calculus notes of m t nair

Calculus - I

M.Thamban Nair

Department of MathematicsIndian Institute of Technology Madras

June 2006/Aug-Dec. 2010/Aug-Dec.2011

Contents

Preface vi

1 Sequence and Series of Real Numbers M.T. Nair 1

1.1 Sequence of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Convergence and divergence . . . . . . . . . . . . . . . . . . . 2

1.1.2 Monotonic sequences . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.3 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.4 Further examples . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.5 Cauchy sequence . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.1.6 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . 19

1.2 Series of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.2.1 Convergence and divergence of series . . . . . . . . . . . . . . 21

1.2.2 Some tests for convergence . . . . . . . . . . . . . . . . . . . 23

1.2.3 Alternating series . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.2.4 Absolute convergence . . . . . . . . . . . . . . . . . . . . . . 27


2 Limit, Continuity and Differentiability of Functions M.T. Nair 32

2.1 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.1.1 Limit point of a set D R . . . . . . . . . . . . . . . . . . . 322.1.2 Limit of a function f(x) as x approaches a . . . . . . . . . . 33

Limit of a function in terms of a sequences . . . . . . . . . . 35

2.1.3 Some properties . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.1.4 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 37

ii

Contents iii

2.1.5 Left limit and right limit . . . . . . . . . . . . . . . . . . . . 39

2.1.6 Limit at and at . . . . . . . . . . . . . . . . . . . . . 402.1.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 42

2.2 Continuity of a Function . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.2.1 Definition and some basic results . . . . . . . . . . . . . . . . 43

2.2.2 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.2.3 Exponential and logarithm functions . . . . . . . . . . . . . . 46

2.2.4 Some properties of continuous functions . . . . . . . . . . . . 51

Attaining max f and min f . . . . . . . . . . . . . . . . . . . 52

Intermediate value theorem . . . . . . . . . . . . . . . . . . . 52

2.2.5 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 53

2.3 Differentiability of functions . . . . . . . . . . . . . . . . . . . . . . . 54

2.3.1 Some properties of differentiable functions . . . . . . . . . . . 55

2.3.2 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 57

2.3.3 Maxima and minima . . . . . . . . . . . . . . . . . . . . . . . 59

2.3.4 Some important theorems . . . . . . . . . . . . . . . . . . . . 61

Rolles theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Lagranges mean value theorem . . . . . . . . . . . . . . . . . 62

Cauchys generalized mean value theorem . . . . . . . . . . . 63

LHospitals rules . . . . . . . . . . . . . . . . . . . . . . . . . 64

Taylors formula . . . . . . . . . . . . . . . . . . . . . . . . . 66

2.3.5 Increasing and decreasing functions . . . . . . . . . . . . . . . 70

2.3.6 More about local maxima and local minima . . . . . . . . . . 71


3 Definite Integral M.T. Nair 74

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.2 Lower and Upper Sums . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.3 Integrability and Integral . . . . . . . . . . . . . . . . . . . . . . . . 76

3.3.1 Some necessary and sufficient conditions for integrability . . . 78

iv Contents

3.4 Integral of Continuous Functions . . . . . . . . . . . . . . . . . . . . 83

3.5 Some Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.6 Some Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

3.6.1 First fundamental theorem . . . . . . . . . . . . . . . . . . . 88

3.6.2 Second fundamental theorem . . . . . . . . . . . . . . . . . . 88

3.6.3 Applications of fundamental theorem . . . . . . . . . . . . . . 90

3.7 Appendix M.T. Nair . . . . . . . . . . . . . 92

4 Improper Integrals M.T. Nair 93

4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

4.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4.2 Integrability by Comparison . . . . . . . . . . . . . . . . . . . . . . . 96

4.2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

4.2.2 Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . 98

4.3 Integrability Using Limits . . . . . . . . . . . . . . . . . . . . . . . . 99

4.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5 Geometric and Mechanical Applications of Integrals M.T. Nair101

5.1 Computing Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.1.1 Using Cartesian Coordinates . . . . . . . . . . . . . . . . . . 101

5.1.2 Using Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 102

5.1.3 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.2 Computing Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.2.1 Using Cartesian Coordinates . . . . . . . . . . . . . . . . . . 103

5.2.2 Using Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 104

5.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.3 Computing Volume of a Solid . . . . . . . . . . . . . . . . . . . . . . 106

5.4 Computing Volume of a Solid of Revolution . . . . . . . . . . . . . . 107

5.5 Computing Area of Surface of Revolution . . . . . . . . . . . . . . . 108

5.6 Centre of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5.6.1 Centre of gravity of a material line in the plane . . . . . . . . 109

Contents v

5.6.2 Centre of gravity of a material planar region . . . . . . . . . 110

5.7 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.7.1 Moment of inertia of a material line in the plane . . . . . . . 111

5.7.2 Moment of inertia of a circular arc with respect to the centre 112

5.7.3 Moment of inertia of a material sector in the plane . . . . . . 113


6 Sequence and Series of Functions M.T. Nair116

6.1 Sequence of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.1.1 Pointwise Convergence and Uniform Convergence . . . . . . . 116

6.1.2 Continuity and uniform convergence . . . . . . . . . . . . . . 121

6.1.3 Integration and uniform convergence . . . . . . . . . . . . . . 122

6.1.4 Differentiation and Uniform Convergence . . . . . . . . . . . 122

6.2 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123


7 Power Series M.T. Nair128

7.1 Convergence and Absolute convergence . . . . . . . . . . . . . . . . . 128

7.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 130

7.3 Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . 130

7.3.1 Series that can be converted into a power series . . . . . . . . 132


8 Fourier Series M.T. Nair133

8.1 Fourier Series of 2pi-Periodic functions . . . . . . . . . . . . . . . . . 133

8.1.1 Fourier series and Fourier coefficients . . . . . . . . . . . . . . 133

8.1.2 Even and odd expansions . . . . . . . . . . . . . . . . . . . . 135

8.2 Fourier Series of 2`-Periodic Functions . . . . . . . . . . . . . . . . . 138

8.2.1 Fourier series of functions on arbitrary intervals . . . . . . . . 138


Preface

This is based on a course that the author gave to B.Tech. students of IIT Madrasmany times since 1996 for the use of students and teachers of IIT Madras.

Comments and suggestions from the readers are welcome.

June 2006 M. T. Nair

[email protected]

Revised: August-November 2010

Revised: August-November 2011

vi

1Sequence and Series of Real Numbers

1.1 Sequence of Real Numbers

Suppose for each positive integer n, we are given a real number an. Then, the listof numbers,

a1, a2, . . . , an, . . . ,

is called a sequence, and this ordered list is usually written as

(a1, a2, . . . , . . .) or (an).

More precisely, we define a sequence as follows:

Definition 1.1 A sequence of real numbers is a function from the set N of naturalnumbers to the set R of real numbers. If f : N R is a sequence, and if an = f(n)for n N, then we write the sequence f as (an) or (a1, a2, . . .).

A sequence of real numbers is also called a real sequence.

Remark 1.1 (a) It is to be born in mind that a sequence (a1, a2, . . . , . . .) is differentfrom the set {an : n N}. For instance, a number may be repeated in a sequence(an), but it need not be written repeatedly in the set {an : n N}. As an example,(1, 1/2, 1, 1/3, . . . , 1, 1/n, . . .) is a sequence (an) with a2n1 = 1 and a2n = 1/(n+ 1)for each n N, where as the set {an : n N} is same as the set {1/n : n N}.

(b) Instead of sequence of real numbers, we can also talk about a sequence ofelements from any nonempty set S, such as sequence of sets, sequence of functionsand so on. Thus, given a nonempty set S, a sequence in S is a function f : N S.For example, for each n N, consider the set An = {j N : j n}. Then weobtain a sequence of subsets of N, namely, (A1, A2, . . .).

In this chapter, we shall consider only sequence of real numbers. In some of thelater chapters we shall consider sequences of functions as well.

EXAMPLE 1.1 (i) (an) with an = 1 for all n N a constant sequence withvalue 1 throughout.

(ii) (an) with an = n for all n N.(iii) (an) with an = 1/n for all n N.

1

2 Sequence and Series of Real Numbers M.T. Nair

(iv) (an) with an = n/(n+ 1) for all n N.(v) (an) with an = (1)n for all n N the sequence takes values 1 and 1

alternately.

Question: Consider a sequence (a1, a2, . . .). Is (a2, a3, . . .) also a sequence? Why?

1.1.1 Convergence and divergence

A fundamental concept in mathematics is that of convergence. We consider conver-gence of sequences.

Consider the sequences listed in Example 1.1 and observe the way an vary as nbecomes larger and larger:

(i) an = 1: every term of the sequence is same.

(ii) an = n: the terms becomes larger and larger.

(iii) an = 1/n: the terms come closer to 0 as n becomes larger and larger.

(iv) an = n/(n+ 1): the terms come closer to 1 as n becomes larger and larger.

(v) an = (1)n: the terms of the sequence oscillates between 1 and 1, and doesnot come closer to any number as n becomes larger and larger.

Now, we make precise the statement an comes closer to a number a as nbecomes larger and larger.

Definition 1.2 A sequence (an) in R is said to converge to a real number a if forevery > 0, there exists a natural number N (in general depending on ) such that

|an a| < n N,

number a is called the limit of the sequence (an).

Remark 1.2 (a) Note that, different can result in different N , i.e., the numberN may vary as varies. We shall illustrate this in Example 1.2.

(b) In Definition 1.2, the relation |an a| < can be replaced by |an a| < c0for any c0 > 0 which is independent of n. In other words, a sequence (an) in Rconverges to a R if and only if for any c0 > 0 and for every > 0, there existsN N such that |an a| < c0 for all n N .

If (an) converges to a, then we write

limn an = a or an a as n

or simply as

an a.

Sequence of Real Numbers 3

Note that|an a| < n N

if and only ifa < an < a+ n N.

Thus, limn an = a if and only if for every > 0, there exists N N such that

an (a , a+ ) n N.

Thus, an a if and only if for every > 0, an belongs to the open interval(a, a+) for all n after some finite stage, and this finite stage may vary accordingas varies.

Remark 1.3 Suppose (an) is a sequence and a R. The to show that (an) doesnot converge to a, we should be able to find an > 0 such that infinitely may ansare outside the interval (a , a+ ). Exercise 1.1 Show that a sequence (an) converges to a if and only if for every openinterval I containing x, there exits N N such that an I for all n N . J

EXAMPLE 1.2 The sequences (1/n), ((1)n/n), (1 1n) are convergent withlimit 0, 0, 1 respectively:

For the sake of illustrating how to use the definition to justify the above state-ment, let us provide the details of the proofs:

(i) Let an = 1/n for all n N, and let > 0 be given. We have to identify anN N such that 1/n < for all n N . Note that

1

n< n > 1

.

Thus, if we take N =

1

+ 1, then we have

|an 0| = 1n< n N.

Hence, (1/n) converges to 0.

Here dxe denotes the integer part of x.

(ii) Next, let an = (1)n/n for all n N. Since |an| = 1/n for all n N, in thiscase also, we see that

|an 0| < n N :=

1

+ 1.

Hence, ((1)n/n) converges to 0.


(iii) Now, let an = 1 1n for all n N. Since, |an 1| = 1/n for all n N, wehave

|an 1| < n N :=

1

+ 1.

Hence, (1 1/n) converges to 0.

EXAMPLE 1.3 Every constant sequence is convergent to the constant term inthe sequence.

To see this, let an = a for all n N. Then, for every > 0, we have

|an a| = 0 < n N := 1.

Thus, (an) converges to a.

EXAMPLE 1.4 For a given k N, let Let an = 1/n1/k for all n N. Then an 0as n.

To see this, first let > 0 be given. Note that

1

n1/k< n > 1

k.

Hence,1

n1/k< n N :=

1

k

+ 1.

Thus, 1/n1/k 0. Exercise 1.2 Corresponding to a sequence (an) and k N, let (bn) be defined bybn = ak+n for all n N. Show that, for a R, an a if and only if bn a. JExercise 1.3 A sequence (an) is said to be eventually constant if there existsk N such that ak+n = ak for all n 1. Show that every eventually constantsequence converges. J

Theorem 1.1 Limit of a convergent sequence is unique. That, is if an a andan a as n, then a = a.

Proof. Suppose an a and an a as n, and suppose that a 6= a. Now,for > 0, suppose N1, N2 N be such that

an (a , a+ ) n N1, an (a , a + ) n N2.

In particular,

an (a , a+ ), an (a , a + ) n N := max{N1, N2}.

If we take < |a a|/2, then we see that (a , a + ) and (a , a + ) aredisjoint intervals. Thus the above observation leads to a contradiction.


An alternate proof. Note that

|a a| = |(a an) + (an a)| |a an|+ |an a|.Now, for > 0, let N1, N2 N be such that

|a an| < /2 for all n N1, |a an| < /2 n N2.Then it follows that

|a a| |a an|+ |an a| < n N := max{N1, N2}.Since this is true for all > 0, it follows that a = a.

Exercise 1.4 Prove the following.

1. Let b 0 such that b < for all > 0. Then b = 0.2. Let an 0 for all n N such that an a. Then a 0.

J

The following theorem can be easily proved.

Theorem 1.2 Suppose an a, bn b as n. Then we have the following :(a) an + bn a+ b as n,(b) For every real number c, can c a as n(c) If an bn for all n N, then a b.(d) (Sandwitch theorem) If an cn bn for all n N, and if a = b, then

cn a as n.Exercise 1.5 Prove Theorem 1.2. J

Exercise 1.6 If an a and there exists b R such that an b for all n N, thenshow that a b. JExercise 1.7 If an a and a 6= 0, then show that there exists k N such thatan 6= 0 for all n k. J

Exercise 1.8 Consider the sequence (an) with an =

(1 +

1

n

)1/n, n N. Then

show that limn an = 1.

[Hint: Observe that 1 an (1 + 1/n) for all n N.] J

Exercise 1.9 Consider the sequence (an) with an =1

nk, n N. Then show that

for any given k N, limn an = 0.

[Hint: Observe that 1 an 1/n for all n N.] J


Definition 1.3 A sequence which does not converge is called a divergentsequence.

Definition 1.4 (i) If (an) is such that for every M > 0, there exists N N suchthat

an > M n N,then we say that (an) diverges to +.

(ii) If (an) is such that for every M > 0, there exists N N such that an < Mfor all n N , then we say that (an) diverges to .

Definition 1.5 If (an) is such that anan+1 < 0 for every n N, that is an changessign alternately, then we say that (an) is an alternating sequence .

An alternating sequence converge or diverge. For example, (Verify that) thesequence ((1)n) diverges, whereas ((1)n/n) converges to 0.

Definition 1.6 A sequence (an) is said to be bounded above if there exists areal number M such that an M for all n N; and the sequence (an) is said to bebounded below if there exists a real number M such that an M for all n N.A sequence which is bound above and bounded below is said to be a boundedsequence.

Exercise 1.10 Show that a sequence (an) is bounded if and only if there existsM > 0 such that |an| M for all n N. JExercise 1.11 Prove the following.

1. If (an) is not bounded above, then there exists a strictly increasing sequence(kn) of natural numbers such that akn + as n.

2. If (an) is not bounded below, then there exists a strictly increasing sequence(kn) of natural numbers such that akn as n.

J

Theorem 1.3 Every convergent sequence is bounded. The converse is not true.

Proof. Suppose (an) converges to x. Then there exists N N such that |anx| 1 for all n N . Hence

|an| = |(an a) + a| |an a|+ |a| 1 + |a| n N,so that

|an| max{1 + |a|, |a1|, |a2|, . . . , |aN1|} n N.To see that the converse of the theorem is not true, consider the sequence ((1)n).It is a bounded sequence, but not convergent.


The above theorem can be used to show that certain sequence is not convergent,as in the following example.

EXAMPLE 1.5 For n N, let

an = 1 +1

2+

1

3+ . . .+

1

n.

Then (an) diverges: To see this, observe that

a2n = 1 +1

2+

1

3+ . . .+

1

2n

= 1 +1

2+

(1

3+

1

4

)+

(1

5+

1

6+

1

7+

1

8

)+

. . .+

(1

2n 1 + . . .+1

2n

) 1 + n

2.

Hence, (an) is not a bounded sequence, so that it diverges.

Using Theorem 1.3, the following result can be deduced

Theorem 1.4 Suppose an a and bn b as n. Then we have the following.(i) anbn ab as n.(ii) If bn 6= 0 for all n N and b 6= 0, then an/bn a/b as n.

Exercise 1.12 Prove Theorem 1.4. J

Exercise 1.13 If (an) converges to a and a 6= 0, then show that there exists k Nsuch that |an| |a|/2 for all n k, and (1/an+k) converges to 1/a. J

1.1.2 Monotonic sequences

We can infer the convergence or divergence of a sequence in certain cases by observ-ing the way the terms of the sequence varies.

Definition 1.7 Consider a sequence (an).

(i) (an) is said to be monotonically increasing if an an+1 for all n N.(ii) (an) is said to be monotonically decreasing if an an+1 for all n N.If strict inequality occur in (i) and (ii), then we say that the sequence is strictly

increasing and strictly decreasing, respectively.

Also, a monotonically increasing (respectively, monotonically decreasing) se-quence is also called an increasing (respectively, a decreasing) sequence.

We shall make use of some important properties of the set R of real numbers.


Definition 1.8 Let S be a subset of R. Then

(i) S is said to be bounded above if there exists b R such that x b for allx S, and in that case b is called an upper bound of S;

(ii)S is said to be bounded below if there exists a R such that a x for allx S, and in that case a is called a lower bound of S.

Definition 1.9 Let S be a subset of R.

(i) A number b0 R is called a least upper bound (lub) or supremum of S Rif b0 is an upper bound of S and for any upper bound b of S, b0 b.

(ii) A number a0 R is called a greatest lower bound (glb) or infimum of S Rif a0 is a lower bound of S and for any lower bound a of S, a a0.

Thus, we have the following:

b0 R is supremum of S R if and only if b0 is an upper bound of S and if < b0, then is not an upper bound of S, i.e., there exists x S such that < x. a0 R is infimum of S R if and only if a0 is a lower bound of S and if

> a0, then is not a lower bound of S, i.e., there exists x S such that x < .

The above two statements can be rephrased as follows:

b0 R is supremum of S R if and only if b0 is an upper bound of S and forevery > 0, there exists x S such that b0 < x b0. a0 R is infimum of S R if and only if a0 is a lower bound of S and for

every > 0, there exists x S such that a0 x < a0 + .Exercise 1.14 Show that supremum (respectively, infimum) of a set S R, ifexists, is unique. JExercise 1.15 Prove the following:

(i) If b0 is the supremum of S, then there exists a sequence (xn) in S whichconverges to b0.

(ii) If a0 is the infimum of S, then there exists a squence (xn) in S whichconverges to a0. J

EXAMPLE 1.6 (i) If S is any of the intervals (0, 1), [0, 1), (0, 1], [0, 1], then 1 isthe supremum of S and 0 is the infimum of S.

(ii) If S = { 1n : n N}, then 1 is the supremum of S and 0 is the infimum of S.(iii) For k N, if Sk = {n N : n k}, then k is the infimum of Sk, and Sk

has no supremum.

(iv) For k N, if Sk = {n N : n k}, then k is the supremum of Sk, and Skhas no infimum.


The above examples show the following:

The supremum and/or infimium of a set S, if exists, need not belong to S. If S is not bounded above, then S need not have supremum. If S is not bounded below, then S need not have infimum.However, we have the following properties for R:

Least upper bound property: If S R is bounded above, then S has a leastupper bound. We may write this least upper bound as lub(S) or sup(S).

Greatest lower bound property: If S R is bounded below, then S has agreatest lower bound, and we write it as glb(S) or inf(S).

Theorem 1.5 (i) Every sequence which is monotonically increasing and boundedabove is convergent.

(ii) Eery sequence which is monotonically decreasing and bounded below is con-vergent.

Proof. Suppose (an) is a monotonically increasing sequence of real numberswhich is bounded above. Then the set S := {an : n N} is bounded above.Hence, by the least upper bound property of R, S has a least upper bound, say b.Now, let > 0 be give. Then, by the definition of the least upper bound, thereexists N N such that aN > b . Since an aN for every n N , we get

b < aN an b < b+ n N.

Thus we have proved that an b as n.To see the last part, suppose that (bn) is a monotonically decreasing sequence

which is bounded below. Then, it is seen that the sequence (an) defined by an = bnfor all n N is monotonically increasing and bounded above. Hence, by the firstpart of the theorem, an a for some a R. Then, bn b := a.

Note that a convergent sequence need not be monotonically increasing or mono-tonically decreasing. For example, look at the sequence ((1)n/n).

1.1.3 Subsequences

Definition 1.10 A sequence (bn) is called a subsequence of a sequence (an) if thereis a strictly increasing sequence (kn) of natural numbers such that bn = akn for alln N.

Thus, subsequences of a real sequence (an) are of the form (akn), where (kn) isa strictly increasing sequence natural numbers.


For example, given a sequence (an), the sequences (a2n), (a2n+1), (an2), (a2n) aresome of its subsequences. As concrete examples, (1/2n), and (1/(2n + 1)), (1/2n)are subsequences of (1/n).

A sequence may not converge, but it can have convergent subsequences. Forexample, we know that the sequence ((1)n) diverges, but the subsequences (an)and (bn) defined by an = 1, bn = 1 for all n N are convergent subsequences of((1)n).

However, we have the following result.

Theorem 1.6 If a sequence (an) converges to x, then all its subsequences convergeto the same limit x.

Exercise 1.16 Prove Theorem 1.6. J

What about the converse of the above theorem? Obviously, if all subsequencesof a sequence (an) converge to the same limit x, then (an) also has to converge tox, as (an) is a subsequence of itself.

Suppose every subsequence of (an) has at least one subsequence which convergesto x. Does the sequence (an) converges to x? The answer is affirmative, as thefollowing theorem shows.

Theorem 1.7 If every subsequence of (an) has at least one subsequence which con-verges to x, then (an) also converges to x.

Proof. Proof is left as an exercise.

We have seen in Theorem 1.3 that every convergent sequence is bounded, but abounded sequence need not be convergent.

Question: For every bounded sequence, can we have a convergent subsequence?

The answer is in affirmative:

Theorem 1.8 (Bolzano-Weirstrass theorem). Every bounded sequence of realnumbers has a convergent subsequence.

Proof: For a first reading this proof can be omitted. 1 Let (an) be a bounded se-quence in R. For each k N, consider the set

Ek := {an : n k},

and let bk := supEk, k N. Clearly, b1 b2 . . ..1From the book: Mathematical Analysis: a straight forward approach by K.G. Binmore.


We consider the following two mutually exclusive cases:

(i) For every k N, bk Ek(ii) There exists k N such that bk 6 Ek.In case (i), we may write (bk) is a monotonically decreasing subsequence of (an),

and since (an) is bounded, (bk) converges.

Now, suppose that case (ii) holds, and let k N such that bk 6 Ek. Then forevery n k, there exists n > n such that an > an. Take n1 = k and n2 = n.Then we have n1 < n2 and an1 < an2 . Again, since n2 Ek, there exists n3 > n2such that an21 < an3 . Continuing this way, we obtain an increasing subsequence(anj )

j=1. Again, since (an) is bounded, (anj )

j=1 is bounded, so that it converges.

Remark 1.4 We may observe that the proof of Theorem 1.8 can be slightly mod-ified so that we, in fact, have the following: Every sequence in R has a monotonicsubsequence.

Exercise 1.17 Prove the statement in italics in Remark 1.4. J

1.1.4 Further examples

EXAMPLE 1.7 Let a sequence (an) be defined as follows :

a1 = 1, an+1 =2an + 3

4, n = 1, 2, . . . .

We show that (an) is monotonically increasing and bounded above.

Note that

an+1 =2an + 3

4=an2

+3

4 an an 3

2.

Thus it is enough to show that an 3/2 for all n N.Clearly, a1 3/2. If an 3/2, then an+1 = an/2 + 3/4 < 3/4 + 3/4 = 3/2.

Thus, we have proved that an 3/2 for all n N. Hence, by Theorem 1.5, (an)converges. Let its limit be a. Then taking limit on both sides of an+1 =

2an+34 we

have

a =2a+ 3

4i.e., 4a = 2a+ 3 so that a =

3

2.

EXAMPLE 1.8 Let a sequence (an) be defined as follows :

a1 = 2, an+1 =1

2

(an +

2

an

), n = 1, 2, . . . .

It is seen that, if the sequence converges, then its limit would be

2.


Since a1 = 2, one may try to show that (an) is monotonically decreasing andbounded below.

Note that

an+1 :=1

2

(an +

2

an

) an a2n 2, i.e., an

2.

Clearly a1

2. Now,

an+1 :=1

2

(an +

2

an

)

2 a2n 2

2an + 2 0,

i.e., if and only if (an

2)2 0. This is true for every n N. Thus, (an) is mono-tonically decreasing and bounded below, so that by Theorem 1.5, (an) converges.

EXAMPLE 1.9 Consider the sequence (an) with an = (1+1/n)n for all n N. We

show that (an) is monotonically increasing and bounded above. Hence, by Theorem1.5, (an) converges.

Note that

an =

(1 +

1

n

)n= 1 + n.

1

n+n(n 1)

2!

1

n2+ + n(n 1) . . . 2.1

n!

1

nn

= 1 + 1 +1

2!

(1 1

n

)+

1

3!

(1 1

n

)(1 2

n

)+

+ 1n!

(1 1

n

) (

1 n 1n

) an+1.

Also

an = 1 + 1 +1

2!

n(n 1)n2

+ + 1n!

n(n 1) . . . 2.1nn

1 + 1 + 12!

+1

3!+ + 1

n!

1 + 1 + 12

+1

22+ + 1

2n1< 3.

Thus, (an) is monotonically increasing bounded above.

Exercise 1.18 Show that (n1/n)n=3 is a monotonically decreasing sequence. J

In the four examples that follow, we shall be making use of Theorem 1.2 withoutmentioning it explicitly.


EXAMPLE 1.10 Consider the sequence (bn) with

bn = 1 +1

1!+

1

2!+

1

3!+ + 1

n!n N.

Clearly, (bn) is monotonically increasing, and we have noticed in the last examplethat it is bounded above by 3. Hence, by Theorem 1.5, it converges.

Let (an) and (bn) be as in Examples 1.9 and 1.10 respectively, and let a and btheir limits. We show that a = b.

We have observed in last example that 2 an bn 3. Hence, taking limits, itfollows that a b. Notice that

an = 1 + 1 +1

2!

(1 1

n

)+

1

3!

(1 1

n

)(1 2

n

)+

. . .+1

n!

(1 1

n

). . .

(1 n 1

n

).

Hence, for m,n with m n, we have

an 1 + 1 + 12!

(1 1

n

)+

1

3!

(1 1

n

)(1 2

n

)+

+ 1m!

(1 1

n

). . .

(1 m 1

n

).

Taking limit as n, we get (cf. Theorem 1.2 (c))

x 1 + 11!

+1

2!+

1

3!+ + 1

m!= bm.

Now, taking limit as m, we get a b. Thus we have proved a = b.The common limit in the above two examples is denoted by the letter e.

EXAMPLE 1.11 Let a > 0. We show that, if 0 < a < 1, then the sequence (an)converges to 0, and if a > 1, then (an) diverges to infinity.

(i) Suppose 0 < a < 1. Then we can write a = 1/(1 + r), r > 0, and we have

an = 1/(1 + r)n = 1/ (1 + nr + + rn) < 1/(1 + nr) 0 as n.

Hence, an 0 as n.

An alternate way: Let xn = an. Then (xn) is monotonically decreasing and bounded

below by 0. Hence (xn) converges, to say x. Then xn+1 = an+1 = axn ax. Hence,

x = ax. This shows that x = 0.

(ii) Suppose a > 1. Then, since 0 < 1/a < 1, the sequence (1/an) converges to0, so that (an) diverges to infinity. (Why ?)


EXAMPLE 1.12 The sequence (n1/n) converges and the limit is 1.

Note that n1/n = 1 + rn for some sequence (rn) of positive reals. Then we have

n = (1 + rn)n n(n 1)

2r2n,

so that r2n 2/(n 1) for all n 2. Since 2/(n 1) 0, by Theorem 1.2(c), thatrn 0, and hence by Theorem 1.2(c), n1/n = 1 + rn 1. Remark 1.5 The proof of the result in Example 1.12 does not require the knowledgethat n1/n 1, but uses the facts that 2/(n 1) 0 (which can be shown easily)and Theorem 1.2. But, if one is asked to show that n1/n 1, then we can giveanother proof of the same by using only the definition, as follows:

Another proof without using Theorem 1.2. Let > 0 be given. To find n0 N suchthat n1/n 1 < for all n n0. Note that

n1/n 1 < n1/n < 1 + n < (1 + )n = 1 +n+ n(n 1)2

2 + . . .+ n.

Hence

n1/n 1 < if i.e., if n > 2

2.

So, we may take any n0 N which satisfies n0 2(1 + /2).

EXAMPLE 1.13 For any a > 0, (a1/n) converges to 1.

If a > 1, then we can write a1/n = 1+rn for some sequence (rn) of positive reals.Then we have

a = (1 + rn)n nrn so that rn a/n.

Since a/n 0, by Theorem 1.2(c), rn 0, and hence by Theorem 1.2(c), a1/n =1 + rn 1.

In case 0 < a < 1, then 1/a > 1. Hence, by the first part, 1/a1/n = (1/a)1/n 1,so that an 1. Exercise 1.19 Give another proof for the result in Example 1.13 without usingTheorem 1.2. J

EXAMPLE 1.14 Let (an) be a bounded sequence of non-negative real numbers.Then (1 + an)

1/n 1 as n:This is seen as follows: Let M > 0 be such that 0 an M for all n N. Then,

1 (1 + an)1/n (1 + M)1/n for all n N. By Example 1.13, (1 + M)1/n 0.Hence the result follows by making use of part (d) of Theorem 1.2.

EXAMPLE 1.15 As an application of some of the results discussed above, considerthe sequence (an) with an = (1+1/n)

1/n, n N. We already know that limn an = 1.

Now, another proof for the same: Note that 1 an 21/n, and 21/n 1 as n.


EXAMPLE 1.16 Consider the sequence (an) with an = (1 + n)1/n. Then an 1

as n. We give two proofs for this result.(i) Observe that an = n

1/n (1 + 1/n)1/n. We already know that n1/n 1, and(1 + 1/n)1/n 1 as n.

(ii) Observe that n1/n (1 + n)1/n (2n)1/n = 21/nn1/n, where n1/n 1 and21/n 1 as n. EXAMPLE 1.17 Suppose (an) is a sequence of positive terms such that lim

nan+1an

=

` < 1. Then an 0.Since lim

nan+1an

= ` < 1, there exists q such that ` < q < 1 and N N suchthat

an+1an q for all n N . Hence,

0 an qnNaN n N.

Now, since qnN 0 as n, it follows that an 0 as n.

Exercise 1.20 Suppose Let (an) is a sequence of positive terms such that

limn

an+1an

= ` > 1. Then (an) diverges to . J

EXAMPLE 1.18 Let 0 < a < 1. Then nan 0 as n.To see this let an := na

n for n N. Then we have

an+1an

=(n+ 1)an+1

nan=

(n+ 1)a

n n N.

Hence, limn

an+1an

= a < 1. Thus, the result follows from the last example.

Exercise 1.21 Obtain the the result in Example 1.18 by using the arguments inExample 1.11. J

Remark 1.6 Suppose for each k N, a(k)n 0, b(k)n 1 as n , and alsoa(n)n 0, b(n)n 1 as n. In view of Theorems 1.2 and 1.4, one may think that

a(1)n + a(2)n + + a(n)n 0 as n

and

b(1)n b(2)n b(n)n 1 as n.

Unfortunately, that is not the case. To see this consider

a(k)n =k

n2, b(k)n = k

1/n k, n N.


Then, for each k N, a(k)n 0, b(k)n 1 as n, and also a(n)n 0, b(n)n 1 asn. But,

a(1)n + a(2)n + + a(n)n =

1

n2+

2

n2+ + n

n2=n+ 1

2n 1

2as n

andb(1)n b

(2)n b(n)n = 11/n21/n n1/n = (n!)1/n 6 1 as n.

In fact, for any k N, if n k, then

(n!)1/n (k!)1/n(knk)1/n = (k!)1/nk1k/n = k( k!kk

)1/n.

Since for any given k N, k(k!kk

)1/n k as n, it follows that (n!)1/n 6 1. Infact, there exists n0 N such that

(n!)1/n k2

n max{n0, k}.

Thus, the sequence is((n!)1/n

)is unbounded.

EXAMPLE 1.19 Let an = (n!)1/n2 , n N. Then an 1 as n. We give two

proofs for this.

(i) Note that, for every n N,1 (n!)1/n2 (nn)1/n2 = n1/n.

Now, since n1/n 1, we have (n!)1/n2 1.(ii) By GM-AM inequality, for n N,

(n!)1/n = (1.2. . . . .n)1/n 1 + 2 + . . .+ nn

=n+ 1

2 n.

Thus,1 (n!)1/n2 n1/n.

Since n1/n 1 we have (n!)1/n2 1.

1.1.5 Cauchy sequence

Theorem 1.9 If a real sequence (an) converges, then for every > 0, there existsN N such that

|an am| < n,m N.

Proof. Suppose an a as n , and let > 0 be given. Then we know thatthere exists N N such that |an a| < /2 for all n N . Hence, we have

|an am| |an a|+ |a am| < n,m N.This completes the proof.


Definition 1.11 A a sequence (an) is said to be a Cauchy sequence if for every > 0, there exists N N such that

|an am| < n,m N.

Theorem 1.9 show that every convergent sequence is a Cauchy sequence. Inparticular, if (an) is not a Cauchy sequence, then (an) does not converge to anya R. Thus, Theorem 1.9 may help us to show that certain sequence is notconvergent. For example, see the follwing exercise.

Exercise 1.22 Let sn = 1 +12 + . . . +

1n for n N. Then show that (sn) is not a

Cauchy sequence. [Hint: For any n N, note that s2n sn 12 .] J

Let us observe certain properties of Cauchy sequences.

Theorem 1.10 Let (an) be a Cauchy sequence. The we have the following.

(i) (an) is a bounded sequence.

(ii) If (an) has a convergent subsequence with limit a, then the sequence (an)itself will converge to a.

Proof. Let (an) be Cauchy sequence and > 0 be given.

(i) Let N N be such that |an am| < for all n,m N . In particular,

|an| |an aN |+ |aN | + |aN | n N.

This proves (i).

(ii) Suppose (ank) is a convergent subsequence of (an), and let a = limk

ank . Let

n0 N be such that |ank a| < for all k n0. Hence, we have

|ak a| |ak ank |+ |ank a| < 2 k max{N,n0}.

Thus, an a as n.

In fact, the converse of Theorem 1.9 also holds:

Theorem 1.11 Every Cauchy sequence of real numbers converges.

Proof of Theorem 1.11. Let (an) be a Cauchy sequence. By Theorem 1.10(i),(an) is bounded. Then, by Bolzano-Weierstrass theorem (Theorem 1.8), (an) has aconvergent subsequence (ank). Let a = lim

kank . Now, by Theorem 1.10(ii), an a

as n.


An alternate proof without using Bolzano-Weierstrass theorem (Theorem 1.8). Let(xn) be Cauchy sequence, and > 0 be given. Then there exists n N such that|xn xm| < for all n,m n. In particular, |xn xn | < for all n n. Hence,for any 1, 2 > 0,

xn1 1 < xn < xn2 + 2 n max{n1 , n2}. ()

From this, we see that the set {xn : > 0} is bounded above, and

x := sup{xn : > 0}

satisfies

xn < x < xn + n n. ()From () and (), we obtain

|xn x| < n n.

This completes the proof.

Here is an example of a general nature.

EXAMPLE 1.20 Let (an) be a sequence of real numbers. Suppose there exists apositive real number < 1 such that

|an+1 an| |an an1| n N, n 2.

Then (an) is a cauchy sequence. To see this first we observe that

|an+1 an| n1|a2 a1| n N, n 2.

Hence, for n > m,

|an am| |an an1|+ . . .+ |am+1 am| (n2 + . . .+ m1)|a2 a1| m1(1 + + . . .+ nm3)|a2 a1|

m1

1 |a2 a1|.

Since m1 0 as m, given > 0, there exists N N such that |anam| < for all n,m N . Exercise 1.23 Given a, b R and 0 < < 1, let (an) be a sequence of real numbersdefined by a1 = a, a2 = b and

an+1 = (1 + )an an1 n N, n 2.

Show that (an) is a Cauchy sequence and its limit is (b+ a)/(1 ). J


Exercise 1.24 Suppose f is a function defined on an interval J . If there exists0 < < 1 such that

|f(x) f(y)| |x y| x, y J,then the for any a J , the sequence (an) defined by

a1 = f(a), an+1 := f(an) n N,is a Cauchy sequence. Show also that the limit of the sequence (an) is independentof the choice of a. J

1.1.6 Additional exercises

1. Prove that xn l as n if and only if |xn l| 0 as n , and inthat case |xn| |l| 0 as n.

2. Establish convergence or divergence of the sequence (an) each of the followingsequences, where an is:

(i)n

n+ 1(ii)

(1)nnn+ 1

(iii)2n

3n2 + 1, (iv)

2n2 + 3

3n2 + 1.

3. Suppose (an) is a real sequence such that an 0 as n . Show thefollowing:

(a) The sequence (a2n) converges to 0.

(b) If an > 0 for all n, then the sequence (1/an) diverges to infinity.

4. Let {xn} be a sequence defined recursively by xn+2 = xn+1 + xn, n 1 withx1 = x2 = 1. Show that {xn} diverges to .

5. Let xn =n+ 1 n for n N. Show that {xn}, {nxn} and {

nxn} are

convergent. Find their limits.

6. Let xn =n

k=11

n+k , n N. Show that {xn} is convergent.7. If (an) converges to x and an 0 for all n N, then show that x 0 and

(an) converges to

x.

8. Prove the following:

(a) If {xn} is increasing and unbounded, then xn + as n .(b) If {xn} is decreasing and unbounded, prove that xn as n .

9. Let a1 = 1, an+1 =

2 + an for all n N. Show that (an) converges. Also,find its limit.

10. Let a1 = 1 and an+1 =14(2an + 3) for all n N. Show that {an} is monotoni-

cally increasing and bounded above. Find its limit.


11. Let a1 = 1 and an+1 =an

1 + anfor all n N. Show that {an} converges. Find

its limit.

12. Prove that if (an) is a Cauchy sequence having a subsequence which convergesto a, then (an) itself converges to a.

13. Suppose (an) is a sequence such that the subsequences (a2n1) and (a2n) con-verge to the same limit, say a. Show that (an) also converges to a.

14. Let {xn} be a monotonically increasing sequence such that {x3n} is bounded.Is {xn} convergent? Justify your answer.

15. Show that, if a sequence (an) has the property that an+1 an 0 as n,then (an) need not converge.

16. Let a1 = 1 and an+1 =14(2an + 3) for all n N. Show that (an) is monotoni-

cally increasing and bounded above. Find its limit.

17. Give an example in support of the statement: If a sequence {an} has theproperty that an+1 an 0 as n, then {an} need not converge.

18. Let a > 0 and xn =ananan+an , n N. Discuss the convergence of the sequence

(xn).

19. If 0 < a < b and an = (an+bn)1/n for all n N, then show that (an) converges

to b. [Hint: Note that (an + bn)1/n = b(1 +(ab

)n)1/n.]

20. If 0 < a < b and an+1 = (anbn)1/2 and bn+1 =

an+bn2 for all n N with a1 = a,

b1 = b, then show that (an) and (bn) converge to the same limit. [Hint: Firstobserve that an an+1 bn+1 bn for all n N.]

21. Let a1 = 1/2 and b = 1, and let an+1 = (anbn)1/2 and bn+1 =

2anbnan+bn

for alln N. Show that (an) and (bn) converge to the same limit. [Hint: Firstobserve that bn bn+1 an+1 an for all n N.]

22. Let a1 = 1/2, b = 1, an = (an1bn1)1/2 and 1bn =12(

1an

+ 1bn1 ). Prove that

an1 < an < bn < bn1 for all n N. Deduce that both the sequences {an}and {bn} converge to the same limit , 1/2 < < 1.

23. Prove that if {an} is a Cauchy sequence having a subsequence which convergesto a, then {an} itself converges to a.

Series of Real Numbers 21

1.2 Series of Real Numbers

Definition 1.12 A series of real numbers is an expression of the form

a1 + a2 + a3 + . . . ,

or more compactly as

n=1 an, where (an) is a sequence of real numbers.

The number an is called the n-th term of the series and the sequence sn :=ni=1

ai

is called the n-th partial sum of the series

n=1 an.

1.2.1 Convergence and divergence of series

Definition 1.13 A series

n=1 an is said to converge (to s R) if the sequence{sn} of partial sums of the series converge (to s R).

If

n=1 an converges to s, then we write

n=1 an = s.

A series which does not converge is called a divergent series.

A necessary condition

Theorem 1.12 If

n=1 an converges, then an 0 as n. Converse does nothold.

Proof. Clearly, if sn is the n-th partial sum of the convergent series

n=1 an,then

an = sn sn1 0 as n.

To see that the converse does not hold it is enough to observe that the seriesn=1 an with an =

1n for all n N diverges whereas an 0.

The proof of the following corollary is immediate from the above theorem.

Corollary 1.13 Suppose (an) is a sequence of positive terms such that an+1 > anfor all n N. Then the series n=1 an diverges.

The above theorem and corollary shows, for example, that the series

n=1nn+1

diverges.

EXAMPLE 1.21 We have seen that the sequence (sn) with sn =n

k=11k! con-

verges. Thus, the series

n=11n! converges. Also, we have seen that the sequence

(n) with n =n

k=11k diverges. Hence,

n=1

1n diverges.


EXAMPLE 1.22 Consider the geometric series series

n=1 aqn1, where a, q R.

Note that sn = a+ aq + . . .+ aqn1 for n N. Clearly, if a = 0, then sn = 0 for all

n N. Hence, assume that a 6= 0. Then we have

sn =

{na if q = 1,a(1qn)1q if q 6= 1.

Thus, if q = 1, then (sn) is not bounded; hence not convergent. If q = 1, then wehave

sn =

{a if n odd,0 if n even.

Thus, (sn) diverges for q = 1 as well. Now, assume that |q| 6= 1. In this case, wehave sn a1 q

= |a||1 q| |q|n.This shows that, if |q| < 1, then (sn) converges to a1q , and if |q| > 1, then (sn) isnot bounded, hence diverges.

Theorem 1.14 Suppose (an) and (bn) are sequences such that for some k N,an = bn for all n k. Then

n=1 an converges if and only if

n=1 bn converges.

Proof. Suppose sn and n be the n-th partial sums of the series

n=1 an andn=1 bn respectively. Let =

ki=1 ai and =

ki=1 bi. Then we have

sn =n

i=k+1

ai =

ni=k+1

bi = n n k.

From this it follows that the sequence (sn) converges if and only if (n) converges.

From the above theorem it follows if

n=1 bn is obtained from

n=1 an byomitting or adding a finite number of terms, then

n=1

an converges n=1

bn converges.

In particular,n=1

an converges n=1

an+k converges

for any k N.The proof of the following theorem is left as an exercise.

Theorem 1.15 Suppose

n=1 an converges to s and

n=1 bn converges to . Thenfor every , R, n=1(an + bn) converges to s+ .


1.2.2 Some tests for convergence

Theorem 1.16 (Comparison test) Suppose (an) and (bn) are sequences of non-negative terms, and an bn for all n N. Then,

(i)

n=1 bn converges =

n=1 an converges,

(ii)

n=1 an diverges =

n=1 bn diverges.

Proof. Suppose sn and n be the n-th partial sums of the series

n=1 an andn=1 bn respectively. By the assumption, we get 0 sn n for all n N, and

both (sn) and (n) are monotonically increasing.

(i) Since (n) converges, it is bounded. Let M > 0 be such that n M for alln N. Then we have sn M for all n N. Since (sn) are monotonically increasing,it follows that (sn) converges.

(ii) Proof of this part follows from (i) (How?).

Corollary 1.17 Suppose (an) and (bn) are sequences of positive terms.

(a) Suppose ` := limn anbn exists. Then we have the following:

(i) If ` > 0, then

n=1 bn converges

n=1 an converges.

(ii) If ` = 0, then

n=1 bn converges

n=1 an converges.

(b) Suppose limn

anbn

=. Then n=1 an converges n=1 bn converges.Proof. (a) Suppose limn anbn = `.

(i) Let ` > 0. Then for any > 0 there exists n N such that ` < anbn < `+ for all n N . Equivalently, (` )bn < an < (`+ )bn for all n N . Had we taken = `/2, we would get `2bn < an 0, there exists n N such that < anbn < for all n N . In particular, an < bn for all n N . Hence, we get the result byusing comparison test.

(b) By assumption, there exists N N such that anbn 1 for all n N . Hence theresult follows by comparison test.

EXAMPLE 1.23 We have already seen that the sequence (sn) with sn =n

k=11k!

converges. Here is another proof for the same fact: Note that 1n! 12n1 for alln N. Since n=1 12n1 converges, it follows from the above theorem that n=1 1n!converges.


EXAMPLE 1.24 Since 1n 1n for all n N, and since the series

n=1

1n diverges,

it follows from the above theorem that the series

n=11n

also diverges.

Theorem 1.18 (deAlemberts ratio test) Suppose (an) is a sequence of positiveterms such that limn

an+1an

= ` exists. Then we have the following:

(i) If ` < 1, then the series

n=1 an converges.

(ii) If ` > 1, then the series

n=1 an diverges.

Proof. (i) Suppose ` < q < 1. Then there exists N N such thatan+1an

< q n N.

In particular,an+1 < q an < q

2an1 < . . . < qn a1, n N.Since

n=1 q

n converges, by comparison test,

n=1 an also converges.

(ii) Let 1 < p < `. Then there exists N N such thatan+1an

> p > 1n N.

From this it follows that (an) does not converge to 0. Hence

n=1 an diverges.

Theorem 1.19 (Cauchys root test) Suppose (an) is a sequence of positive termssuch that limn an1/n = ` exists. Then we have the following:

(i) If ` < 1, then the series

n=1 an converges.

(ii) If ` > 1, then the series

n=1 an diverges.

Proof. (i) Suppose ` < q < 1. Then there exists N N such thatan

1/n < q n N.Hence, an < q

n for all n N . Since the n=1 qn converges, by comparison test,n=1 an also converges.

(ii) Let 1 < p < `. Then there exists N N such thatan

1/n > p > 1 n N.Hence, an 1 for all n N . Thus, (an) does not converge to 0. Hence,

n=1 an

also diverges.

Remark 1.7 We remark that both dAlemberts test and Cauchy test are silentfor the case ` = 1. But, for such case, we may be able to infer the convergence ordivergence by some other means.


EXAMPLE 1.25 For every x R, the series n=1 xnn! converges:Here, an =

xn

n! . Hence

an+1an

=x

n+ 1 n N.

Hence, it follows that limnan+1an

= 0, so that by dAlemberts test, the seriesconverges.

EXAMPLE 1.26 The series

n=1

(n

2n+1

)nconverges: Here

an1/n =

n

2n+ 1 1

2< 1.

Hence, by Cauchys test, the series converges.

EXAMPLE 1.27 Consider the series

n=11

n(n+1) . In this series, we see that

limnan+1an

= 1 = limn an1/n. However, the n-th partial sum sn is given by

sn =nk=1

1

k(k + 1)=

nk=1

(1

k 1k + 1

)= 1 1

n+ 1.

Hence {sn} converges to 1. EXAMPLE 1.28 Consider the series

n=1

1n2

. In this case, we see that

1

(n+ 1)2 1n(n+ 1)

n N.

Since

n=11

n(n+1) converges, by comparison test, the given series also converges.

EXAMPLE 1.29 Since

n=11n2

converges and

n=11n diverges, by comparison

test, we see that the series

n=11np converges for p 2 and diverges for p 1.

EXAMPLE 1.30 Consider the series

n=11np for p > 1. To discuss this example,

consider the function f(x) := 1/xp, x 1. Then we see that for each k N,

k 1 x k = 1kp 1xp

= 1kp kk1

dx

xp.

Hence,nk=2

1

kp

nk=2

kk1

dx

xp=

n1

dx

xp=n1p 1

1 p 1

p 1 .

Thus,

sn :=

nk=1

1

kp 1p 1 + 1.

Hence, (sn) is monotonically increasing and bounded above. Therefore, (sn) con-verges.


Now, we have a more general result.

Theorem 1.20 (Integral test) Suppose f(x) is a continuous, non-negative anddecreasing function for x [1,). For each n N, let an :=

n1 f(x)dx. Then

n=1

f(n) converges (an) converges.

Proof. First we observe that an = n1 f(x) dx =

nk=2

kk1 f(x) dx. Now, since

f(x) is a decreasing function for x [1,), we have for each k N,k 1 x k = f(k) f(x) f(k 1).

Hence, for k = 2, 3, . . .,

f(k) kk1

f(x) dx f(k 1)

so thatnk=2

f(k) nk=2

kk1

f(x) dx nk=2

f(k 1).

Thus,nk=2

f(k) n1f(x) dx

nk=2

f(k 1).

Now, let sn :=

nk=1

f(k) for n N. Then from the above inequalities, togetherwith the fact that (sn) is a monotonically increasing sequence, it follows that (an)converges if and only if (sn) converges.

1.2.3 Alternating series

Definition 1.14 A series of the form

n=1(1)n+1un where (un) is a sequence ofpositive terms is called an alternating series.

Theorem 1.21 (Leibnizs theorem) Suppose (un) is a sequence of positive termssuch that un un+1 for all n N, and un 0 as n . Then the alternatingseries

n=1(1)n+1un converges.

Proof. Let sn be the n-th partial sum of the alternating series

n=1(1)n+1un.We observe that

s2n+1 = s2n + u2n+1 n N.Since un 0 as n , it is enough to show that (s2n) converges (Why?). Notethat

s2n = (u1 u2) + (u3 u4) + . . .+ (u2n1 u2n),


s2n = u1 (u2 u3) . . . (u2n2 u2n1) u2nfor all n N. Since ui ui+1 0 for each i N, (s2n) is monotonically increasingand bounded above. Therefore (s2n) converges. In fact, if s2n s, then we haves2n+1 = s2n + u2n+1 s, and hence sn s as n.

By the above theorem the series

n=1(1)n+1

n converges. Likewise, the seriesn=1

(1)n+12n1 and

n=1

(1)n+12n also converge.

Remark 1.8 The series

n=1(1)n+12n1 appear in the work of a Kerala mathemati-

cian Madhava presented in the year around 1550 by another Kerala mathematicianNilakantha. The discovery of the above series is normally attributed to Leibniz andJames Gregory after nearly 300 years of its discovery.

Suppose (un) is as in Leibnizs theorem (Theorem 1.21), and let s R be suchthat sn s, where sn is the nth partial sum of

n=1

(1)n+1un.

How fast (sn) converges to s?

In the proof of Theorem 1.21, we have shown that {s2n} is a monotonicallyincreasing sequence. Similarly, it can be shown that {s2n1} is a monotonicallydecreasing sequence.

Since (s2n1) is monotonically decreasing and (s2n) is monotonically increasing,we have

s2n1 = s2n + u2n s+ u2n, s s2n+1 = s2n + u2n+1.Thus,

s2n1 s u2n, s s2n u2n+1.Consequently,

|s sn| un+1 n N.

1.2.4 Absolute convergence


n=1 an is said to converge absolutely, if

n=1 |an| con-verges. Theorem 1.22 Every absolutely convergent series converges.

Proof. Suppose

n=1 an is an absolutely convergent series. Let sn and n bethe n-th partial sums of the series

n=1 an and

n=1 |an| respectively. Then, for

n > m, we have

|sn sm| =

nj=m+1

an

n

j=m+1

|an| = |n m|.


Since, {n} converges, it is a Cauchy sequence. Hence, form the above relation itfollows that {sn} is also a Cauchy sequence. Therefore, by the Cauchy criterion, itconverges.

Another proof without using Cauchy criterion. Suppose

n=1 an is an absolutelyconvergent series. Let sn and n be the n-th partial sums of the series

n=1 an and

n=1 |an| respectively. Then it follows that

sn + n = 2pn,

where pn is the sum of all positive terms from {a1, . . . , an}. Since {n} converges,it is bounded, and since pn n for all n N, the sequence {pn} is also bounded.Moreover, {pn} is monotonically increasing. Hence {pn} converge as well. Thus,both {n}, {pn} converge. Now, since sn = 2pn n for all n N, the sequence{sn} also converges.


n=1 an is said to converge conditionally if

n=1 anconverges, but

n=1 |an|diverges.


n=1(1)n+1

n is conditionally convergent.


n=1(1)n+1

n2is absolutely convergent.


n=1(1)n+1

n! is absolutely convergent.

EXAMPLE 1.34 For any R, the series n=1 sin(n)n2 is absolutely convergent:Note that sin(n)n2

1n2 n N.Since

n=1

1n2

converges, by comparison test,

n=1

sin(n)n2 also converges. EXAMPLE 1.35 The series

n=3

(1)n log nn log(log n)

is conditionally convergent. To see

this, let un =log n

n log(log n). Since n log n log(log n) we have

1

n log nn log(log n)

1log(log n)

()

so that un 0. It can be easily seen that un+1 un. Hence, by Leibnitz theorem,the given series converges. Inequality () also shows that the series n=3 un doesnot converge.

Here are two more results whose proofs are based on some advanced topics inanalysis



n=1 an is an absolutely convergent series and (bn) is asequence obtained by rearranging the terms of (an). Then

n=1 bn is also absolutely

convergent, and

n=1 an =

n=1 bn.


n=1 an is a conditionally convergent series. The forevery R, there exists a sequence (bn) whose terms are obtained by rearrangingthe terms of (an) such that

n=1 bn = .

To illustrate the last theorem consider the conditionally convergent series

n=1(1)n+1

n .Consider the following rearrangement of this series:

1 12 1

4+

1

3 1

6 1

8+ + 1

2k 1 1

4k 2 1

4k+ .

Thus, if an =(1)n+1

n for all n N, the rearranged series is

n=1 bn, where

b3k2 =1

2k 1 , b3k1 =1

4k 2 , b3k =1

4k

for k = 1, 2, . . .. Let sn and n be the n-th partial sums of the series

n=1 an andn=1 bn respectively. Then we see that

3k = 1 12 1

4+

1

3 1

6 1

8+ + 1

2k 1 1

4k 2 1

4k

=

(1 1

2 1

4

)+

(1

3 1

6 1

8

)+ +

(1

2k 1 1

4k 2 1

4k

)=

(1

2 1

4

)+

(1

6 1

8

)+ +

(1

4k 2 1

4k

)=

1

2

[(1 1

2

)+

(1

3 1

4

)+ +

(1

2k 1 1

2k

)]=

1

2s2k.

Also, we have

3k+1 = 3k +1

2k + 1, 3k+2 = 3k +

1

2k + 1 1

4k + 2.

We know that {sn} converge. Let limn sn = s. Since, an 0 as n, it thenfollows that

limk

3k =s

2, lim

k3k+1 =

s

2, lim

k3k+2 =

s

2.

Hence, we can infer that n s/2 as n.


1.2.5 Additional exercises

(These problems were prepared for the students of MA1010, Aug-Nov, 2010.)

1. Using partial fractions, prove thatn=1

3n 2n(n+ 1)(n+ 2)

= 1

2. Test the following series for convergence:

(a)n=1

(n!)2

(2n)!(b)

n=1

(n!)2

(2n)!5n (c)

n=1

(n

n+ 1

)n2

(d)

n=1

(n

n+ 1

)n24n (e)

n=1

1

n

(n+ 1n) (f)

n=1

(1)(n1)n

(g)n=1

1

n2 + 2n+ 3(h)

1

2+

2

22+

3

23+ ...+

n

2n+ ...

(i)110

+120

+130

+ ...+110n

+ ... (j) 2 +3

2+

4

3+ ...+

n+ 1

n+ ...

(k)13

7+

13

8+

13

9+ ...+

13n+ 6

+ ...

(l)1

2+

(2

3

)4+

(3

4

)9+...+

(n

n+ 1

)n2+... (m)

1

2+

2

5+

3

10+...+

n

n2 + 1+...

3. Is the Leibniz Theorem applicable to the series:

12 1

12 + 1

+1

3 1 1

3 + 1+ ...+

1n 1

1n+ 1

+ ...

Does the above series converge?

Justify your answer.

4. Find out whether (or not) the following series converge absolutely or condi-tionally:

(a) 1 132

+1

52 1

72+

1

92+ ...+ (1)n+1 1

(2n 1)2 + ...

(b)1

ln 2 1

ln 3+

1

ln 4 1

ln 5+ ...+ (1)n 1

lnn+ ...

5. Find the sum of the series:1

1.2.3+

1

2.3.4+ ...+

1

n.(n+ 1).(n+ 2)+ ...

6. Test for the convergence of the following series:

(a)

n=1

1

(n+ 1)n

(b)

n=1

1

n!(c)

n=1

(n4 + 1

n4 1

)


(d)

n=1

1

(a+ n)p(b+ n)p, a, b, p, q > 0 (e)

n=1

tan

(1

n

)

(f)n=1

(3n3 + 1 n

)(g)

n=1

2n!

n!(h)

n=1

1

n(n+ 1)

(i)

n=1

(n+ 6)1/3 (j)n=1

(log n)n (k)n=1

(1 +

1

n

)n2

(l)

n=1

(nx

1 + n

)n(m)

n=1

(1)n1 nn+ 1

(n)

n=1

(1)n1n(n+ 1)(n+ 2)

(o)n=1

(1)n+1 (n+ 1n) (p) n=1

(2)nn2

7. Examine the following series for absolute / conditional convergence:

(a)n=1

(1)n+1(2n 1)2 (b)

n=1

(1)n+1n3n1

(c)n=1

(1)nn(log n)2

(d)n=1

(1)n log nn log log n

(e)n=1

(1)n+1 1n

(f)n=1

(1)n+1 1n2n

8. Let (an) be a sequence of non-negative numbers and (akn) be a subsequence(an). Show that, if

n=1 an converges, then

n=1 akn also converges. Is the

converse true? Why?

2Limit, Continuity and Differentiabilityof Functions

In this chapter we shall study limit and continuity of real valued functions definedon certain sets.

2.1 Limit of a Function

Suppose f is a real valued function defined on a subset D of R. We are going todefine limit of f(x) as x D approaches a point a, not necessarily in D. First wehave to be clear about what x D approaches a point a means.

2.1.1 Limit point of a set D R

Definition 2.1 Let D R and a R. Then a is said to be a limit point of D ifthere exists a sequence (an) in D which is not eventually constant such that an aas n.

We may recall that a sequence (an) is said to be eventually constant if thereexists k N such that an = ak for all n k.Theorem 2.1 A point a R is a limit point of D R if and only if for every > 0,

(a , a+ ) (D \ {a}) 6= .

Proof. Suppose a R is a limit point of D. Then we know that there exists asequence (an) in D which is not eventually constant such that an a. Hence, forevery > 0, there exists N N such that an (a , a + ) for all n N . Inparticular, there exists n N such that an (a , a+ ) (D \ {a}).

Conversely, suppose for every > 0, (a, a+)(D\{a}) 6= . Then, for eachn N, taking = 1/n, there exists an D \ {a} such that an (a 1/n, a+ 1/n).Hence, 0 |an a| < 1/n for all n N, showing that an a. Clearly, (an) is noteventually constant.

32

Limit of a Function 33

EXAMPLE 2.1 (i) Every point in an interval is its limit point.

(ii) If I is an open interval of finite length, then both the end points of I arelimit points of I.

(iii) If D = {x R : 0 < |x| < 1}, then every point in the interval [1, 1] is alimit point of D.

Remark 2.1 (i) For a R, an open interval of the form (a , a + ) for some > 0 is called a neighbourhood of a, or sometimes, a -neighbourhood of a.

(ii) By a deleted neighbourhood of a point a R we mean a set of the formD := {x R : 0 < |x a| < } for some > 0, i.e., the set (a , a+ ) \ {a}.

With the terminologies in the above remark,

a point a R is a limit point of D R if and only if every deleted neighbour-hood of a contains at least one point of D.

Now, we define limit of f(x) as x approaches a.

2.1.2 Limit of a function f(x) as x approaches a

Definition 2.2 Let f be a real valued function defined on a set D R, andlet a R be a limit point of D. We say that f(x) has the limit b R as xapproaches a if for every > 0, there exists > 0 such that

|f(x) b| < whenever x D, 0 < |x a| < ,

and in that case we write

limxa f(x) = b or f(x) b as x a.

Thus, limxa f(x) = b if and only if for every open interval Ib containing b thereexists an open interval Ia containing a such that

x Ia (D \ {a}) = f(x) Ib.

In the following, whenever we talk about limit of a function f as xapproaches a R, we assume that f is defined on a set D R and a isa limit point of D.

Exercise 2.1 Show that, if limit of a function, if exists, is unique. J

Exercise 2.2 Show that, if limxa f(x) = b, then there exists a deleted neighbourhood

D of a and M > 0 such that |f(x)| M for all x D. J

34 Limit, Continuity and Differentiability of Functions M.T. Nair

Remark 2.2 Suppose f is a real valued function defined on an interval I and a I.What do we mean by the statement that

limxa f(x) does not exist?

If you see the definitiion of existence of a limit, then you see that, the above state-ment means the following:

For any b R, there exists > 0 such that for any > 0, there is atleastone x (a , a+ ) such that f(x) 6 (b , b+ ).

How do we show, by first principle, i.e., from the definition, that a function f doesnot have a limit as x approaches to a particular point x0?

Assume that the limxx0 f(x) exists and it is equal to for some R.Then we know that for every > 0, there exists := such that

0 < |x x0| < = |f(x) | < .

Then, for a particular choice of , say = 0, one must be able to findan x such that |x x0| < 0 but |f(x) | 0.

Let us illustrate this by a simple example.

Let f : [1, 1] R be defined by f(x) ={

0, 1 x 0,1, 0 < x 1. We show that

limx 0f(x) does not exist. Suppose limx0 f(x) = for some R and let > 0 be given. Then, we know from the definition of the limit that there exists := such that

x 6= 0, x (, ) = f(x) ( , + ).

We know that, f(x) takes the values only 0 and 1 and the length of the interval( , + ) is 2. Hence, if we take 0 such that 20 < 1, then f(x) cannot belongto (0, +0) for all x in a neighbourhood of 0. To see this consider the followingcases:

case (i): = 0. In this case, f(x) 6 ( 0, + 0) for any x > 0.case (ii): = 1. In this case, f(x) 6 ( 0, + 0) for any x < 0.case (iii): 6= 0, 6= 1. In this case, f(x) 6 ( 0, + 0) for any x0.Thus, we arrive at a a contradiction to our assumption.


Limit of a function in terms of a sequences

Let a be a limit point of D R and f : D R.Suppose lim

xa f(x) = b. Since a is a limit point of D, we know that there existsa sequence (xn) in D which is not eventually constant such that xn a. Doesf(xn) b? The answer is yes. In fact, we have more!Theorem 2.2 If lim

xa f(x) = b, then for every sequence (xn) in D such that xn a,we have f(xn) b.

Proof. Suppose limxa f(x) = b. Let (xn) be a sequence in D such that xn a. Let

> 0 be given. We have to show that there exists n0 N such that |f(xn) b| < for all n n0.

Since limxa f(x) = b, we know that there exists > 0 such that

x D \ {a}, |x a| < = |f(x) b| < . ()Also, since xn a, there exists n0 N such that |xna| < for all n n0. Hence,from (), we have |f(xn) b| < for all n n0.

What about the converse of the above theorem? The converse is also true, in aslight restricted sense.

Theorem 2.3 If for every sequence (xn) in D which is not eventually constant, thesequence (f(xn)) converges to b, then lim

xa f(x) = b.

Proof. Suppose for every sequence (xn) in D which is not eventually constant,the sequence (f(xn)) converges to b. Assume for a moment that f does not have thelimit b as x approaches a. Then, by the definition of the limit, there exists 0 > 0such that for every > 0, there exists at least one x D such that

x 6= a and |x a| < , but |f(x) b| > 0.In particular, for every n N, there exists xn D \ {a} such that

xn 6= a and |xn a| < 1n, but |f(xn) b| > 0.

Thus, (xn) is not eventually constant such that xn a but f(xn) 6 b. Thus wearrive at a contradiction to our hypothesis.

Remark 2.3 The advantage of Theorem 2.2 is that, if we can find a sequence (xn)in D such that xn a, but (f(xn)) does not converge, then we can assert thatlimn f(x) does not exist. Similarly, by Theorem 2.3, we can determine limn f(x) ifwe are able to show convergence of (f(xn)) to some b for any arbitrary (not for aspecific) non-eventually constant sequence (xn) in D which converges to a.


2.1.3 Some properties

The following two theorems can be proved using Theorems 2.2 and 2.3, and theresults on convergence of sequences of real numbers.

Theorem 2.4 We have the following.

(i) If limxa f(x) = b and limxa g(x) = c, then

limxa[f(x) + g(x)] = b+ c, limxa f(x)g(x) = bc.

(ii) If limxa f(x) = b and b 6= 0, then f(x) 6= 0 in a deleted neighbourhood of a

and

limxa

1

f(x)=

1

b.

Theorem 2.5 If f and g have the same limit b as x approaches a, and if h is afunction such that f(x) h(x) g(x) for all x deleted neighbourhood of a, thenlimxah(x) = b.

The following two corollaries are immediate from Theorem 2.4.

Corollary 2.6 If limxa f(x) = b, limxa g(x) = c, and c 6= 0, then g is nonzero in a

deleted neighbourhood of c and

limxa

f(x)

g(x)=b

c.

Corollary 2.7 If limxa f(x) = b, limxa g(x) = c and f(x) g(x) for all x in a deleted

neighbourhood of a, then b c.Theorem 2.8 Suppose lim

xa f(x) = b and limybg(y) = c. Then lim

xa g(f(x)) = c.

Proof. Let > 0 be given. Then there exists 1 > 0 such that

|y b| < 1 = |g(y) c| < .

Also, let 2 > 0 be such that

|x a| < 2 = |f(x) b| < 1.

Hence,|x a| < 2 = |f(x) b| < 1 = |g(f(x)) c| < .

This completes the proof.

Exercise 2.3 Write details of the proof of Theorem 2.4 and Corollary 2.6 andCorollary 2.7. J


Exercise 2.4 Suppose is a function defined in a neighbourhood of a point x0such that lim

xx0(x) = x0. If f is also a function defined in a neighbourhood of x0

and limxx0

f(x) exists, then prove that limxx0

f((x)) exists and

limxx0

f((x)) = limxx0

f(x).

J

2.1.4 Some Examples

EXAMPLE 2.2 If f(x) is a polynomial, say f(x) = a0 + a1x + . . . + akxk, then

for any a R,limxa f(x) = f(a).

We obtain this by using Theorem 2.4.

Let us show the same by using the definition, i.e., using arguments: Letb = f(a) and let > 0 be given. We have to find > 0 such that |x a| < =|f(x) b| < . Note that

f(x) f(a) = a1(x a) + a2(x2 a2) + . . .+ ak(xk ak),where

xn an = (x a)[xn1 + xn2a+ . . .+ xan2 + an1].Now, suppose |x a| < 1. Then we have |x| < 1 + |a| so that

|xnjaj1| < (1 + |a|)n1

and hence,|xn an| < |x a|n(1 + |a|)n1.

Thus, |x a| < 1 implies

|f(x) f(a)| |x a|(|a1|+ |a2|2(1 + |a|) + . . .+ |ak|k(1 + |a|)k1

),

Therefore, taking := |a1|+ |a2|2(1 + |a|) + . . .+ |ak|k(1 + |a|)k1, we have|f(x) f(a)| < whenever |x a| < := min{1, /}.

EXAMPLE 2.3 Let D = R \ {2} and f(x) = x24x2 . Then limx2 f(x) = 4.

Note that, for x 6= 2,

f(x) =(x+ 2)(x 2)

x 2 = (x+ 2).

Hence, for > 0, |f(x) 4| < whenever |x 2| < := .


EXAMPLE 2.4 Let D = R\{0} and f(x) = 1x . Then limx0 f(x) does not exist. Tosee this consider the sequence (xn) with xn = 1/n for n N. Then we have xn 0but (f(xn) diverges to infinity. Therefore, by Theorem 2.2, lim

x0f(x) does not exist.

Alternatively, for any b R,|f(x) b| |f(x)| |b| > 1 whenever |f(x)| > 1 + |b|.

But,

|f(x)| > 1 + |b| |x| < 11 + |b| .

Thus, for any b R,

|f(x) b| > 1 whenever |x| < 11 + |b| .

Thus, we have proved that it is not possible to find a > 0 such that |f(x) b| < 1for all x with |x| < .

EXAMPLE 2.5 We show that (i) limx0

sin(x) = 0 and (ii) limx0

cos(x) = 1.

From the graph f the function sinx, it is clear that

0 < x 0 such that

|f(x) b| < whenever x D, a < x < a.

(ii) We say that f(x) has the right limit b R as x approaches a fromright if for every > 0, there exists > 0 such that

|f(x) b| < whenever x D, a < x < a+ .

If f(x) has the left limit b R as x approaches a from left we writelimxa

f(x) = b or f(x) b as x a,

and if f(x) has the right limit b R as x approaches a from right we writelimxa+

f(x) = b or f(x) b as x a+.

The following theorem can be proved easily.

Theorem 2.9 Let f be a real valued function defined on a set D R, and let a Rbe a limit point of D. Then lim

xa f(x) exists if and only if limxaf(x) and lim

xa+f(x)

exist and limxa

f(x) = limxa+

f(x), and in that case

limxa f(x) = limxa

f(x) = limxa+

f(x).


In view of the above theorem, if (i) limxa

f(x) does not exist or (ii) limxa+

f(x)

does not exist or (iii) both limxa

f(x) and limxa+

f(x) exist but limxa

f(x) 6= limxa+

f(x),

then limxa f(x) does not exist.

Exercise 2.7 Give examples to illustrate the above statements. J

2.1.6 Limit at and at

Definition 2.4 Suppose a function f is defined on an interval of the form (a,) forsome a > 0. Then we say that f has the limit b as x and write lim

x f(x) = b,if for every > 0, there exits M > a such that

|f(x) b| < whenever x > M.

Definition 2.5 Suppose a function f is defined on an interval of the form (, a)for some a < 0. Then we say that f has the limit b as x and write

limx f(x) = b, if for every > 0, there exits M < a such that

|f(x) b| < whenever x < M.

Exercise 2.8 Prove the following:

1. If f is defined in (a,) with a > 0, then the function g(x) := f(1/x) is definedon a (0, 1/a), and

limx f(x) exists limx0 g(x) exists

and in that case limx f(x) = limx0

g(x).

2. If f is defined in (, a) with a < 0, then the function g(x) := f(1/x) isdefined on a (1/a, 0), and

limx f(x) exists limx0 g(x) exists

and in that case limx f(x) = limx0

g(x).

J

Definition 2.6 1. limxa f(x) = + if for every M > 0, there exists > 0 such

that0 < |x a| < = f(x) > M.


2. limxa f(x) = if for every M > 0, there exists > 0 such that

0 < |x a| < = f(x) < M.

3. limx+ f(x) = + if for every M > 0, there exists > 0 such that

x > = f(x) > M.

4. limx+ f(x) = if for every M > 0, there exists > 0 such that

x > = f(x) < M.

5. limx f(x) = + if for every M > 0, there exists > 0 such that

x < = f(x) > M.

6. limx f(x) = if for every M > 0, there exists > 0 such that

x < = f(x) < M.

For the next example, we require the following definition.

Definition 2.7 For positive numbers a and b, and p, q N, we defineap/q := (ap)1/q.

If b is a positive real number and (bn) is s sequence of positive rational numberssuch that bn b, then we define

ab := limn a

bn .

(Of course, one has to show that this limit exists; which is true!).

EXAMPLE 2.7 Recall that limn

(1 +

1

n

)nexists, and we denoted it by e. Now

we show that

limx

(1 +

1

x

)x= e.

Let > 0 be given. We have to find an M > 0 N such that

e M. ()

Now, we can see that, for every n N, if x R is such that n x n+ 1, then

1 +1

n+ 1 1 + 1

x 1 + 1

n


so that (1 +

1

n+ 1

)n (1 + 1x

)x (1 + 1n

)n+1.

Thus is is same as

n (

1 +1

x

)x n,where

n :=(

1 +1

n+ 1

)1(1 +

1

n+ 1

)n+1, n :=

(1 +

1

n

)n(1 +

1

n

).

We know that n e and n e as n . Therefore, there exists n0 N suchthat for all n n0,

e < n < e+ , e < n < e+ .Now, take M = n0 and let x > M . Take n n0 such that n+ 1 x n. For thisn and x, we have

e < n (

1 +1

x

)x n < e+ .Thus, we obtained an M > 0 such that

e M.

Thus, we have proved (). Using the arguments used in the above example, we obtain a more general result.

Theorem 2.10 Suppose (n) and (n) are sequences of positive real numbers andf is a (real valued) function defined on (0,) having the following property: Forn N, x R,

n < x < n+ 1 = n f(x) n.If (n) and (n) converge to the same limit, say b, then lim

x f(x) = b.

2.1.7 Additional Exercises

1. Using the definition of limit, show that limx3

x

4x 9 = 1.

2. Show that the function f defined by f(x) =

{x, if x < 1,1 + x, if x 1 does not

have the limit as x 1.

3. Let f be defined by f(x) =

3 x, if x > 1,1, if x = 1,2x, if x, 1.

Find limx1

f(x). Is it f(1)?

4. Let f be defined on a deleted neighbourhood D0 of a point x0 and limxx0

f(x) =

b. If b 6= 0, then show that there exists > 0 such that f(x) 6= 0 for everyx (x0 , x0 + ) D0.

Continuity of a Function 43

5. Let f be defined by f(x) =

{1, if x Q,0, if x 6 Q. Show that

(i) limx0

f(x) does not exist, and

(ii) limx0

xf(x) = 0.

6. Suppose limx f(x) = and limx g(x) = b. Show that limx g(f(x)) = b.

7. Let f : (0,) R be such that limx0

f(x) = b. Show that limx f(x

1) = b.

2.2 Continuity of a Function

2.2.1 Definition and some basic results

Definition 2.8 Let f be a real valued function defined on an interval I and a I.The f is said to be continuous at a if lim

xa f(x) = f(a).

Definition 2.9 Let f be a real valued function defined on an interval I. Then f issaid to be continuous on I if f is continuous at every a I.

Thus, f is continuous at a I if and only if for every > 0, there exists > 0such that

|f(x) f(a)| < whenever x I, |x a| < .

By Theorems 2.2 and 2.3, we have the following:

Theorem 2.11 A function f : I R is continuous at a I if and only if for everysequence (xn) in I with xn a, we have f(xn) f(a).

Also, using Theorem 2.4, we obtain the following.

Theorem 2.12 Suppose f and g are defined on an interval I and both f and g arecontinuous at a I. Then we have the following.

(i) f + g and fg are continuous at a.

(ii) If g(a) 6= 0, then there exists 0 > 0 such that g(x) 6= 0 for every x I with|x a| < 0 and f/g is continuous at a.

From Theorem 2.8, we have the following.

Theorem 2.13 Suppose f is continuous at a point x0 and g is continuous at thepoint y0 := f(x0). Then g f is continuous at x0.

The following property of a continuous function is worth noticing.


Theorem 2.14 Suppose f is a continuous function defined on an interval I andx0 I is such that f(x0) > 0. Then there exists > 0 such that f(x) > 0 for allx I (x0 , x0 + ).

Proof. Suppose the the conclusion in the theorem does not hold. Then for every > 0, there exists x I (x0 , x0 + ) such that f(x) 0. In particular,for each n N, taking n = 1/n, there exists xn I (x0 1/n, x0 + 1/n) suchthat f(xn) 0. Thus, we have xn x0 as n . Hence, by Theorem 2.11,f(xn) f(x0). Since f(xn) 0 for all n N, we have f(x0) 0, which contradictsthe assumption that f(x0) > 0.

2.2.2 Some examples

EXAMPLE 2.8 If f(x) is a polynomial, say f(x) = a0 + a1x+ . . .+ akxk, then f

is continuous on R

EXAMPLE 2.9 For given a R, let f(x) = |x a|, x R. Then f is continuousat every x0 R. To see this, note that

|f(x) f(x0)| = ||x a| |x0 a|| |x x0.Hence, for every > 0, we have

|x x0| < = |f(x) f(x0)| < .

EXAMPLE 2.10 Let f(x) = x24x2 for x R \ {2} and f(2) = 4. Then f is

continuous on R.

EXAMPLE 2.11 The functions f and g defined by f(x) = sinx and g(x) = cosxfor x R are continuous at 0.

Note that for x, y R,

sinx sin y = 2 sin(x y

2

)cos(x+ y

2

)so that

| sinx sin y| |x y| x, y R.Hence, for every > 0 and for every x0 R,

|x x0| < = | sinx sinx0| < .Thus, f(x) = sinx is continuous at every x R. Since cosx1 2 sin2(x/2), x R,it also follows that g(x) = cosx is continuous at every x R. EXAMPLE 2.12 Let f(x) = sinxx for x 6= 0 and f(0) = 1. Then f is continuousat every point in R.


EXAMPLE 2.13 Let f(x) = 1x for x 6= 0. Then there does not exist a continuousfunction g defined on R such that g(x) = f(x) for all x 6= 0. EXAMPLE 2.14 The function f defined by f(x) = 1/x, x 6= 0 is continuous atevery x0 6= 0:

Note that for x 6= 0, x0 6= 0,1x 1x0 = |x x0||xx0| 2|x x0||x20| whenever |x x0| |x0|2

since |x| = |x0 (x0 x)| |x0| |x0 x|. Hence, for any > 0,1x 1x0 < whenever |x x0| < := min{x202 , x02 }.

Thus, f is continuous at every x0 6= 0. EXAMPLE 2.15 Let f be defined by f(x) = 1/x on (0, 1]. Then there does notexist a continuous function g on [0, 1] such that g(x) = f(x) for all x (0, 1].

Suppose g is any function defined on [0, 1] such that g(x) = f(x) for all x (0, 1].Then we have 1/n 0 but g(1/n) = f(1/n) = n. Thus, g(1/n) 6 g(0). EXAMPLE 2.16 The function f defined by f(x) =

x, x 0 is continuous at

every x0 0:Let > 0 be given. First consider the point x0 = 0. Then we have

|f(x) f(x0)| =x < whenever |x| < 2.

Thus, f is continuous at x0 = 0. Next assume that x0 > 0. Since |x x0| =(x+x0)|xx0|, we have

|xx0| = |x x0|x+x0 |x x0|

x0.

Thus,|xx0| < whenever |x x0| < := x0.

More generally, we have the following example.

EXAMPLE 2.17 Let k N. Then the function f defined by f(x) = x1/k, x 0is continuous at every x0 0:

Let > 0 be given. First consider the point x0 = 0. Then we have

|f(x) f(x0)| = x1/k < whenever |x| < k.Thus, f is continuous at x0 = 0. Next assume that x0 > 0. Let y = x

1/k and

y0 = x1/k0 . Since

yk yk0 = (y y0)(yk1 + yk2y0 + . . .+ yyk2 + yk10 ),


so that

x x0 = (x1/k x1/k0 )(yk1 + yk2y0 + . . .+ yyk2 + yk10 ).

Hence,

|x1/k x1/k0 | =|x x0|

yk1 + yk2y0 + . . .+ yyk2 + yk10 |x x0|

yk10.

Thus,

|x1/k x1/k0 | < whenever |x x0| < := yk10 = x11/k0 .

Thus, f is continuous at every x0 > 0.

EXAMPLE 2.18 For a rational number r, let f(x) = xr for x > 0. Then usingthe above example together with Theorem 2.13, we see that f is continuous at everyx0 > 0.

We know that given r R, there exists a sequence (rn) of rational numbers suchthat rn r. For n N, let fn(x) = xrn , x > 0. Since each fn is continuous forx > 0, one may enquire whether the function f defined by f(x) = xr is continuousfor x > 0.

First of all how do we define the xr for x > 0?

We shall discuss this issue in the next subsection, where we shall introduce twoimportant classes of functions, namely, exponential and logarithm functions. In fact,our discussion will also include, as special cases, the Examples 2.16 - 2.18.

2.2.3 Exponential and logarithm functions

We have already come across expression such as ab for a > 0 and b R, though wehave not proved existence of such numbers, and also defined a number denoted by

e as the limit of the sequence

(1 +

1

n

)1/nor the series

n=0

1

n!. Now, we formally

define the following functions:

Exponential function: ex, x R. Natural logarithm function: lnx, x > 0. Exponential function: ax, x R for a given a > 0. Logarithm function with base a > 0: loga x, x > 0.First, we observe that for every x R, the series n=0 xnn! converges absolutely.

This can be seen by using the ratio test. This series plays a very significant role inmathematics.


Definition 2.10 For x R, let

exp(x) :=

n=0

xn

n!.

The function exp(x), x R, is called the exponential function. From the above definition, it is clear that

exp(0) = 1 and exp(1) = e.

In order to derives some of the important properties of the function exp(x), thestudent is urged to do the following exercise.

Exercise 2.9 Suppose that series

n=0 an and

n=0 bn are absolutely convergent.Then, the series

n=0

cn with cn :=nk=0

akbnk ()

is absolutely convergent. Further, show that if =

n=0 an and =

n=0 bn, thenn=0 cn = .

The series

n=0 cn defined in () is called the Cauchy product of the seriesn=0 cn and

n=0 cn. J

Using the conclusion in the above exercise, it can be proved that

exp(x+ y) = exp(x) exp(y) x, y R. ()

Exercise 2.10 Prove () above. J

We observe the following properties:

exp(x) = 1exp(x)

x R. In particular, since exp(x) > 0 for x 0, wehave

exp(x) > 0 x R.[This follows from ()] exp(x) > 1 x > 0 and exp(x) = 1 x = 0.

[From the definition, x > 0 implies exp(x) > 1. Next, suppose x 0. If x = 0, thenexp(x) = exp(0) = 1. If x < 1, then taking y = x, we have y > 1, and hence fromthe first part, exp(y) > 1, i.e., 1/ exp(x) = exp(x) > 1 so that exp(x) < 1. Hence,exp(x) > 1 x > 0. From this, we get exp(x) = 1 x = 0.] x > y exp(x) > exp(y).

[x > y x y > 0 exp(x y) > 1. But, exp(x y) = exp(x)/ exp(y).]


exp(kx) = [exp(x]k x R, k Z. In particular, taking x = 1 and x = 1,

exp(k) = ek k Z.

Since e = exp(1) = exp(k/k) = [exp(1/k]k k N, we have

exp(1/k) = e1/k k N.

exp(m/n) = em/n m,n N. Hence,

exp(r) = er r Q.

We know that every real number is a limit of a sequence of rational numbers.Thus, if r R, there exits a sequence (rn) of rational numbers that rn r. So, itis natural to define

er = limn e

rn

provided the above limit exists. Thus, our next attempt is to show that the functionexp(x), x R, is continuous.

In view

calculus notes of m t nair

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