derivative, multiple integrals
Week 1
1
2
Week 1
3
5-10
Week 1-2
2
5-10
Week 2
3
3-5
Week 2-3
3
5-10
Week 3
2
3-5
Week 3
2
3-5
Sequences
Limits of functions
Continuity
Differentiation
Integration
Functions of several variables
1 Real numbers
1.1
Number Systems
The number systems that we use in calculus are the natural numbers, the integers, the rational numbers,
and the real numbers. Let us describe each of these :
Definition 1.1 (Natural numbers).
The natural numbers are the system of positive counting numbers 1, 2, 3 . . . . We denote the set of all natural
numbers by N.
N = {1, 2, 3, 4, 5, 6, 7, 8, . . . }.
If a number n is a natural number then we use the short-hand n ∈ N.
Definition 1.2 (Integers).
The integers are the positive and negative whole numbers and zero, . . . , −3, −2, −1, 0, 1, 2, 3, . . . . We denote the
set of all integers by Z.
Z = {. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . . }.
If a number n is an integer, then we use the short-hand n ∈ Z. If n is a positive integer, then we use
n ∈ Z+ . Therefore N = Z+ . Numerous other short-hand notions are used for example, non-negative
integers: Z+ ∪ {0} or N0 .
Definition 1.3 (Rational numbers).
p
The rational numbers are quotients of integers or fractions, such as 23 , − 45 . Any number of the form q , with
p, q ∈ Z and q ̸= 0, is a rational number. We denote the set of all rational numbers by Q.
Q=
p
p, q ∈ Z, q ̸= 0 .
q
A decimal number of the form x = 3.16792 is actually a rational number, for it represents
x = 3.16792 =
316792
.
100000
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calculus
A decimal number of the form
m = 4.27519191919 . . . ,
with a group of digits that repeats itself indefinitely, is also a rational number. To see this, notice that
100 · m = 427.519191919 . . .
and therefore we may subtract
100m = 427.519191919 . . .
m = 4.27519191919 . . .
Subtracting, we see that
99m = 423.244
or
m=
423244
99000
So, as we asserted, m is a rational number or quotient of integers. To indicate recurring decimals we
sometimes place dots over the repeating cycle of digits, e.g., m = 4.2751̇9̇,
19
6 = 3.16̇.
Definition 1.4 (Real numbers).
The real numbers are the set of all decimals, both terminating and non-terminating. We denote the set of all real
numbers by R.
Another kind of decimal number is one which has a non-terminating decimal expansion that does not
keep repeating. An example is π = 3.14159265 . . . . Such a number is irrational, that is, it cannot be
expressed as the quotient of two integers. There are two different short-hand for a set of all irrational
numbers namely I or R \ Q. Clearly, we have
N⊂Z⊂Q⊂R
There is a subclass of real numbers called algebraic numbers denoted by A. These are numbers that
√
can be found from solving an algebraic equation with integer coefficients i.e x2 − 2 = 0 therefore 2 is
an algebraic number. Clealry all rational numbers and some irrational numbers are algebraic numbers.
Is there an example of a number that is not algebraic? Yes and these numbers are called transcendental
numbers. The two most important transcendental numbers are π and e.
1.1.1
Field axioms
Definition 1.5 (Axiom of real numbers).
The set R of real numbers together with binary operations +, · and < obey the following properties: Axioms of
addition
real numbers
(A1) If a ∈ R and b ∈ R, then a + b ∈ R.
(A2) a + b = b + a ∀ a, b ∈ R.
(Closure)
(Commutativity)
(A3) a + (b + c) = ( a + b) + c ∀ a, b, c ∈ R.
(Associativity)
(A4) R contains an element called identity denoted by 0 such that a + 0 = a for all R
(Identity)
(A5) For every a ∈ R, there exists an element of R called the inverse and denoted by − a such that a + (− a) =
(− a) + a = 0. (Inverse)
Axioms of multiplication
(M1) If a ∈ R and b ∈ R, then a · b ∈ R.
(M2) a · b = b · a.
(Closure)
(Commutativity)
(M3) a · (b · c) = ( a · b) · c.
(Associativity)
(M4) R contains an element called identity denoted by 1 such that a · 1 = a.
(Identity)
(M5) For every a ∈ R, a ̸= 0 there exists an element of R called the inverse and denoted by 1a such that
a·
1
= 1.
a
(Inverse)
Distributive law
(D1) a · (b + c) = a · b + a · c. distributivity
Oder axioms
(O1) For any a ∈ R, exactly one of the following is true.
a > 0,
(O2) If a < b and b < c, then a < c.
(O3) If a < b, then a + c < b + c.
a=0
or
a<0
(Trichotomy)
(Transitivity)
(Order under addition)
(O4) If 0 < a and 0 < b, then 0 < a · b.
(Order under multiplication)
These axioms (A1)–(A5), (M1)–(M5) define an algebraic structure called a field. If we add (O1)–(O4)
we get an ordered field. From these rules we can build more theory of numbers and they are precisely
the rules we use solve algebraic equations and inequalities. Finally we add the completeness axiom. In
informal terms the completeness axiom says that real numbers form a continuum with no gaps between
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calculus
any two real numbers.A more detailed exposition on the subject of real numbers and application of
the above axioms is be discussed in Real Analysis and is beyond the scope of this module. One direct
consequence of these axioms which we seem “obvious” and will be used extensively in the next chapter
is the Archimedean property.
Theorem 1.1 (Archimedean property).
If x and y are positive real numbers, then there exists a positive integer n such that
n·x > y
Proof: Not required for this module.
1.2
■
Intervals
Definition 1.6.
A subset of the real line is called an interval if it contains at least two numbers and all the real numbers between
any of its elements.
Examples :
1. x > −2 defines an infinite interval. We can also write this as (2, ∞)
2. 3 ≤ x ≤ 6 defines a finite interval. We can also express this as [3, 6]
Definition 1.7 (Finite Intervals).
Let a and b be two finite real numbers such that a < b.
1. Open interval denoted by ( a, b) includes all points between a and b, but not a and b or set of all x ∈ R such
that a < x < b.
2. Closed interval [ a, b] includes all points between a and b as well as a and b i.e set of all x such that a ≤ x ≤ b.
3. Half-open interval we mean an open interval ( a, b) together with one of its endpoints. There are two such
intervals are [ a, b) and ( a, b] respectively denoted by set of all x such that a ≤ x < b and a < x ≤ b.
Definition 1.8 (Infinite Intervals).
Let a, b ∈ R. The following represent infinite intervals.
1. The set of all points x ∈ R such that a < x, denoted by ( a, ∞),
2. The set of all points x ∈ R such that a ≤ x, denoted by [ a, ∞),
3. The set of all points x ∈ R such that x < b denoted by (−∞, b)and (−∞, b] denotes the set of all x such that
x ≤ b,
real numbers
4. The set of all points x ∈ R such that x ≥ b denoted by (−∞, b],
5. The set of all x ∈ R denoted by (−∞, ∞)
1.3
The Absolute Value
It is a quantity that gives the magnitude or size of a real number. The absolute value or modulus of a
real number x, denoted by | x |, is given by
(
|x| =
x,
− x,
if x ≥ 0
if x < 0.
Geometrically, | x | is the distance between x and 0. For example, | − 6| = 6, |5| = 5, |0| = 0.
Th following properties of absolute numbers will be useful through this course.
Theorem 1.2 (Properties of the Absolute Value).
Let x and y be real numbers then,
1. | x | ≥ 0
2. | x | = 0 is zero if and only if x = 0,
3. −| x | ≤ x ≤ | x |,
4. | − x | = | x |,
5. | x − y| = |y − x |,
6. | x | = |y| implies x = ±y,
7. | xy| = | x | · |y|,
8.
x
y
= ||yx|| if y ̸= 0.
9. | x + y| ≤ | x | + |y|. (Triangle inequality),
10. | x | ≤ y if and only if −y ≤ x ≤ y,
11. | x | ≥ y if and only if x ≥ y or x ≤ −y.
12. x2 = | x |2
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1.4
calculus
The Principle of Mathematical Induction
The principle of mathematical induction is used to prove some mathematical statements that are formulated in terms of positive integers. We begin with function called propositional functions. This
function, denoted by P(n) represents some statement about positive integers n. The statement can
either true or false. Our goal in induction is to show that P(n) is true for all values of n > a where a is
fixed number. The steps that we will follow in all induction proofs is given below.
Aim: To prove P(n) is true for all positive integers n ≥ a.
1. Verify the P(a) is true i.e. the statement is true for n = a. This is called the base step.
2. Assume the statement is true for n = k i.e. assume P(k ) is true. This is called the inductive
assumption or inductive hypothesis.
3. Use the inductive assumption to show that P(k + 1) is true.
4. Hence conclude by the Principle Mathematical Induction that P(n) is true for all n ≥ a.
Steps 2 and 3 are called the inductive step.
Example 1.1.
For any positive integer n,
n
∑i = 1+2+···+n =
i =1
n ( n + 1)
.
2
Solution.
Let P(n) be the statement “ ∑in=1 i = n(n2+1) .”
Base Case:
We first show the statement is true for P(1) is true:
LHS =
1(1 + 1)
=1
2
and
RHS =
Hence the statement is true for n = 1 i.e. P(1) is true.
Inductive step:
Next we assume that the statement holds for n = k, that is P(k ) is true,
1+2+···+k =
k ( k + 1)
.
2
2
= 1,
2
real numbers
This is called the inductive assumption of the inductive hypothesis. Next, we use this assumption to prove
the statement holds for n = k + 1. Now,
1 + 2 + · · · + k + ( k + 1) = (1 + 2 + · · · + k ) + ( k + 1)
k ( k + 1)
+ (k + 1) (by inductive hypothesis)
2
k ( k + 1) + 2( k + 1)
=
2
k2 + 3k + 2
=
2
(k + 1)(k + 2)
=
2
(k + 1) [(k + 1) + 1]
=
2
=
Therefore P(k + 1) is true. Hence by the principle of mathematical induction P(n) is true for all n.
Example 1.2.
Prove that for any natural number
1 + 3 + 5 + · · · + 2n − 1 = n2 .
Solution.
Let P(n) be the statement that “1 + 3 + 5 + · · · + 2n − 1 = n2 ”
Base step:
To prove P(1) is true:
LHS = 1
and
RHS = 12 = 1,
Hence P(1) is true.
Inductive step:
Next we assume that P(k ) is true i.e.,
1 + 3 + 5 + · · · + 2k − 1 = k2 .
Next we show the statement holds for n = k + 1. To begin, we have
1 + 3 + 5 + · · · + (2k − 1) + 2(k + 1) − 1 = 1 + 3 + 5 + · · · + (2k − 1) + 2(k + 1) − 1
= k2 + 2(k + 1) − 1 (by inductive hypothesis)
= k2 + 2k + 1
= ( k + 1)2 .
Therefore P(k + 1) is true. Hence by the principle of mathematical induction
1 + 3 + 5 + · · · + 2n − 1 = n2
is true for all natural numbers n.
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calculus
Example 1.3.
Prove that 3n > 2n for all natural numbers n.
Solution.
Let P(n) be the statement “3n > 2n for all natural numbers n.”
Base step:
We first prove P(1) is true.
LHS = 31 = 3
and
RHS = 21 = 2.Thus LHS > RHS
Thus the statement holds true for n = 1.
Inductive step:
Next we assume P(k ) is true, that is,
3k > 2k .
Now, to prove the statement is true for n = k + 1 we have,
3k +1 = 3k · 3
> 2k · 3 by inductive hypothesis
> 2k · 2 since 3 > 2
> 2k +1 ,
which is true. Hence, by thr principle of mathematical induction P(n) is true for all n.
Many induction problems that you will come across in mathematics are disguised, that is the problem
may not precisely ask you to use induction to solve the problem. Here is a good example.
Example 1.4.
Find all positive integers n such that 22n − 1 is prime number.
Solution.
We can do some trial and error to see what’s going on
n
22n − 1
Prime or composite?
1
2
3
4
5
6
3
15
63
255
1023
4095
prime
Composite
Composite
Composite
Composite
Composite
real numbers
It seems that all the values greater than 3 are not prime. In fact, on closer inspection we can see that all these
cases so far we have a multiples of 3. Can we conclude that 22n − 1 is a multiple 3 for all n. We can use induction
to prove this assertion. Let P(n) be the statement that “22n − 1 is a multiple of 3” for n ≥ 2.
Base step:
To prove P(2) is true:
Substituting n = 2 we have 24 − 1 = 15 which is a multiple of 3, hence P(2) is true.
Inductive step:
Next we assume P(k ) is true, that is, 22k − 1 is a multiple of 3 for all k ≥ 2. Another way of expressing this is
to say 22k − 1 = 3m, for some integer m.
Next we prove the statement for n = k + 1.
22(k+1) − 1 = 4 · 22k − 1
= (1 + 3) · 22k − 1
= ·22k − 1 + 3 · 22k
= 3m + 3 · 2
by the inductive hypothesis
2k
Therefore 22(k+1) − 1 is a multiple of 3. Thus the statement holds true for n = k + 1. Hence, by induction 22n − 1
is a multiple of 3 for all n ≥ 2.
To go back to the original question since 22n − 1 is a multiple of 3 for all n ≥ 2 we can conclude there is only one
natural number i.e. n = 1 such that 22n − 1 is prime.
Here is another example
Example 1.5.
Which the two expressions n! and 2n is bigger?
Solution.
Again we can start with some trial and error
n
n!
2n
n! > 2n ?
1
2
3
4
5
6
1
2
6
24
120
720
2
4
8
16
32
64
No
No
No
Yes
Yes
Yes
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calculus
It looks like, initially n! < 2n but n! seems to be growing more rapidly. From this, we claim that n! > 2n for all
n ≥ 4. We can then proceed to prove this by induction.
Let P(n) be the statement “ n! > 2n for all n ≥ 4. The base case is when n = 4.
Base step:
From the above table we can see that 4! > 24 . Therefore P(4) is true.
Inductive step:
Assume P(k ) is true that is k! > 2k for k ≥ 4.
Now to for n = k + 1 we have
(k + 1)! = (k + 1)k!
> ( k + 1 )2k
> 2·2
k
By the inductive hypothesis
Since k + 1 > 4 + 1 > 2
= 2k +1
Therefore P(k + 1) is true. Hence by the principle f mathematical induction we have P(n) is true for all n ≥ 4.
Thus to answer the initial question n! < 2n for n = 1, 2, 3 but n! > 2n for all n ≥ 4.
Exercise 1.1.
Use induction to prove that for all n ∈ N,
1.
n
∑ i2 =
n(n + 1)(2n + 1)
6
n
n2 ( n + 1)2
4
i =1
2.
∑ i3 =
i =1
3.
3 + 11 + · · · + (8n − 5) = 4n2 − n
4. 8n − 2n is a multiple of 6
5.
n
1
1
∑ i (i + 1) = 1 − n + 1
i =1
6.
n
∑ ari−1 =
i =1
where a is a constant and r ̸= 1.
a (1 − r n )
,
1−r
real numbers
7. if f( x ) = x n , then the derivative of f , denoted by f ′ , is given by
f ′ ( x ) = nx n−1
.
8. if f ( x ) = x n , then the n−th derivative of f , denoted by f (n) , is given by
f (n) ( x ) = n!
.
9.
1 1 1
1
+ + + · · · + n −1 < 1
2 4 8
2
17
2 Sequences
2.1
Sequences
Definition 2.1 (Sequences).
A sequence is a set of numbers with a 1-1 correspondence with N i.e. u1 , u2 , · · · , where the first term is u1 .
second term is u2 and the n-th term is un .
Example 2.1.
a)
1, 2, 3, 4, 5, . . . ,
( Natural numbers)
b)
1, 4, 9, 16, 25, . . . ,
( Per f ect squares)
c)
2, 3, 5, 7, 11 . . . ,
( Prime numbers)
d)
3, 3, 3, 3, 3, . . . ,
(Constant sequence)
e)
f)
g)
1 1 1
1, , , , . . . ,
2 4 8
1 1 1 1
1, − , , − , , . . . ,
3 5 7 9
− 3, 9, −15, 33, −63, . . . ,
( Geometric progression)
There are many ways for representing a general sequence (un ), {un }∞
n=1 , { u n } n∈N e.t.c. For example
the sequence in d) above can be expressed as
{un }∞
n=1 = {2, 3, 5, 7, . . . }
In this case u1 = 2, u2 = 3, u3 = 5, u4 = 7, . . . ,.
Some sequences have general formula for the n-th term, (un ). The general formulae for the sequences
in Example 2.1 are given below.
Example 2.2.
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calculus
a)
un = n
b)
u n = n2
c)
None
d)
un = 3
e)
un = 21−n
or
un =
1
2n −1
(−1)n+1
2n − 1
g) 1 + (−1)n · 2n+1
f)
un =
Some sequences can also be described using a recursive formula that is the next term is expressed in
terms of previous terms.
Example 2.3.
A sequence is given by the following formula.
un+1 = 2un + 1,
u1 = 1
We calculate the first 4 terms of this sequences.
u1 = 1
u2 = 2u1 + 1 = 2(1) + 1 = 3
u3 = 2u2 + 1 = 2(3) + 1 = 7
u4 = 2u3 + 1 = 2(7) + 1 = 15
Notice in addition to being defined recursively as above, this sequence can also be defined as
u n = 2n − 1
Another famous sequence that can be described using a recursive formula is the Fibonacci sequence.
Example 2.4 (Fibonacci sequence).
The Fibonacci sequence is given by the following formula.
u n +2 = u n +1 + u n ,
u1 = 1,
u2 = 1
sequences
We calculate the first 7 terms of this sequences.
u1 = 1
u2 = 1
u3 = u2 + u1 = 1 + 1 = 2
u4 = u3 + u2 = 2 + 1 = 3
u5 = u4 + u3 = 3 + 2 = 5
u6 = u5 + u4 = 5 + 3 = 8
u7 = u6 + u5 = 8 + 5 = 13
Sequence can be finite or infinite. Finite sequences have a last term but infinite sequences do not.
Finite sequence are not that interesting so we will mainly concentrate on infinite sequences.
Definition 2.2 (Bounded above).
A sequence (un ) of real numbers is bounded above if there is an M ∈ R such that un ≤ M for all n ∈ N.
Example 2.5.
The sequence (un ) given by un = 1 − n1 is bounded from above. Observe that this sequence is
1 2 3 4
0, , , , , . . . ,
2 3 4 5
All the terms in this sequence are less than or equal to 1 (it seems!), thus any number greater than or equal to 1
can be used as M in Definition (2.2). For example, take M = 7.6, then un ≤ 7.6 for all n.
Definition 2.3 (Bounded below).
A sequence (un ) of real numbers is bounded below if there is an m ∈ R such that un ≥ M for all n ∈ N.
Example 2.6.
The sequence ( xn ) given by xn = n is bounded from below. Observe that this sequence is
1, 2, 3, 4, . . . ,
All the terms in this sequence are greater than or equal to 1, thus any number less than or equal to 1 can be used
as m in Definition (2.3). For example take m = −3 then all the terms of the sequence ( xn ) are greater than 3.
Notice the sequence in Example 2.5 is also bounded below since all terms of the sequence are greater
than or equal to 0.
Definition 2.4 (Bounded sequence).
A sequence (un ) of real numbers is bounded if there is an M ∈ R such that |un | ≤ M for all n ∈ N i.e.
− M ≤ un ≤ M.
In other words a sequence is bounded if it is both bounded from below and bounded from above.
The sequence in Example 2.5 is both bounded above and bounded below therefore it is bounded. For
example, take M = 1, we observe that Other examples of a bounded sequences are
−1, 1, −1, 1, −1, 1, · · ·
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calculus
22
and
1 1 1 1 1
1, − , , − , , − , . . .
2 3 4 5 6
Clearly, all finite sequences are bounded. The converse is not true, that means it’s not true to say “All
bounded sequences are finite”. We can prove this by use of a counterexample. One such counterexample is the sequence
1 1 1 1 1
1, − , , − , , − , . . .
2 3 4 5 6
Clearly, this sequence is bounded but not finite.
Definition 2.5 (Unbounded sequence).
A sequence (un ) of real numbers is unbounded if is not bounded.
Another definition of unbounded is as follows
Definition 2.6 (Unbounded sequence).
A sequence (un ) of real numbers is unbounded if for any M ∈ R there exists an n ∈ N such that |un | ≥ M.
The sequence in Example 2.6 is bounded below but is not bounded above therefore it is not unbounded.
Using Definition 2.6 we see that for any value of M we choose we can always find n such |un | ≥ M.
For example if M = 24.4, then we can choose (say) n = 25 and we observe u25 = 25 and u25 > 24.4. If
we let M = 106 then a choice of (say) n = 106 + 1 again leads to |un | ≥ M.
2.2
Limit of a sequence
For infinite sequences we are interested in knowing the limit of a sequence, i.e. the value that a
sequence is converging to. For example, the sequence
3
,
2
2,
4
,
3
5
,
4
6
,
5
7
,···
6
1
is approaching 1. The n-th term for this sequence is given by un = n+
n . We say limit of { u n } as n
approaches infinity is 1. In shorthand notion, we say
lim un = 1
n→∞
or
lim
n→∞
n+1
=1
n
Notice that that in many cases (certainly the most interesting ones), the sequence does not actually
attain the limit value. For example, the sequence above has the limit of 1, even though the value 1
1
is never in the sequence i.e. there is no n such that n+
n is exactly equal to 1. Compare this with the
following sequence which it hits 3 remains at 3 forever:
−2, 7, −1, 4, 3, 3, 3, 3, 3, . . .
sequences
The limit of the sequence is 3, which is actually attained.
If the limit of sequence {un } exists i.e. is a finite number, then {un } is called a convergent sequence. If
such finite limit does not exist then {un } is called a divergent sequence. The following sequences are
convergent:
3, 3, 3, 3, 3, . . . ,
1, 21 , 41 , 81 , . . .
1, − 13 , 51 , − 17 , 19 , . . . ,
The following sequences are divergent:
1, 2, 3, 4, 5, . . . ,
1, 4, 9, 16, 25, . . . ,
−3, 9, −15, 33, −63, . . . ,
Please note that being bounded and being convergent are not the same thing. We will revisit the topic
of the differences between boundedness and convergence later on.
We finally arrive at the formal definition of a limit which was introduced by the German mathematician
Karl Weierstrass.We will use the definition to give a rigorous proof that a limit of a sequence exist.
Definition 2.7 (Definition of a limit).
A number L is called the limit of an infinite sequence u1 , u2 , u3 , . . . , if for any ε > 0, we can find a positive
number N depending on ε such that
|un − l | < ε for all integers n > N
To get a feel of what this definition is about, lets look at the limit of un = n1 as n approaches in infinity
using the following steps.
1
1. Choose ε = 10
2. Show that we can find N such that for all n > N
|un − l | =
1
1
−0 <
n
10
Take N = 10, and consider some values of n > N, we have
1
1
−0 <
= ε if we choose n = 11
11
10
1
1
|u20 − l | =
−0 <
= ε if we choose n = 20
20
10
1
1
|u100 − l | =
−0 <
= ε if we choose n = 100
100
10
|u11 − l | =
23
calculus
24
Clearly if N = 10, then for all n> N we have |un − l | < ε
1
3. Choose an even smaller ε, say ε = 100
4. Show that we can find N such that for all n > N
|un − l | =
1
1
−0 <
n
100
Take N = 100, and consider some values of n > N, we have
1
1
−0 <
=ε
120
10
1
1
−0 <
=ε
|u200 − l | =
200
10
1
1
|u600 − l | =
−0 <
=ε
600
10
|u120 − l | =
if we choose n = 120
if we choose n = 200
if we choose n = 600
Clearly if N = 100, then for all n> N we have |un − l | < ε
5. If we choose ε to be as small as we like can we always find N such that |un − l | < ε. In this case it
clear to see that if we choose N = 1ε , then for all n > N we have
|un − l | =
1
− 0 < ε.
n
|un − l | =
1
− 0 < ε.
n
Thus by Definition 2.7 lim ( n1 ) = 0
n→∞
Next we write the proof with less words
Example 2.7.
Prove that lim
1
n→∞ n
= 0.
Solution.
Let ε > 0, we need to find N (ε) such that
Now
1
1
−0 =
n
n
1
=
n
We take N > 1ε . Then for all n > N we have
|un − l | =
Hence the limit is proved.
1
1
<
<ε<ε
n
N
sequences
Notice that it possible to always find a natural number N > 1ε no matter how small ε. This statement
is true due to the Archimedean property shown in the previous chapter.
Example 2.8.
Prove that lim
n→∞
n+1
= 1.
n
Solution.
Let ε > 0, we need to find N (ε) such that
|un − l | =
n+1
− 1 < ε.
n
To begin we choose ε > 0. Next we have
n+1
n+1−n
−1 =
n
n
1
=
n
1
=
n
If we take N such that N1 < ε or N > 1ε , then for all n > N we have
n+1
−1
n
1
1
1
1
= <
Notice that n > N implies that <
n
N
n
N
<ε
|un − l | =
Hence the limit is proved.
Example 2.9.
“Guess” the limit of the following sequence and use the definition of a limit to prove this limit.
7 10
4, , , . . .
2 3
Solution.
This sequence can be written as follows
4 + 3(0)
,
1
4 + 3(1)
,
2
4 + 3(2)
, ...
3
Thus general term is given by
un =
4 + 3( n − 1)
3n + 1
1
=
= 3+
n
n
n
We can "observe" that
lim un = 3.
n→∞
We can definitively prove that lim un = 3 using the formal definition of a limit. That is we want to show that
n→∞
for every ε > 0, there exists a natural number N such that for all natural numbers greater than N we have
25
26
calculus
|un − 3| < ε. To begin, let ε > 0. Then, we have
1 + 3n
−3
n
1 + 3n − 3n
=
n
1
=
n
1
=
n
| u n − 3| =
If we take N1 < ε i.e. N > 1ε , then for all n > N we have
| u n − 3| =
1
1
<
<ε
n
N
Hence the limit is proved.
Example 2.10.
Use the definition of a limit to prove that
2n − 1
2
=
n→∞ 3n + 2
3
lim
Solution.
Let ε > 0.Then
3(2n − 1) − 2(3n + 2)
6n − 3 − 6n − 4
2n − 1 2
=
=
−
3n + 2 3
3(3n + 2)
3(3n + 2)
−7
=
3(3n + 2)
7
=
3(3n + 2)
7
<
9n
7
If we take 9N
< ε i.e N > 9ε7 then for all n > N we have
2n − 1 2
7
7
−
<
<
<ε
3n + 2 3
9n
9N
Hence the limit is proved
Example 2.11.
Use the definition of a limit to prove that
lim
2n2
n → ∞ n3 + 2
=0
Solution.
Let ε > 0, Then we have
|un − l | =
2n2
2n2
2n2
2
2
=
=
−
0
<
=
n3 + 2
n3 + 2
n3
n
n
sequences
Take N2 < ε or N > 2ε . Then, for all n > N, we have
|un − l | =
2n2
2
2
−0 < <
<ε
3
n +2
n
N
Hence the limit is proved.
Example 2.12.
Use the definition of a limit to prove that
lim
n→∞
1 + 4n
5 + 6n
=
2
3
Solution.
Let ε > 0. Next we have
3(1 + 4n) − 2(5 + 6n)
3 + 12n − 10 − 12n
1 + 4n 2
=
=
−
5 + 6n 3
3(5 + 6n)
3(5 + 6n)
−7
7
=
=
3(5 + 6n)
3(5 + 6n)
7
=
15 + 18n
7
<
18n
7
7
< ε i.e. N > 18ε
. Then for all n > N we have
Take 18N
1 + 4n 2
7
7
<
−
<
<ε
5 + 6n 3
18n
18N
Hence the limit is proved.
We slightly modify the above to come up with a more complicated example
Example 2.13.
Use the definition of a limit to prove that
lim
n→∞
4n + 1
6n − 5
=
2
3
Solution.
Let ε > 0. Next we have
4n + 1 2
3(4n + 1) − 2(6n − 5)
12n + 3 − 12n + 10
−
=
=
6n − 5 3
3(6n − 5)
3(6n − 5)
13n
13
=
<
3(6n − 5)
3(6n − 5)
13
=
18n − 15
In the previous example we had +15 which led us to the conclusion
7
7
<
15 + 18n
18n
27
28
calculus
In this case we have −15. Unfortunately, for us
13
13
≮
−15 + 18n
18n
7
7
Take 18N
< ε i.e. N > 18ε
. Then for all n > N we have
1 + 4n 2
7
7
<
−
<
<ε
5 + 6n 3
18n
18N
In this case choose any confidence of n less than for example 16n or 15n or 9n. Suppose we choose 9n, we then
proceed to find all n such that
18n − 15 > 9n
Solving this inequality we get
18n − 9n > 15
15
= 1.666 . . .
9
n ≥ 2 since in n is a natural number
n>
Therefore for n ≥ 2 we have
18n − 15 > 9n
By inverting this we have that n ≥ 2
1
1
<
18n − 15
9n
We go back to the original problem and proceed as follows
4n + 1 2
13
1
−
<
<
6n − 5 3
18n − 15
9n
provided n > 2
1
Now we take 9N
< ε i.e N < 9ε1 . Now if we take N > max 2, 9ε1 , then for n > N we have
4n + 1 2
1
1
−
<
<
<ε
6n − 5 3
9n
9N
Hence the limit is proved.
Example 2.14.
Use the definition of a limit to prove that
lim
n→∞
sin(n2 )
√
3
n
=0
sequences
Solution.
Let ε > 0, Then we have
sin(n2 )
√
|un − l | =
−0
3
n
sin(n2 )
√
=
3
n
| sin(n2 )|
√
=
| 3 n|
1
≤ √
since | sin(θ )| < 1 for all θ
3
| n|
1
= √
3
n
1
< ε i.e. N > ε13 .Thus for n > N, we have
Take √
3
N
|un − l | =
sin(n2 )
√
3
n
1
1
< √
−0 ≤ √
<ε
3
3
n
N
Example 2.15.
√
√
Use the definition of a limit to prove that lim n + 4 − n
n→∞
Solution.
Notice that as n → ∞ we have
√
n + 1) −
√
n → ∞−∞
This quantity is undefined, so this approach will not help us solve this problem. Thus we need to try a new
approach.
Whenever we are confronted with a expression of the form
p
p
some expression − another expression,
we multiply by term in a way similar to multiplication by complex conjugates (remember high school maths!) i.e
we rationalize the numerator Thus we have
√
n+4−
√
n=
√
n+4−
√
√
√
n+4+ n
√
n·
√
n+4+ n
The numerator becomes a difference of two squares
n+4−n
√
n+4+ n
4
= √
√
n+4+ n
= √
Now it is clear to see that as n increases, the denominator increases. Therefore a good “guess” for the limit of
√
√
n + 1) − n is 0. Using the definition we need to show that for all ε > 0, there exists N such that for all
n > N we have
√
√
n+4− n−0 < ε
29
30
calculus
To begin we have
√
n+4−
√
n−0 =
√
n+4−
√
n
4
√
n+4+ n
4
√
< √
n+ n
4
= √
2 n
2
= √
n
2
= √
n
= √
<ε
if we take N > ε42 . Thus the limit is proved.
Not all sequence have limits. Consider the following sequence
1, −1, 1, −1, 1, −1 · · ·
We will prove this by contradiction:
Suppose the limit of this sequence is l. Also suppose xn − 1, then xn+1 = −1. Take ε = 21 and since l is
the limit of this sequence, then there exits N such that for all n > N we have
| Xn − l | <
1
2
| Xn +1 − l | <
1
2
and
Therefore we have
|1 − l | <
1
2
| − 1 − l| <
1
2
and
Notice that we can write 2 as (1 − l ) − (−1 − l ). Therefore
2 = |2| = |(1 − l ) − (−1 − l )|
< |(1 − l )| + |(−1 − l )| by the triangle inequality
1 1
< + =1
2 2
Hence we have deduced that 2 < 1 which is a contradiction. Thus our original supposition that the
sequence has limit is incorrect. Hence (un ) has no limit.
sequences
2.3
Properties of limits
It is important to observe that boundedness and convergence of a sequence are different concepts. We
look at the following theorem.
Theorem 2.1.
Let (un ) be a convergent sequence. Then (un ) s a bounded sequence.
Proof: Beyond the scope of Calculus
■
This converse is not true. That means it’s not true to say that every bounded sequence is convergent.
A counterexample is
−1, 1, −1, −1, . . .
This sequence is bounded i.e |un | ≤ 1 for all n but as shown before it’s not convergent.
Theorem 2.2 (Limit of a sequence is unique).
The limit of a sequence is unique.
Proof: Beyond the scope of Calculus.
■
Theorem 2.3.
Let { an }, {bn } and {cn } be sequence with the following limits
lim an = a
n→∞
lim bn = b
n→∞
lim cn = c,
n→∞
c ̸= 0
Then we have the following
1. lim ( an + bn ) = lim an + lim bn = a + b.
n→∞
n→∞
n→∞
2. lim ( an − bn ) = lim an − lim bn = a − b.
n→∞
n→∞
n→∞
3. lim ( an · bn ) = ( lim an ) · ( lim bn ) = ab.
n→∞
n→∞
n→∞
lim an
a
an
= n→∞ =
n→∞ cn
lim cn
c
4. lim
n→∞
5. If k is a constant, we have lim (k ) = k.
n→∞
6. If k is a constant, we have lim (k · an ) = k · a.
n→∞
7. The limit of a convergent sequence {un } of real numbers is unique.
31
32
calculus
For this module we will only prove 1) and 2).
Proof:
1. Let ε > 0. We need to show that there exists N ∈ N such that for all n > N,
|( an + bn ) − ( a + b)| < ε.
First we note that lim an = a. That is, there exists N1 ∈ N such that for n > N1
n→∞
| an − a| <
ε
2
Similarly lim bn = b implies that there exists N2 ∈ N such that for n > N2
n→∞
| bn − b | <
ε
2
Now we have
|( an + bn ) − ( a + b)| = |( an − a) + (bn − b)|
< | an − a| + |bn − b| using the triangle inequality
If we take N = max( N1 , N2 ), then for all n > N we have
|( an + bn ) − ( a + b)| < | an − a| + |bn − b| <
ε
ε
+
2 2
=ε
Thus the limit is proved.
2. To prove lim ( an − bn ) = a − b we proceed as we did in 1) only changing sign from positive to
n→∞
negative where appropriate.
■
These properties are useful at breaking down complicated limits into simpler limits.
We want to be able to evaluate limits, for example, of the form lim
n→∞
Example 2.16.
Evaluate the following limits
1. lim
n→∞
1
3
2− + 2 .
n n
3n2 − 5n
.
n→∞ 5n2 + 2n − 6
2. lim
2−
3
1
+ 2
n n
5 − 2n2
.
n→∞ 4 + 3n + 2n2
or lim
sequences
√
√
3. lim ( n + 1 − n).
n→∞
Solution.
1.
lim
n→∞
2−
1
3
+ 2
n n
1
1
+ 3 lim 2
n→∞ n
n→∞ n
= lim 2 − lim
n→∞
= 2−0+0
=2
2.
lim 3 − n5
3n2 − 5n
n→∞
lim
=
n→∞ 5n2 + 2n − 6
lim 5 + n2 − n62
n→∞
We know that
3−
5
n
lim
2
6
−
n n2
n→∞
5
= 3−0 = 3
n→∞
n
= lim 3 − lim
n→∞
and
5+
lim
n→∞
= lim 5 + lim
n→∞
2
n→∞ n
− lim
6
n → ∞ n2
= 5+0+0
Therefore combining the two limits we have
lim 3 − n5
3
3n2 − 5n
n→∞
=
=
lim
2
n→∞ 5n + 2n − 6
5
lim 5 + n2 − n62
n→∞
3. As before we rationalize the numerator since the
√
n+1−
√
n=
√
n+1−
√
√
√
n+1+ n
n· √
√
n+1+ n
The numerator becomes a difference of two squares
n+1−n
√
n+1+ n
1
= √
√
n+1+ n
= √
Therefore, the limit becomes
√
√
lim ( n + 1 − n) = lim √
n→∞
n→∞
1
√
n+1+ n
=0
Theorem 2.4 (Squeeze Theorem).
If lim an = l = lim bn and there exists an N such that an ≤ cn ≤ bn , for all n > N, then lim cn = l.
n→∞
n→∞
n→∞
33
34
calculus
Example 2.17.
Use the Squeeze theorem to find lim
n→∞
cos n
.
n
Solution.
We know that
−1 ≤ cos n ≤ 1
cos n
1
1
≤
=⇒ − ≤
n
n
n
1
cos n
1
− lim ≤ lim
≤ lim
n→∞ n
n→∞
n→∞ n
n
cos n
0 ≤ lim
≤0
n→∞
n
cos n
=0
lim
n→∞
n
sequences
Exercise 2.1.
1. Use the definition of limit to prove the following limits
(a)
lim
n→∞
(b)
n
2
n +1
3n + 1
2n + 5
lim
lim
n→∞
(d)
lim
n→∞
(e)
lim
n→∞
=0
2n
n+1
n→∞
(c)
=2
=
(−1)n n
n2 + 1
4 + 5n
6 − 7n
3
2
=0
=−
5
7
(f)
lim
n→∞
sin(n2 )
√
3
n
=0
2. Use properties of limits to simplify the following limits
(a)
lim
n→∞
(b)
lim
n→∞
(c)
lim
n→∞
1
√ .
n→∞ n n
3. Use the Squeeze theorem to find lim
n
n2 + 1
2n2 + 3
n2 + 1
3n3 + 2n2 + 5
4n3 + n
35
3 Functions
3.1
Monotone and Bounded Functions
Definition 3.1 (Monotone functions).
Definition 3.1 (a) A function f is monotone increasing on an interval I if for all points x and y in I with
x<y
then
f ( x ) ≤ f (y)
Definition 3.1 (b) A function f is strictly increasing on an interval I if for all points x and y in I with
x<y
then
f ( x ) < f (y)
Definition 3.1 (c) A function f is monotone decreasing on an interval I if for all points x and y in I with
x<y
then
f ( x ) ≥ f (y)
Definition 3.1 (d) A function f is strictly decreasing on an interval I if for all points x and y in I with
x<y
then
f ( x ) > f (y)
Definition 3.1 (e) A function f is monotone on interval I if f is either monotone increasing or monotone
decreasing.
Example 3.1.
Consider the function
f ( x ) = (2x − 1)( x + 5)
We observe that f is increasing on the interval (−9/4, ∞) and is decreasing on the interval (−∞, −9/4).
38
calculus
Definition 3.2 (Bounded Functions).
A function f is
1. bounded above if there is a real number M such that f ( x ) ≤ M for all points x in its domain. The number
M is then called an upper bound of f .
2. bounded below if there is a real number m such that f ( x ) ≥ m for all points x in its domain. The number
m is then called a lower bound of f .
3. bounded if f is bounded above and below, that is, there exist real numbers M and m such that m ≤ f ( x ) ≤ M
for all points x in its domain.
Example 3.2.
The function f ( x ) = x + 3 is bounded in −1 ≤ x ≤ 1. An upper bound is 4 (or any number greater than 4). A
lower bound is 2 (or any number less than 2).
Observe that boundedness depends on the domain. For example, the function h( x ) = tan x is bounded
on the interval 0, π4 but not bounded on the interval 0, π2
3.2
Types of Functions
Elementary Functions
Definition 3.3 (Polynomial Function).
These are functions of the form
f ( x ) = a 0 x n + a 1 x n −1 + · · · + a n −1 x + a n
(3.1)
where a0 , a1 , . . . , an are constants and n is a positive integer called the degree of the polynomial provided a0 ̸= 0.
Example.
x5 + 10x3 − 2x + 1 is a polynomial of degree 5.
Definition 3.4 (Rational Functions).
These are functions of the form
f (x) =
P( x )
Q( x )
(3.2)
where P( x ) and Q( x ) are polynomial functions and Q( x ) ̸= 0.
Example.
An example of a rational function is
x3 + x + 5
x2 − 3x − 4
is a rational function. Since ( x + 1)( x − 4) = 0 for x = −1 and x = 4, the domain of f is the set of all real
numbers except −1 and 4.
f (x) =
functions
Definition 3.5 (Power Function).
These are functions of the form
f ( x ) = [ P( x )]n ,
(3.3)
n a real number..
Example.
Examples of power functions are
y=
1
x
1
y = x2
2
y = x3
A function need not be defined by a single formula.
Piecewise Defined Functions
Definition 3.6.
A piecewise defined function is a function described by using different formula on different parts of its domain.
Example.
1.
f (x) =


−1,

x<0
0,
x=0



x + 2,
x>0
The graph of this function is given by
y
x
39
40
calculus
2.
g( x ) =



− x + 2,
2



x<0
x ,
0≤x≤1
4,
x > 1.
The graph of this function is given by
y
x
Transcendental Functions
The following are sometimes called elementary transcendental functions.
1. Exponential function, f ( x ) = a x , a ̸= 0, 1.
10 y
8
6
4
2
x
−3
−2
−1
2. Logarithmic function, f ( x ) = loga x, a ̸= 0, 1.
1
2
3
y = 2x
functions
6
4
y = log2 ( x )
2
1
2
3
5
4
−2
−4
−6
3. Trigonometric functions
sin x, cos x, tan x =
sin x
, csc x, cot x, sec x
cos x
. The graph of y = sin x is given below
y
y = sin x
1
x
200
100
300
−1
4. Inverse trigonometric functions,
y = sin−1 x, y = cos−1 x
y
π/2
y = arcsin x
x
−1
1
−π/2
41
calculus
42
5. Hyperbolic Functions,
sinh x, cosh x, tanh x, coth x
y
1
y = tanh x
x
−3
−2
−1
1
2
3
−1
Definition 3.7 (Even and Odd Functions).1. Let f ( x ) be a real-valued function of a real variable. Then f is
even if
f ( x ) = f (− x )
.
y
y = cos x
1
x
−300 −200 −100
100
200
−1
2. Let f ( x ) be a real-valued function of a real variable. Then f is even if
f ( x ) = f (− x )
300
functions
y
y = sin x
1
x
−300 −200 −100
100
200
300
−1
Example.1. Examples of odd functions are
| x |, x2 , x4 , cos x, cosh x
2. Examples of even functions are
x, x3 , sin x, sinh x
Definition 3.8.
Determine whether the following function is odd or even
f (x) =
Solution.
Observe that
f (− x ) =
3x
x2 + 1
3(− x )
3x
=− 2
= − f ( x ).
2
(− x ) + 1
x +1
The function is odd.
3.3
Combining Functions
A function f can be combined with another function g by means of arithmetic operations to form other
f
functions, the sum f + g, difference f − g, product f g and quotient g are defined as :
Let f and g denote functions, then
1. ( f + g)( x ) = f ( x ) + g( x ).
2. ( f − g)( x ) = f ( x ) − g( x ).
3. ( f g)( x ) = f ( x ) g( x ).
f
f (x)
4. g ( x ) = g(x) , g( x ) ̸= 0.
43
44
calculus
f
Example: If f ( x ) = 2x2 − 5 and g( x ) = 3x + 4. Find f + g, f − g, f g, g .
Solution:
( f + g)( x ) = (2x2 − 5) + (3x + 4) = 2x2 + 3x − 1.
( f − g)( x ) = (2x2 − 5) − (3x + 4) = 2x2 − 3x − 9.
( f g)( x ) = (2x2 − 5)(3x + 4) = 6x3 + 8x2 − 15x − 20
f
2x2 − 5
(x) =
.
g
3x + 4
4 Limits of functions
4.1
Limit of a function
Let f be a function, then we say
lim f ( x ) = A,
x→a
if the value of f ( x ) gets arbitrarily closer to L as x gets closer and closer to a. For example,
lim x2 = 9
x →3
since x2 gets arbitrarily close to 9 as x approaches as close as one wishes to 3. This leads us to a formal
definition of limit of a function.
Definition 4.1 (The ε − δ definition for limit of a function).
Let f be a function. Then a real number L is called the limit of f at a if for any given real number ε > 0, there
exists a positive number δ (possibly depending on ε), such that
| f ( x ) − L| < ε whenever | x − a| < δ
We denote this limit by
lim f ( x ) = L
x→a
Example 4.1.
Use the ε − δ definition to show that
lim(4x − 3) = 5
x →2
Solution.
Looking at figure below we see that as x → 2 we have f ( x ) → 5.
46
calculus
10 y
5
x
−3
−2
−1
1
2
y = (4x − 3)
3
−5
However on a rigorous definition is able to definitively prove that such a limit exist and is useful in cases where
in might not be easy to just “eye-ball” the limit.
To be certain that lim f ( x ) = 5, we need to find δ such that, for a given ε, | f ( x ) − 5| < ε whenever | x − 2| < δ.
x →2
| f ( x ) − L| = |(4x − 3) − 5|
= |4x − 8|
= 4| x − 2|
< 4δ
Take δ = 4ϵ . Then whenever| x − 2| < δ, we have
| f ( x ) − L| = |(4x − 3) − 5| < 4δ = 4 ·
ϵ
=ε
4
Hence the limit is proved
Example 4.2.
Show that lim( x2 + 1) = 2.
x →1
Solution.
Let ε > 0. Our goal is to find δ such that
| f ( x ) − 2| < ε whenever | x − 1| < δ.
To proceed
| f ( x ) − 2| = | x 2 + 1 − 2|
= | x 2 − 1|
= |( x + 1)( x − 1)|
= | x + 1|| x − 1|
limits of functions
In order to determine how small the δ we choose can be, we need to put a bound on the term x + 1. To do this,
we take an initial value δ = 1 (it could be any number but the smaller the better!).
| x − 1| < δ = 1
−1 < x − 1 < 1
0<x<2
1 < x+1 < 3
1 < | x + 1| < 3
Therefore going back to the previous calculation we have
| f ( x ) − 2| = | x 2 + 1 − 2|
= | x + 1|| x − 1|
< 3δ since 1 < x + 1 < 3
Take δ = min 1, 3ε . Then whenever | x − 1| < δ, we have
|( x2 + 1) − 2| < 3δ = 3 ·
ε
= ε.
3
Hence the limit is proved.
Example 4.3.
Show that lim( x2 + 3x ) = 10.
x →2
Solution.
Let ε > 0, our goal is to find δ such that
| f ( x ) − 10| < ε whenever | x − 2| < δ.
To proceed
| f ( x ) − 10| = | x2 + 3x − 10|
= |( x + 5)( x − 2)|
= | x + 5|| x − 2|
We need to put a bound on the term | x + 5|. To do this, we take an initial value δ = 1. Then
| x − 2| < δ = 1
−1 < x − 2 < 1
1 < x < 3.
6 < x+5 < 8
6 < | x + 5| < 8
. Therefore,
| f ( x ) − 10| = | x2 + 3x − 10|
= | x + 5|| x − 2|
< 8δ
47
48
calculus
Take δ = min 1, 8ε . Then whenever | x − 2| < δ, we have
|( x2 + 3x ) − 10| < 8δ = 8 ·
ε
= ε.
8
Hence the limit is proved.
Example 4.4.
Prove that
lim
2
x →5 x
=
2
5
Solution.
Let ε > 0, our goal is to find δ such that
2
| f ( x ) − | < ε whenever | x − 5| < δ.
5
To proceed
f (x) −
2
2
2
=
=
5
x
5
|2(5 − x )|
=
5| x |
|2( x − 5)|
=
5| x |
2| x − 5|
=
5| x |
2δ 1
<
·
5 |x|
We need to put a bound on the term |1x| . Taking an initial value δ = 1. Then
| x − 5| < 1
−1 < x − 5 < 1
4<x<6
1
1
1
< <
6
x
4
1
1
1
<
<
6
|x|
4
Therefore
f (x) −
2
|2( x − 5)|
=
5
5| x |
2δ 1
·
<
5 |x|
2δ 1
<
·
5 4
1
·δ
<
10
limits of functions
Take δ = min {1, 10ϵ}. Then whenever | x − 5| < δ, we have
2 2
1
1
<
−
·δ =
· 10ϵ = ε
x 5
10
10
ence the limit is proved.
Example 4.5.
Prove that
lim
x →−6
x+4
2−x
Solution.
Let ε > 0, our goal is to find δ such that
1
<ε
f (x) − −
4
=−
whenever
1
4
| x − (−6)| < δ.
To proceed
x+4 1
−
2−x 4
4( x + 4) − (2 − x )
=
4(2 − x )
3 x+6
=
4 x−2
3 | x + 6|
=
4 | x − 2|
3 δ
<
4 | x − 2|
| f ( x ) − L| =
Take an initial value of δ = 1. Then
| x − (−6)| = | x + 6| < 1
−1 < x + 6 < 1
−7 < x < −5
−9 < x − 2 < −7
7 < | x − 2| < 9
1
1
1
<
<
9
| x − 2|
7
Therefore
x+4 1
−
2−x 4
3 δ
<
4 | x − 2|
3 δ
< ·
4 7
3
= δ
28
| f ( x ) − L| =
49
50
calculus
Take δ = min 1, 28
3 ε . Then whenever | x − (−6)| < δ, we have
x+4 1
3
3 28
< δ=
−
· ε=ε
2−x 4
28
28 3
Hence the limit is proved.
Example 4.6.
Use ϵ − δ definition to prove that
√
lim ( 19 − x ) = 3
x →10
Solution.
Let ε > 0, our goal is to find δ such that
| f ( x ) − 3| < ε whenever | x − 10| < δ.
To proceed
√
| f ( x ) − L| = | 19 − x − 3|
Using the same approach as in 2.16 (c), we have
√
This for
√
x+a−
√
√
√
√
√
( x + a − b) · ( x + a + b)
√
b=
√
x+a+ b
x+a−b
√
= √
x+a+ b
19 − x − 3 this becomes
√
19 − x −
√
9
√
√
19 − x + 9
√
= ( 19 − x − 9) · √
19 − x + 9
√
√
√
√
( 19 − x − 9) · ( 19 − x + 9)
√
√
=
19 − x + 9
19 − x − 9
√
= √
19 − x + 9
10 − x
√
= √
19 − x + 9
√
√
Therefore
√
| f ( x ) − L| = | 19 − x − 3|
= √
10 − x
19 − x +
| x − 10|
√
9
√
= √
| 19 − x + 9
δ
√
< √
| 19 − x + 9
limits of functions
As before, we have to look for a bound for √19−1x+√9 . We begin by taking a preliminary δ = 1. This implies
−1 < | x − 10| < 1
9 < x < 11
8 < 19 − x < 10
√
√
8 < 19 − x < 10
√
√
√
8 + 3 < 19 − x + 3 < 10 + 3
√
√
1
10 + 3
< √
1
1
< √
19 − x + 3
8+3
Therefore, we have
√
| f ( x ) − L| = | 19 − x − 3|
δ
< √
| 19 − x + 3
1
< √
δ
8+3
o
n √
Take δ = min 1, ( 8 + 3)ε . Then whenever | x − 10| < δ, we have
√
√
1
1
| 19 − x − 3| < √
δ= √
δ · ( 8 + 3) ε = ε
8+3
8+3
Hence the limit is proved.
Exercise 4.1.
Use the definition of a limit to prove the following limits
1. lim (4x + 3) = −5
x →−2
2. lim(−1 − 3x ) = −10
x →3
3. lim( x2 − 3x + 1) = 11
x →5
4. lim ( x2 + 2x + 7) = 7
x →−2
3
5
x − 2x − 4
5. lim
=
x →2
x2 − 4
2
4.1.1
Properties of limits
Theorem 4.1.
If lim f ( x ) = A,
x→a
lim g( x ) = B, and h( x ) = k. Then
x→a
51
52
calculus
(a) lim h( x ) = k.
x→a
(b) lim[ f ( x ) + g( x )] = lim f ( x ) + lim g( x ) = A + B.
x→a
x→a
x→a
(c) lim[ f ( x ) − g( x )] = lim f ( x ) − lim g( x ) = A − B.
x→a
x→a
x→a
h
i h
i
(d) lim [ f ( x ) · g( x )] = lim f ( x ) · lim g( x ) = AB.
x→a
x→a
x→a
lim f ( x )
f (x)
A
= x→a
= , provided B ̸= 0.
x→a g( x )
lim g( x )
B
(e) lim
x→a
(f) lim c f ( x ) = cA, c being any constant.
x→a
n
(g) lim [ f ( x )] = An , where n is a positive integer.
x→a
q
√
n
(h) lim n f ( x ) = A, where n is a positive integer.
x→a
(i) lim b f (x) = b A where b is a constant.
x→a
Proof:
We will only prove (b) and (c) in this module.
(b) To show that lim [ f ( x ) + g( x )] = ( A + B), we need to show that for each ε > 0 there exists δ > 0
x→a
such that
[ f ( x ) + g( x )] − ( A + B) < ε
We know that lim f ( x ) = A means that for each ε > 0 there exists δ1 > 0 such that
x→a
g( x ) − B <
ε
2
whenever | x − a| < δ1 . Also lim g( x ) = B means that for each ε > 0 there exists δ2 > 0 such that
x→a
g( x ) − B <
ε
2
whenever | x − a| < δ2 . Now, taking δ = min{δ1 , δ2 } then
[ f ( x ) + g( x )] − ( A + B) = [ f ( x ) − A] + [ f ( x ) − B]
< [ f ( x ) − A] + [ f ( x ) − B]
ε
ε
= +
2 2
=ε
Thus lim [ f ( x ) + g( x )] = ( A + B).
x→a
(c) Proof that lim [ f ( x ) − g( x )] = ( A − B) and is left as an exercise
x→a
limits of functions
The rest of the proofs require slightly more work will not be covered in this module.
Example 4.7.
Evaluate the following limits
(a) lim( x + 3)
x →1
1
,
x →1 x + 2
(b) lim
(c) lim( x2 − 7x + 5).
x →8
x2 − 4
x →2 x − 2
(d) lim
Solution.
1. lim( x + 3) = lim x + lim 3 = 1 + 3 = 4,
x →1
x →1
x →1
lim 1
1
1
1
1
x →1
=
=
=
= ,
x →1 x + 2
limx→1 ( x + 2)
lim x + lim 2
1+2
3
2. lim
x →1
2
x →1
2
3. lim( x − 7x + 5) = 8 − 7(8) + 5 = 13
x →8
( x + 2)( x − 2)
x2 − 4
= lim
= lim( x + 2) = 4.
x →2
x →2
x →2 x − 2
x−2
4. lim
We can use properties of limits to simplify more complicated limits.
Exercise 4.2.
Simplify the following limits
1. lim( x4 + 3x − 5)
x →2
x4 + x2 − 1
x →1
x2 + 5
√
3. lim 2x2 − 1
2. lim
x →5
4. lim (3x + 4)3
x →−2
2
5. lim ex +3x−2
x →1
Not all limits can be found using the above properties. Some examples are given below
Example 4.8.
(a) lim
x−2
x →2 x 2 − 4
■
53
54
calculus
sin θ − 12
θ − π6
θ→ 6
(b) limπ
x − 8x2
x →∞ 12x 2 + 5x
(c) lim
1
(d) lim (ln x ) x
x →∞
1
(e) lim(ex + x ) x
x →0
(f) lim x x
x →0+
√
(g) lim
h →0
5+h−
h
√
5
In the table below, the second column shows the “limits” if we naively apply the properties above. The
third column shows the correct limit.
Example
lim
“Limit” using limit properties
Correct limit
0
0
0
0
1
4
√
3
2
−∞
∞
− 23
∞0
1
1∞
e2
00
1
0
0
1
√
2 5
x−2
x →2 x 2 − 4
sin θ − 21
θ − π6
θ→ 6
x − 8x2
lim
x →∞ 12x 2 + 5x
1
lim (ln x ) x
limπ
x →∞
lim (ln x )
x →∞
1
x
lim x x
x →0+√
lim
h →0
5+h−
h
√
5
0
These limits of the form 00 , ∞
∞ , 0 , etc. are said to be of the indeterminate form. We can immediately
solve (a) and (h) by modifying the problem.
(a)
x−2
x−2
= lim
2
x →2 x − 4
x →2 ( x − 2)( x + 2)
1
= lim
x →2 x + 2
1
=
4
lim
limits of functions
(h)
√
lim
h →0
5+h−
h
√
5
√ √
√
5+h− 5
5+h+ 5
√
= lim
·√
h →0
h
5+h+ 5
√ √
√
√
( 5 + h − 5)( 5 + h + 5)
√
√
= lim
h →0
h ( 5 + h + 5)
5+h−5
√
= lim √
h →0 h ( 5 + h +
5)
h
√
= lim √
h →0 h ( 5 + h +
5)
1
√
= lim √
h →0
5+h+ 5
1
√
= √
( 5 + 5)
1
= √
2 5
√
We will revisit the evaluation of the rest of the limits in Section 6.3 on the L’Hôpitals Rule.
4.1.2
Right-hand and Left-hand Limits
So far we have considered the limit of a function f as x → a from either side of a. We can also compute
limits as x → a from from only one side. If the approach is from the right, the limit is a right-hand
limit or limit from the right and is denoted by lim f ( x ). Similarly, a left-hand limit is also called a
x → a+
limit from the left and is denoted by lim f ( x ). The definitions of such limits are
x → a−
Definition 4.2 (Right-hand limit).
Let f be a function. Then a real number L is called the right-hand limit of f at a if for any chosen real number
ε > 0, there exists a positive number δ (possibly depending on ε), such that
| f ( x ) − L| < ε whenever a < x < a + δ
Definition 4.3 (left-hand limit).
Let f be a function. Then a real number L is called the left-hand limit of f at a if for any chosen real number
ε > 0, there exists a positive number δ (possibly depending on ε), such that
| f ( x ) − L| < ε whenever a − δ < x < a
Example 4.9.
Consider the following function.
f (x) =
The graph of this function is given by

− x2 + 1,
x≤0
 x + 2,
x>0
55
56
calculus
y
x
From the graph we can see that
lim f ( x ) = lim (− x2 + 1) = 0 + 1 = 1
x →0−
x →0−
lim f ( x ) = lim ( x + 2) = 0 + 2 = 2
x →0+
x →0+
Example 4.10.
Consider the following function.
g( x ) =

 x2 − 1,
x<0
 x − 1,
x>0
The graph of this function is given by
y
x
limits of functions
In this case we have
lim g( x ) = lim ( x2 − 1) = 0 − 1 = −1
x →0−
x →0−
lim g( x ) = lim ( x − 1) = 0 − 1 = −1
x →0+
x →0+
In Example 4.9 the left and right limits are unequal and in Example 4.10 the limits are equal. Clearly,
when the left-hand limit and right-hand limits are equal, the limit exist and where the left-hand and
right-hand limits are unequal the limit does not exist. We put this result in the following theorem.
Theorem 4.2.
Let f be a function. Then we have
lim f ( x ) = L
x→a
if, and only if, both one-sided limits exist and be equal to L, that is,
lim f ( x ) = L
x → a−
lim f ( x ) = L
x → a+
Example 4.11.
|x|
does not exist.
Show that lim
x →0 x
Solution.
Notice that

 x = 1,
if
x≥0
x
if
x < 0.
|x|
= x
 −x = −1,
x
Thus
|x|
= lim (−1) = −1.
x
x →0−
|x|
= lim (1) = 1
lim
x →0+
x →0+ x
lim
x →0−
Since
lim
x →0+
it means that lim
x →0
|x|
|x|
̸= lim
x
x →0− x
|x|
does not exist.
x
We now look at a theorem used to find some complicated limits.
Theorem 4.3 (Squeeze Theorem).
Let f , g and h be functions such that f ( x ) ≤ g( x ) ≤ h( x ) for all x near a, and suppose
lim f ( x ) = L
x→a
and
lim h( x ) = L
x→a
Then,
lim g( x ) = L
x→a
57
58
calculus
This theorem is sometimes called the Sandwich Theorem. It states that if a a function h is sandwiched
between two functions which are approaching L then h must also approach L.
Example 4.12.
1. Suppose g is a function such that for all x
2x ≤ g( x ) ≤ x4 − x2 + 2
Evaluate lim g( x )
x →1
2. Evaluate the following limits.
(a) lim sinθ θ
θ →0
(b) lim
√
x →0+
xesin( x )
π
Solution.1. Observe that
lim 2x = 2(1) = 2
x →1
4
lim x − x2 + 2 = (1)4 − (1)2 + 2 = 2
x →1
Therefore by the Squeeze theorem we have lim g( x ) = 2
x →1
2. Prove that
(a) Consider the two sectors given below. Triangle ABC is right-angled with AB = 1 and angle θ is in radians.
(b) Observe that
thus
√
x ≥ 0 for all x ≥ 0. Therefore,
√
xesin( x ) ≥ 0. On the other hand we have sin
π
esin( x ) ≤ e1
π
π
x
≤ 1,
limits of functions
Therefore
√
√
π
xesin( x ) ≤ e x
√
√
π
xesin( x ) ≤ e x
and this leads to
0≤
Set f ( x ) = 0, g( x ) =
observe that
√
√
π
xesin( x ) and h( x ) = e x. Clearly, we have f ( x ) ≤ g( x ) ≤ h( x ). Next, we
lim f ( x ) = lim 0 = 0
x →0+
x →0+
and
√
lim h( x ) = lim e x = 0
x →0+
x →0+
Therefore by the Squeeze theorem we have
lim g( x ) = lim
x →0+
√
x →0+
xesin( x ) = 0
π
A direct consequence of Example 4.12 2(a) we have the following result
Example 4.13.
Evaluate the following limit
lim
θ →0
1 − cos θ
θ
Solution.
lim
θ →0
1 − cos θ
θ
= lim
θ →0
1 − cos θ 1 + cos θ
·
θ
1 + cos θ
1
1 − cos2 θ
·
θ →0
θ
1 + cos θ
2
sin θ
1
= lim
·
θ →0
θ
1 + cos θ
sin θ
sin θ
= lim
·
θ →0
θ
1 + cos θ
sin θ
sin θ
· lim
By properties of limits
= lim
θ →0
θ →0
θ
1 + cos θ


lim sin θ
sin θ
 By properties of limits
= lim
·  θ →0
θ →0
θ
lim(1 + cos θ )
= lim
θ →0
= 1·0
=0
sin θ
Since lim
=1
θ →0 θ
59
5 Continuity
We now go on to look at at one of the most important concept regarding real numbers: That is the
subject of continuity.
Notice in the previous chapter that the definition of limit does not require the function to be defined
at the point of the limit for example if
f (x) =
sin x
x
Then
lim f ( x ) = lim
x →0
x →0
sin x
=1
x
but f is undefined at x = 0. In some cases the function is defined at point but lim f ( x ) ̸= f (c) An
x →c
example of this given below
Example 5.1.
Consider the following function.
h( x ) =
The graph of this function is given by

2

x ,

x<1
2,
x=1



x,
x>1
62
calculus
y
x
Notice that
lim h( x ) = lim ( x2 ) = 12 = 1
x →2−
x →1−
lim h( x ) = lim ( x ) = 1
x →2+
x →1+
Since left-hand and right-hand limits exist and are both equal to 1. It means the
lim h( x ) = 1
x →1
However, h( x ) ̸= 1 i.e. limx→1 h( x ) ̸= h(1). This is because there is a discontinuity at x = 1.
This brings us to the following definition of continuity.
Definition 5.1 (Continuity).
A function f ( x ) is continuous at a point x = c if the following conditions are satisfied.
(a) f (c) is defined.
(b) lim f ( x ) exists.
x →c
(c) lim f ( x ) = f (c).
x →c
The function in Example 5.1 passes at Condition [(a)] since h(1) is defined. It also passes Condition
[(b)] since lim h( x ) exists, but fails at Condition [(c)] since lim h( x ) = 1 ̸= h(1) since h(1) = 2
x →1
x →1
Example 5.2.
Determine whether the following functions are continuos at the point x = −1.
continuity
1. f ( x ) = x
2. f ( x ) = | x + 1|
3.

1
x < −1
 −1
x ≥ −1

−| x |
x ̸ = −1
1
x = −1
f (x) =
4.
f (x) =
5. f ( x ) = x+1 1
Solution.1. f ( x ) = x
y
x
All three conditions pass so f is continuous.
2. f ( x ) = | x + 1|
y
x
All three conditions pass so f is continuous.
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calculus
3. f ( x ) =

1
x < −1
 −1
x ≥ −1
y
x
Condition [(a)] passes but conditions [(b)] and [(c)] fail. Therefore f is not continuous.

−| x |
x ̸ = −1
4. f ( x ) =
1
x = −1
y
x
Condition [(a)] and [(b)] pass but conditions [(c)] fails. Therefore f is not continuous.
5. f ( x ) = x+1 1 All three conditions fail. Therefore f is not continuous.
Definition 5.2 (Discontinuous functions).
A function f is discontinuous at x = c if one or more of the conditions for continuity in Definition 5.1 fails at
c.
The conditions in Definition 5.1 allow us to use ε − δ notions rephrase the definition of continuity.
continuity
Definition 5.3 (ε − δ definition of continuity).
A function f is continuous at point c, for each ε > 0, there exist δ > 0 such that if
| x − c| < δ then | f ( x ) − f (c)| < ε
Example 5.3.
Use the ε − δ definition to show that the following functions are continuous at the given points c.
(a) f ( x ) = 3x + 2,
(b) f ( x ) = x2 ,
c=2
c=1
(c) f ( x ) = x2 + 2x + 1,
√
(d) f ( x ) = x, c = 4
c=2
Solution.
1. First, note that f (2) = 3(2) + 2 = 8. We need to show that for each ε > 0 there exist δ > 0 such that
| f ( x ) − f (2)| < ε whenever | x − 2| < δ
To proceed, we have
| f ( x ) − f (2)| = |(3x + 2) − 8|
= |3x − 6|
= 3| x − 2|
< 3δ
Take δ = 3ε . Then whenever | x − 2| < δ, we have
| f ( x ) − f (2)| = |(3x + 2) − 8| < 3δ = 3 ·
ε
=ε
3
Thus f ( x ) = 3x + 2 is continuous at x = 2.
2. We want to show that for each ε > 0 there exist δ > 0 such that if
| x − 1| < δ then | f ( x ) − f (1)| < ε.
To begin, we have
| f ( x ) − f (1)| = | x2 − (1)2 |
= | x 2 − 1|
= |( x − 1)( x + 1)|
= | x − 1|| x + 1|
< | x + 1| δ
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calculus
To find a bound for | x + 1| we proceed as follows; we first take δ = 1 then
| x − 1| < 1
−1 < x − 1 < 1
0<x<2
1 < x+1 < 3
1 < | x + 1| < 3
Therefore, using the above, we have
| f ( x ) − f (1)| = | x2 − (1)2 |
< | x + 1| δ
< 3δ
Take δ = min
ε
3, 1
. Then whenever | x − 1| < δ, we have
| f ( x ) − f (1)| = | x2 − (1)2 | < 3δ = 3 ·
ε
=ε
3
Therefore f ( x ) = x2 is continuous at x = 1.
3. We want to show that for each ε > 0 there exist δ > 0 such that if
| x − 2| < δ then
f ( x ) − f (2)| < ε.
To begin, we have
| f ( x ) − f (2)| = | x2 + 2x + 1 − (2)2 + 2(2) + 1 |
= | x2 + 2x + 1 − 9|
= | x2 + 2x − 8|
= |( x + 4)( x − 2)|
= | x + 4|| x − 2|
< | x + 4| δ
To find a bind for | x + 4| we proceed as follows; we first take δ = 1 i.e. | x − 2| < 1 then
−1 < x − 2 < 1
1<x<3
5 < x+4 < 7
5 < | x + 4| < 7
Therefore, using the above, we have
| f ( x ) − f (2)| = | x + 4|δ
< 7δ
Take δ = min
ε
7, 1
. Then whenever | x − 2| < δ, we have
| f ( x ) − f (2)| < 7δ = 7 ·
Therefore f ( x ) = x2 + 2x + 1 is continuous at x = 2.
ε
=ε
7
continuity
4. We want to show that for each ε > 0 there exist δ > 0 such that if
| x − 4| < δ then | f ( x ) − f (4)| < ε.
To begin, we have
√
√
| f ( x ) − f (4)| = | x − 4|
√
√
√
√
( x − 4) · ( x + 4)
√
=
√
x+ 4
x−4
√
x+ 4
| x − 4|
√
= √
x+ 4
δ
√
< √
| x + 4|
= √
To find a bind for √x+1 √4 we proceed as follows; we first take δ = 1 i.e. | x − 4| < 1 then
−1 < x − 4 < 1
3<x<5
√
√
3< x< 5
√
√
√
√
√
√
3+ 4 < x+ 4 < 5+ 4
√
√
1
1
1
√ < √
√ < √
√
5+ 4
x+ 4
3+ 4
1
1
1
√
< √
< √
| x + 2|
5+2
3+2
Therefore, using the above, we have
| f ( x ) − f (4)| = √
< √
Take δ =
√
δ
x+2
δ
3+2
3 + 2 ε. Then whenever | x − 4| < δ, we have
√
1
δ
= √
·
| f ( x ) − f (4)| < √
3+2 ε = ε
| 3 + 2|
3+2
Therefore f ( x ) =
√
x is continuous at x = 4.
Theorem 5.1 (Properties of continuity).
Let f and g are continuous at x = a. Then the following are also continuous at x = a
1. k · f , where k is a constant.
2. f + g,
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calculus
3. f − g,
4. f · g,
5.
f
g provided g ( a ) ̸ = 0.
Proof: Prove (b) and (c) as an exercise. The rest will ot be proved in this module.
■
Example.
Let f ( x ) = 3x + 2 and g( x ) = x2 + 2x + 1. Both functions are continuous at x = 2 [see Examples (5.3) and
(5.3)]. Then the following are also continuous at x = 2
1. (3x + 2) − ( x2 + 2x + 1) = − x2 + x + 1
2. (3x + 2)( x2 + 2x + 1)
3.
3x +2
x2 +2x +1
Theorem 5.2 (Composite function of continuous functions).
Let f and g be continuous functions at x = a. Then the composition of two continuous functions i.e f ◦ g or
g ◦ f are both continuous at x = a.
Recall that ( f ◦ g)( x ) = f ( g( x ))
Example.
Let f ( x ) = 3x + 2 and g( x ) = x2 + 2x + 1. Both functions are continuous at x = 2 [see Examples (5.3) and
(5.3)]. Then
( f ◦ g)( x ) = f ( x2 + 2x + 1) = 3( x2 + 2x + 1) + 2 = 3x2 + 6x + 5 is continuous as x = 2
( g ◦ f )( x ) = g(3x + 2) = (3x + 2)2 + 2(3x + 2) + 1 = · · ·
is continuous as x = 2
Theorem 5.3 (Continuity on an interval).
A function f is continuous on the interval I if f is continuous at every point a ∈ I
Example.
1. The function f ( x ) = x2 is continuous on the interval [−3, 5] since it is continuous at every point on the
interval l [−3, 5]. In fact f is continuous on R.
2. The function g( x ) = x12 is not continuous on the interval [−3, 5] since it is not continuous at x = 0. However,
g is continuous on R \ {0}.
Theorem 5.4 (Intermediate value theorem).
Suppose that f is continuous on the closed interval [ a, b]and let M be any number between f ( a) and f (b) where
f ( a) ̸= f (b). Then there exists a number c in ( a, b) such that f (c) = M
In simple terms the intermediate value theorem states that a continuous function takes on every intermediate value between the function values f ( a) and f (b).
continuity
Example 5.4.
Show that
ln x = x −
√
x
has a solution on the interval (2, 3).
Solution.
√
Let f ( x ) = ln x − x + x. The function f is continuous on the interval [2, 3]. Further,
√
f (2) = ln 2 − 2 + 2 = 0.10736 . . .
√
f (3) = ln 3 − 3 + 3 = −0.16933 . . .
Therefore
f (2) ̸ = f (3)
Take M between f (2) and f (3). Then by the Intermediate value theorem there exist c ∈ (2, 3) such that
√
f (c) = M. Setting M = 0 implies there exists c ∈ (2, 3) such that f (c) = 0 i.e. ln x = x − x has a root in
(2, 3)
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calculus
70
Exercise 5.1.
1. Use the ε − δ definition to prove the following limits
(a)
lim
x →0+
√
x=0
(b)
lim 3x2 − 7x + 1 = 7
x →3
2. Suppose the f is defined as follows


x2 ,





 x − 1.

√
f (x) =
x




−1,



1,
x = −1
−1 < x < 0
0<x<1
x=1
x>1
Draw the graph of this function and use it to analyze the following limits if they exist
(a) lim f ( x )
x →−1
(b) lim f ( x )
x →1
(c) lim f ( x )
x →0+
(d) lim f ( x )
x →2
3. Use the ε − δ definition to prove that the following functions are continuous at the given value of c
(a) f ( x ) = x2 − 2x + 3,
2
(b) f ( x ) = x + x − 11,
c = −1,
c = 4.
4. For what values of a is the function f continuous for all x

 a2 x − 2a,
f (x) =
12,
x≥2
x<2
5. Use the intermediate value theorem to show that following functions have roots on the given interval
(a) x3 − x + 1,
[−1, 2],
(b) x3 − 2x + 2,
[−2, 0]
sin x
= 1 to prove the following limits
x →0 x
6. Use the property lim
sin 2x
5x
sin x (1 − cos x )
(b) lim
x →0
x
(a) lim
x →0
continuity
sin x
√
x →0 sin
x
(c) lim
7. Suppose f and g be continuous at x = c. Prove that
(a) f + g is continuous at x = c,
(b) f − g is continuous at x = c.
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