analysis note

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Lecture Note

Introduction to Mathematical Analysis

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FIRST SEMESTER 2010

Department of Mathematics

The College of Natural Sciences

Kookmin University

COPYRIGHT2010 DEPARTMENT OF MATHEMATICS, KOOKMIN UNIVERSITY. ALL RIGHTS RESERVED.


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Lecture note for

Introduction to Mathematical Analysis

Department of MathematicsThe College of Natural Sciences

Kookmin University861-1, Jeongneung-dong, Seongbuk-guSeoul, 136-702, Koreahttp://math.kookmin.ac.kr


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TABLE DES MATIRES 3

Table des matires

1 The Real Number System 4

1.1 Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . . 4

1.2 The Algebraic Properties of Real Number R . . . . . . . . . . . . . . . . 5

1.3 The Order Properties of Real Number R . . . . . . . . . . . . . . . . . . 6

1.4 The Completeness Property of Real Number R . . . . . . . . . . . . . . . 10

1.5 Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Sequences 14

2.1 Convergent Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Subsequences and the Cauchy criterion . . . . . . . . . . . . . . . . . . . 27

2.5 Upper and Lower Limits of Bounded and Unbounded Sequences . . . . . 35


3 Limits of Functions 42

3.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2 Some Properties of Limits of Functions . . . . . . . . . . . . . . . . . . . 50


4 Continuous Functions 62

4.1 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.2 Properties of Continuous Functions . . . . . . . . . . . . . . . . . . . . . 68

4.3 Uniformly Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 734.4 Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

References 85


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4 The Real Number System


The pain purpose of this chapter is presentation of basic background for the studyof mathematical analysis.

1.1 Principle of Mathematical Induction

Mathematical Induction is one of powerful method of proof that is frequently usedto establish the validity of statements that are given in terms of the natural numbers.Although its utility is restricted to this rather special context, mathematical inductionis an indispensable tool in all branches of mathematics. In this section, we state theprinciple and give various examples to illustrate how inductive proofs proceed.

Let us denote N be the set of natural numbers :

N = {1, 2, 3, } ,

with the usual operations of addition and multiplication, and with the meaning of a na-tural number being less than another one. We will also assume the following fundamentalproperty of natural number.

Axiom 1.1 (Well-ordering property) Every nonempty subset ofN has a least ele-ment.

A more detailed statement of this property is as follows : If S is a subset ofN and ifS = , then there exists m S such that m k for all k S. Based on this property,the principle of mathematical induction can be expressed in terms of subsets ofN.

Theorem 1.2 (Principle of Mathematical Induction) Let S be a subset ofN thatsatisfies the following two properties :

1. The number 1 S.2. For every k

N, if k

S then k + 1

S.

Then S = N.

Now, let us generalize the principle of mathematical induction. Let us denote P(n)be a meaningful statement about n N. Then P(n) may be true for some values n andfalse for others. With this statement, the principle of mathematical induction can bestated as follows :

Theorem 1.3 For each n N, let P(n) be a statement about n. Suppose that1. P(1) is true.

2. For every k N, if P(k) is true then P(k + 1) is true.Then P(n) is true for all n N.


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1.2 - The Algebraic Properties of Real Number R 5

Example 1.4 Use induction to show that

12 + 22 + 32 + + n2 = 16

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

for every n N.In fact, it may happen that statement P(n) are false for some n N but then are true

for every n n0 for some particular n0. Then the principle of mathematical inductioncan be modified to deal with this situation as follows :

Theorem 1.5 (Principle of Mathematical Induction (second version)) Letn0 N and P(n) be a statement for each natural number n n0. Suppose that

1. P(n0) is true.

2. For every k(N)

n0, P(k) is true implies P(k + 1) is true.

Then P(n) is true for all n n0.

It is worth mentioning that another version of the principle of mathematical induc-tion so called Principle of Strong Induction is sometimes quite useful. It can be statedas follows :

Theorem 1.6 (Principle of Strong Induction) LetS be a subset ofN that satisfiesthe following two properties :

1. 1 S.2. For every k N, if {1, 2, , k} S then k + 1 S.

Then S = N.

1.2 The Algebraic Properties of Real Number R

In this section, we shall give the algebraic structure of the real number system. Brieflyexpressed, the real numbers form a field in the sense of abstract algebra. We shall nowexplain what that means. We begin with a definition of binary operation.

Definition 1.7 (Binary operation) A binary operation(or simply, operation) Bon a set F is a function from F F into F.

In the set R of real numbers, there are two binary operations (denoted by + and and called addition and multiplication, respectively) satisfying the following familiarproperties :

(A1) a + b = b + a for all a, b R (commutative property of addition)(A2) (a + b) + c = a + (b + c) for all a,b,c R (associative property of addition)(A3) There exists an element 0

R such that a +0 = 0 + a = a for all a

R (existence

of a zero element)(A4) For each a R, there exists an element a R such that a + (a) = 0 and

(a) + a = 0 (existence of negative element)


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(M1) a b = b a for all a, b R (commutative property of multiplication)(M2) (a b) c = a (b c) for all a,b,c R (associative property of multiplication)(M3) There exists an element 1 R such that a 1 = 1 a = a for all a R (existence

of a unit element)

(M4) For each nonzero a R, there exists an element 1a

R such that a 1a

= 1 and

1

a

a = 1 (existence of reciprocals)(D) a (b + c) = (a b) + (a c) and (b + c) a = (b a) + (c a) for all a,b,c R

(distributive property of multiplication over addition).

From now on, we will obtain some corresponding results of them. First, we will showthat 0 and 1 is the only element ofR that satisfies (A3) and (M3), respectively.

Theorem 1.8 Leta, u and z are elements ofR

1. If a + z = a then z = 0.2. For a = 0, if a u = a then u = 1.

Second, we will show that a and 1a

(when a = 0) are uniquely determined by theproperties given in (A4) and (M4), respectively.

Theorem 1.9 Leta and b are elements ofR,

1. If a + b = 0 then b = a.2. For a = 0, if a b = 1 then b = 1

a.

Third, now we can obtain the following uniqueness of solution of equations :

Theorem 1.10 Let a and b are elements ofR,

1. The equation a + x = b has the unique solution x = (a) + b.2. For a = 0, the equation a x = b has the unique solution x = 1

a

b.Last, we would like introduce some properties :

Theorem 1.11 Let a and b are elements ofR then1. a 0 = 0.2. (a) (b) = ab.

Note that one can explore more algebraic properties of real number. We recommendsome references [2, 3, 4, 5, 7].

1.3 The Order Properties of Real Number R

In this section, we introduce the important order properties of real number R, whichwill play a very important role in subsequent sections. The simplest way to introducethe notion of order is to make use of the notion ofstrict positivity, which we now explain


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1.3 - The Order Properties of Real NumberR 7

Axiom 1.12 (Axiom of order) A relation < defined onRR satisfies the followingaxiom of order

1. For a, b R, exactly one of the following holds (property of trichotomy) :

a = b, a < b or b < a.

2. For a, b R, if 0 < a and 0 < b then 0 < a + b and 0 < ab.3. For a,b,c R, if a < b then a + c < b + c.

If a R, we say that a is a strictly positive real number and write a > 0. Ifa is either in R or is 0, we say that a is a positive real number and write a 0. Ifa R, we say that a is a strictly negative real number and write a < 0. If a iseither in R or is 0, we say that a is a negative real number and write a 0.

Now, we introduce some well-known properties

Theorem 1.13 Leta,b,c R then1. Ifa < b then b < a.2. Ifa < b and b < c then a < c.

3. a2 0 therefore 1 > 0.4. Ifa < b and c < 0 then bc < ac.

5. If0 < a then 0 < 1a

.

6. If0 < a < b then 0 < 1b

< 1a

.

Based on these properties, one can prove following :

Theorem 1.14 For a, b R, if a < b then

a 0,which is equivalent to saying that a and b have the same sign. There are many variations

of the triangle inequality. Herein, we consider two of them.

Corollary 1.20 If a, b R then1. ||a| |b| | |a b|.2. |a b| |a| + |b|.

The following corollary is the generalized triangle inequality :

Corollary 1.21 If a1, a2, , an R then|a1 + a2 + + an| |a1| + |a2| + + |an|.

Now, let us mention a simple but important thing. We will later need precise languageto discuss the notion of one real number being close to another. Ifa is given real number,then saying that a real number b is close to a should mean that the distance |a b|between them is small. A context in which this area can be discussed is provided by theterminology of neighborhoods, which we now define.

Definition 1.22 Leta R and > 0. The neighborhood of a is the setN(a) := {x R : |x a| < } .


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1.3 - The Order Properties of Real NumberR 9

With this definition and corollary 1.16, we can obtain the following important theo-rem.

Theorem 1.23 Leta, b R. For arbitrary > 0, if |a b| < then a = b.

The order relation on real number R determines a natural collection of subsets calledintervals. The following notations and terminology for these special sets will be familiarfrom earlier courses.

Definition 1.24 Let a, b R satisfy a < b1. The open interval determined by a and b is the set

(a, b) := {x R : a < x < b} .

2. The closed interval determined by a and b is the set

[a, b] := {x R : a x b} .

3. The half-open (or half-closed) intervals determined by a and b is the set

[a, b) := {x R : a x < b}(a, b] := {x R : a < x b} .

Notice that the points a and b are called the endpoints of the interval.

There are five types of unbounded intervals for which the symbols

(or +

) and

1 are used as notational convenience in place of the endpoints. The infinite openintervals are the sets of the form

(a, ) := {x R : a < x}(, b) := {x R : x < b} .

Notice that the first and second sets have no upper and lower bounds, respectively.Adjoining endpoint gives the infinite closed intervals as

[a, ) := {x R : a x}(

, b] :={

xR : x

b}

.

It is often convenient to think of the entire set R as an infinite interval. In this case, wewrite

(, ) := R.

An obvious property of intervals is that if two points a, b with a < b belong to aninterval I then any point lying between them also belongs to I. In other words, if a andb belongs to I then the interval [a, b] is contained in I.

Theorem 1.25 (Characterization theorem) If I is a subset ofR that contains at

least two points a and b and a < b. If every t satisfies a < t < b belongs to I then I isan interval.

1It must be emphasized that and are not elements ofR, but only convenient symbols.


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1.4 The Completeness Property of Real Number R

In this section we shall present an important property of the real number systemwhich is often called the completeness property since it guarantees the existence of

elements in R when certain hypotheses are satisfied.We now introduce the notion of an upper bound of a set of real numbers.

Definition 1.26 LetX be a nonempty subset ofR.

1. The set X is said to be bounded above if there exists a number a R such thatx a for all x X. Each number a is called an upper bound of X.

2. The set X is said to be bounded below if there exists a number b R such thatb x for all x X. Each number b is called an lower bound of X.

3. The set X is said to be bounded if it is both bounded above and bounded below.

Example 1.27 The set

A =

1 1

n: n = 1, 2, 3,

is bounded below. The number 0 and any number smaller than 0 is a lower bound of A.This set is also bounded above. The number 1 and any number larger than 1 is an upperbound.

If a set has one upper bound then it has infinitely many upper bounds, because ifa is an upper bound of X then the numbers a + 1, a + 2, are also upper boundsof X (similarly, lower bound is also). So, in the set of upper bounds of X and set oflower bounds ofX, we focus on their least and greatest elements, respectively, for specialattention in the following definition.

Definition 1.28 LetX be a nonempty subset ofR

1. If X is bounded above then a number a is said to be a supremum or a leastupper bound of X if it satisfies the following conditions :

(a) a is an upper bound of X(b) if b is any upper bound of X then a b.

2. If X is bounded above then a number b is said to be a infimum or a greatestlower bound of X if it satisfies the following conditions :

(a) b is an lower bound of X

(b) if a is any lower bound of X then a b.

If the supremum or the infimum of a set X exists, we will denote them by supX andinfX. Let us note the for a nonempty subset X ofR,

1. a X is the maximum of X if x a for every x X and denote

a = max X.


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1.4 - The Completeness Property of Real Number R 11

2. b X is the minimum of X if b x for every x X and denote

b = min X.

If X contains maximum then maxX=supX. Similarly, if X contains minimum thenminX=infX.

Theorem 1.29 LetA be a bounded above, nonempty subset ofR anda R is an upperbound of A. Then the following statements are equivalent :

1. a is the supremum of A

2. for any b R satisfying b < a, there exists x A such that b < x a.

It is impossible to prove on the basis of the field and order properties of real number

that every nonempty subset ofR that is bounded above has a supremum in R. However,it is a deep and fundamental property of the real number system that this is indeedthis case. We will make frequent and essential use of this property, especially in ourdiscussion of limiting processes. The following statement concerning the existence ofsuprema is our final assumption about R.

Axiom 1.30 (Completeness property of real number) Every nonempty set ofreal numbers which has an upper bound also has a supremum inR.

This property is also called the supremum property of real number. The analogousproperty for infima can be deduced from the completeness property as follows :

Theorem 1.31 Every nonempty set of real numbers which has a lower bound has aninfimum inR.

So, based on the completeness property ofR, we can say that R is a complete or-dered (field). From now on, we will give some important applications in order to derivefundamental properties ofR.

One important consequence of the supremum property is that the set of natural

numbers N is not bounded above in R.

Theorem 1.32 (Archimedean property) If x R, there is a natural number nx N such that

x < nx.

This property induces following corollary.

Corollary 1.33 Let x, y be real numbers.

1. Ifx > 0 then there exist n N such thaty < nx.


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2. For any x > 0, there exist n N such that

0 0, there exist n N

uniquely such thatn x < n + 1.

One important property of the supremum property if that it assures the existence ofcertain real numbers. We shall make use of it many times in this way. At the moment wewill show that is guarantees the existence of a positive real number x such that x2 = 2,that is, a positive square root of 2.

Theorem 1.34 There exists a positive number x R such that x2 = 2.

From the above theorem, we now know that there exists at least one irrational realnumber, namely

2. Actually, there are more irrational numbers than rational numbers

in the sense that the set of rational numbers is countable, while the set of irrationalnumbers is uncountable (as shown in the Set Theory). However, we will show that inspite of this apparent disparity, the set of rational numbers is dense in R in the sensethat given any two real numbers there is a rational number between them (in fact, thereare infinitely many such rational numbers).

Theorem 1.35 (The density theorem) If x andy are any real numbers with x < y.

1. Then there exists a rational number r such that x < r < y.

2. Then there exists a irrational number z such that x < z < y.

Another method of completing the rational numbers to obtain R was revised byDedekind. It is based on the notion of a cut.

Definition 1.36 An ordered pair (A, B) of non-empty subset ofR is said to form acutif

A B = , A B = R and a < bfor all a A and b B.

Example 1.37 A typical example of a cut inR is obtained for a fixed element Rby defining

A = {x R } and B = {x R > } .Alternatively, we could take

A = {x R < } and B = {x R } .

Actually, what Dedekind did was, in essence, to define a real number to be a cut inthe rational number system. This procedure enables one to construct the real numbersystem R from the set of rational numbers.

Theorem 1.38 (Dedekind cur theorem) If (A, B) is a cut inR then there exists aunique number R such that a for all a A and b for all b B.


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1.5 - Exercises for Chapter 1 13

1.5 Exercises for Chapter 1

1. Prove that n! > 2n for all n 4, n N.2. Ifa R and a = 0, prove that

1a

=1

a .aa

= 1.

3. Ifa,b,c,d R, prove that(a) ifb = 0 and d = 0 then a

b

cd

=

ac

bd.

(b) ifb = 0 and d = 0 thena

b+

c

d=

ad + bc

bd.

4. Ifa1, a2, , an R then

|a1 + a2 + + an| |a1| + |a2| + + |an|.

5. Prove the Bernoullis inequality : If x > 1 then

(1 + x)n 1 + nx.

6. Obtain the supremum and infimum of following sets :

S1 = 1n (1)n : n N .S2 =

1 +

(1)nn

: n N

.

S3 = {x R : |2x 1| < 11} .

S4 =

(1)nn2n + 1

: n N

.

7. Prove corollary 1.16 by using the completeness property of real number.


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14 Sequences

2 Sequences

This chapter will deal primarily with sequences of real numbers. We shall begin witha study of the convergence of sequences. Some of the results in this chapter may be

familiar to the students from other courses, e.g. Calculus, but the study here is intendedto be rigorous and to give certain more profound results than are usually discussed inearlier courses.

2.1 Convergent Sequences

We begin our study with the introduction of a sequence of real numbers.

Definition 2.1 (A sequence of real numbers) A sequence of real numbers (or

a sequence inR

) is a function defined on the setN

= {1, 2, } of natural numberswhose range is contained in the setR of real numbers.

In other words, a sequence in R assigns to each natural number n = 1, 2, auniquely determined real number. If f : N R is a sequence, we will usually denotethe value off at n by the symbol f(n) := xn. The values xn are called the (nth) termsor the elements of the sequence. We will denote this sequence by the notations {xn}n=1or simply {xn}.

Example 2.2 Let us consider the sequence

xn = (1)n

.

This sequence has infinitely many terms that alternate between 1 and 1, whereas theset of values {xn} is equal to the set{1, 1}.

Let us consider the sequence whose nth terms is defined by the formula

xn = 1 +1

2n.

The first four terms of this sequence are

3

2 ,

5

4 ,

9

8 ,

17

16and the terms corresponding to n = 40, 41, 42 are

1099511627777

1099511627776,

2199023255553

2199023255552,

4398046511105

4398046511104

which are close to 1. For example, x40 =1099511627777

1099511627776differs from 1 by only 1

1099511627776

9.1 1013. It is clear that xn is close to 1 for all large enough positive integers n. Forthis reason we can say that the sequence xn has limit 1, refer to Fig. 2.1.

Generally, we say that a sequence {xn} has limit L if xn is close to L for all largepositive integers n. To define the limit of a sequence, we need to make the concepts close

to and for all large positive integers n precise. In fact, there are a number of differentlimit concepts in real analysis. In this chapter, we introduce the following definition oflimit by using theorem 1.23 in chapter 1.


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2.1 - Convergent Sequences 15

0 5 10 15 20 25 30 35 400.9

1

1.1

1.2

1.3

1.4

1.5

1.6

Fig. 2.1 First 10 values of sequence xn = 1 +1

2n. xn is getting close to 1 when n is

increasing.

Definition 2.3 (Convergent and limit) A sequence {xn} inR is said to convergeto L R or L is said to be a limit of {xn}, if for every > 0 there exists a naturalnumber N() such that for all n N(), the terms xn satisfy

|xn L| < .

If a sequence has a limit, we say that the sequence if convergent; if it has no limit, we

say that the sequence is divergent.

1. Let us notice that the notation N() is used to emphasize that the choice of Ndepends on the value of . However, it is often convenient to write N instead ofN(). For the sake of simplicity, we will use N instead of N().

2. When a sequence {xn} has limit L, we will use the notation

limn

xn = L, limn

xn = L or lim xn = L.

3. Sometimes, the symbolism xn L is used in order to indicate the intuitive ideathat the values xn approach the number L as n .

Example 2.4 A sequence

{xn} =

1

n: n N

is converges to 0. Because, if > 0 is given then 1

> 0. By the archimedean property (seetheorem 1.32 in chapter 1), there exists a natural number N = N() such that 1

N< .

Then, if n N, we have 1n

1N

< . Consequently, if n N then

1n 0 =1

n 1

N < .

Therefore, we can say that the sequence {xn} converges to 0.


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16 Sequences

Example 2.5 A sequence

{xn} =

2 +1

2n: n N

is converges to 2.

Proof. Let > 0 be given. In order to find N, we first note that if n N and a > 1then by applying Bernoullis inequality,

1

(1 + a)n 1

1 + na 0 is an arbitrary positive number, we conclude that L1 = L2 by theorem 1.23.

Let us notice that, above theorem can be argued by contradiction. A more detaineddescription, see [4, Theorem 10.3].

Now, we will consider some results that enable us to evaluate the limits of certain se-quences of real numbers. These results will expand our collection of convergent sequencesrather extensively. We begin by establishing an important property of convergent se-

quences that will be needed in this and later sections.

Definition 2.7 (Bounded sequences) Let{xn} be a sequence of real numbers.


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2.2 - Limit Theorems 17

1. {xn} is said to be bounded above if there exists a real number M > 0 such thatfor all n N,

xn M.2.

{xn

}is said to be bounded below if there exists a real number M > 0 such that

for all n N,xn M.

3. {xn} is said to be bounded when it is both bounded above and bounded below, i.e.,if there exists a real number M > 0 such that for all n N,

|xn| M.

Note that, the sequence {xn} is bounded if and only if the set {xn : n N} of its valueis a bounded subset ofR.

Theorem 2.8 A convergent sequence of real numbers is bounded.

Proof. Suppose thatlimn

xn = L

and = 1. Then there exists a natural number N such that for all n N,

|xn L| < 1.

By applying the triangle inequality (theorem 1.19 in chapter 1), we can obtain for n N|xn| = |xn L + L| |xn L| + |L| < 1 + |L| .

Now, if we setM := sup {|x1| , |x2| , , |xN1| , 1 + |L|} ,

then it follows that |xn| M for all n N.

Example 2.9 The sequence {xn} defined by

xn := 0 if n is odd

1 if n is even

is bounded but has no limit. This example shows that the converse of theorem 2.8 doesnot hold.

2.2 Limit Theorems

In this section, we collect some miscellaneous theorems which are often useful in

proving limits. Before starting, we will examine how the limit process interacts with thealgebraic operations of addition, substraction, multiplication and division of sequences.

Let X = {xn} and Y = {yn} are sequences of real numbers. Then we define :


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18 Sequences

1. Sum of X and Y :

X+ Y = {xn + yn : n N} .2. Difference of X and Y :

X Y = {xn yn : n N} .

3. Product of X and Y :

XY = {xnyn : n N} .4. Multiple of X by k R :

kX = {kxn : n N} .

5. Quotient of X and Y :

XY

=xn

yn: n N

with yn = 0 for all n N.We now show that sequences obtained by applying these operations to convergent

sequences give rise to new sequences whose limits can be predicted.

Theorem 2.10 Let{xn} and{yn} be sequences of real numbers that converges to x andy, respectively. Then

1. For k

R,

{kxn

}converges to kx.

2. {xn + yn} converges to x + y.3. {xnyn} converges to xy.4. If{yn} is a sequence of nonzero numbers that converges to nonzero number y then

xnyn

converges to x

y.

Proof. Proof of 1. is very easy. So, we will prove remaining properties.

2. By hypothesis, for given > 0 there exists a natural number N1 such that ifn

N1 then

|xn x| < 2

.

Similarly, there exists a natural number N2 such that if n N2 then

|yn y| < 2

.

Hence, if N = max {N1, N2}, it follows that if n N then

|(xn + yn) (x + y)| |xn x| + |yn y| < 2

+

2= .

Therefore,

limn

(xn + yn) = x + y.


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2.2 - Limit Theorems 19

3. In order to prove this property, we will consider the following estimation :

|xnyn xy| = |(xnyn xny) + (xny xy)| |xn(yn y)| + |(xn x)y|= |xn| |yn y| + |y| |xn x| .

Since {xn} is a convergent sequence, according to theorem 2.8, there exists a realnumber M1 > 0 such that for all n N,

|xn| M1.

If we set M := max {M1, |y|} then we can obtain the following estimation

|xnyn xy| M|yn y| + M|xn x| .

From the convergence of {xn} and {yn}, we can say that if > 0 is given thenthere exist natural numbers N1 and N2 such that if n N1 and n N2 then

|xn x| < 2M

and |yn y| < 2M

,

respectively. Now, by taking N = max {N1, N2}, we can infer that if n N then

|xnyn xy| M|yn y| + M|xn x| < M 2M

+ M

2M= .

Therefore,limn

xnyn = xy.

4. By 3., it is enough to show that

limn

1

yn=

1

y.

Since {yn} converges, there exists a natural number N1 such that if n N1 then

|yn y| < |y|2

.

From corollary 1.20,

|y|2

|yn y| |yn| |y|for n N1, whence it follows that

|y|2

= |y| |y|2

< |y| |y yn| |y (y yn)| = |yn|

for n N1. Therefore1

|yn| 2

|y|for n

N1 so we have the following estimation 1yn

1

y

=y ynyny

= 1|yn| |y| |y yn| 2

|y|2 |y yn| .


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20 Sequences

Now, if > 0 is given then there exists a natural number N2 such that if n N2then

|y yn| < 12

|y|2 .

By taking N = max {N1, N2} then for n N 1yn 1

y

2|y|2 |y yn| 0, there exists a naturalnumbers N1 and n2 such that if n N1 and n N2 then

|xn L| < and |yn L| < ,respectively. From the hypothesis, we can say that for all n N,

xn L zn L yn Lit follows that

|zn

L

| max

{|xn

L

|,

|yn

L

|}.

Hence, by taking N := max {N1, N2}, we can deduce that|zn L| max {|xn L| , |yn L|} < .


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22 Sequences

Example 2.16 Compute

limn

n

10n.

Proof. Since n

2

< 10n

,0 > xn > xn+1 > .

5. {xn} is (strictly) monotone if it is either (strictly) increasing or (strictly) decrea-sing.

Example 2.19 The following sequences are increasing

{an} = {n : n N} , {bn} = {3n : n N} , {cn} =

1 +1

n

: n N

.

The following sequences are decreasing

{dn} =

1

n: n N

, {en} = {2n : n N} .

The following sequences are not monotone

{fn} = {(1)n : n N} , {gn} = {cos n : n N} .


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24 Sequences

Now, we will introduce an important theorem.

Theorem 2.20 (Monotone convergence theorem) A monotone sequence of realnumbers is convergent if and only if it is bounded. Further :

1. If{xn} is bounded increasing sequence then

limn

xn = sup {xn : x N} .

2. If{xn} is bounded decreasing sequence then

limn

xn = inf{xn : x N} .

3. Bounded monotone sequence is convergent.

Proof. We will prove 1. only. Proof of 2. is a homework.

Let {xn} be a bounded increasing sequence and set S = {xn : x N}. Since {xn} isbounded, there exists a real number M such that

xn M

for all n N. According to the completeness property of real number (see axiom 1.30),the supremum

x = sup {xn : x N}exists in R.

In order to show that lim xn = sup {xn : x N} let > 0 be given. Then x is notan upper bound of set S and hence there exists N N such that

x < xN.

The fact that {xn} is increasing sequence implies that xN xn whenever n N, sothat for all n N,

x < xN xn x < x + .Therefore we have

|xn x| < for all n N. Therefore, we can conclude that

limn

xn = sup {xn : x N} .

The monotone convergence theorem establishes the existence of the limit of a boun-ded monotone sequence. It also gives us a way of calculating the limit of the sequence

provided we can evaluate the supremum (in case 1.) or the infimum (in case 2.). So-metimes it is difficult to evaluate this supremum or infimum, but once we know that itexists, it is often possible to evaluate the limit by other methods.


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2.3 - Monotone Sequences 25

Example 2.21 (Recurrence formula) Let{yn} be defined inductively by

y1 = 3, yn+1 =yn2

+3

yn

for n 1. Show that{yn} is convergent andlimn

yn =

6.

Proof. Since y1 = 3 > 0, yn > 0 for all n N and,

yn+1 yn = yn2

+3

yn yn = 6 (yn)

2

2yn.

It is clear that yn is decreasing. So, in order to apply theorem 2.20, we now show, by

induction, that yn > 6 for all n N.The truth of this assertion can be verified for n = 1 since y1 = 3 >

6. Now suppose

that yk >

6 for some k then

6yk > 6 and

1

2>

36yk

implies1

2(yk

6) >

36yk

(yk

6).

So one can obtainyk2

6

2>

36

3yk

.

Therefore,yk+1 =

yk2

+3

yk>

62

+3

6=

6.

We have shown that the sequence yn is decreasing and bounded below by

6. Itfollows from the theorem 2.20, yn is convergent sequence.

Unfortunately, in this case, it is not so easy to evaluate the lim yn by calcula-ting inf{yn : x N}. However, there is another way to evaluate. Let lim yn = L thenlim yn+1 = L also. By applying theorem 2.10, we can say

L =L

2 +3

L = L = 6, 6.

Since yn > 0 for all n N, L =

6 implies

L = limn

yn =

6.

We end this section by introducing a sequence that converges to one of the mostimportant transcendental numbers in mathematics.

Example 2.22 (Eulers number e) Let{

xn}

be a sequence of real numbers such thatfor all n N,

xn =

1 +

1

n

n.


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26 Sequences

We will show that this sequence is bounded and increasing ; hence it is convergent. Thelimit of this sequence is the famous Eulers number e, whose approximate value is

e

2.718281828459045

,

which is taken as the base of the natural logarithm, refer to Fig. 2.3.

Proof. If we apply the binomial theorem, we have

xn = nC01n + nC11

n1 1

n+ + nCk1nk

1

n

k+ + nCn

1

n

n

= 1 +

nk=1

nCk1nk 1nk

:= 1 +

nk=1

yk.

Similarly,

xn+1 = 1 +n+1k=1

n+1Ck1n+1k

1

n + 1

k:= 1 +

n+1k=1

zk.

Then for k = 1, 2, ,

zk = n+1Ck1n+1k 1

n + 1

k

=(n + 1)n(n 1) (n + 1 k + 1)

k!

1

n + 1

k

=n + 1

n + 1 n

n + 1 n 1

n + 1 n + 1 k + 1

n + 1

1

k!

= 1

1 1n + 1

1 2n + 1

1 k 1

n + 1

1

k!

1

1 1

n 1 2

n 1 k 1

n 1

k!

=n

n n 1

n n 2

n + 1 n + 1 k

n + 1

1

k!

=n(n 1)(n 2) (n + 1 k)

k!

1

n

k

= nCk1nk

1

n + 1

k= yk.

Therefore, xn xn+1 for all n N, so that {xn} is an increasing sequence. In orderto show {xn} is bounded, we will apply the following inequality (see exercise 1.1.11 ofmain textbook)

2n1 n!


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2.4 - Subsequences and the Cauchy criterion 27

for all n N. Then nth term of{xn} is

xn =nC01n + nC11

n1 1

n+ + nCk1nk

1

n

k+ + nCn

1

n

n

=1 + n 1n

+ n(n 1)2!

1n2 + + n(n 1) (n k + 1)

k! 1

nk + + 1

nn

=1 + 1 +1

2!

1 1

n

+ + 1

n!

1 1

n

1 2

n

1 n 1

n

1 + 11!

+1

2!+ 1

k!+ + 1

n!

1 + 120

+1

21+ 1

2k1+ + 1

2n1

=1 +1 1

2

n1

1

2

< 1 +1

1

1

2

= 3.

Hence, we deduce that {xn} is bounded sequence, so that {xn} converges by the mono-tone convergence theorem.

0 20 40 60 80 100 120 140 160 180 2002

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Fig. 2.3 First 200 values of sequence xn =

1 + 1n

n. xn is getting close to e when n

is increasing.

2.4 Subsequences and the Cauchy criterion

In this section we will introduce the notion of a subsequence of a sequence of realnumbers. Informally, a subsequence of a sequence is a selection of terms from the givensequence such that the selected terms form a new sequence. Usually, subsequences arevery useful in establishing the convergence or the divergence of sequence. We will alsoprove the important existence theorem known as the Bolzano-Weierstrass theorem, whichwill be used to establish a number of significant results.

Definition 2.23 (Subsequence) Let{xn} be a sequence of real numbers and let n1 0 be given and let N N be such that if n N then

|xn L| < .

Since n1 < n2 < < nk < is an increasing sequence of natural numbers, it can beproved (by induction) that nk k. Hence if k N, we also have nk k N so that

|xnk L| < .

Therefore, the subsequence {xnk} converges to L.Conversely, since {xn} is a subsequence of itself2 and any subsequence of {xn}

converges to L, {xn} converges to L.

Corollary 2.26 Let{xn} be a sequence of real numbers1. If{xn} converges and there exists a subsequence which converges to L then {xn}

converges to L.

2. If

{xn

}has two convergent subsequences whose limits are not equal then

{xn

}diverges.3. If a subsequence of {xn} diverges then {xn} diverges.

Now, we will prove the important existence theorem known as the Bolzano-Weierstrass theorem : a bounded sequence of real numbers has a convergent subsequence.For that purpose, we will also prove the nested interval theorem.

Definition 2.27 We say that a sequence of intervals {In : n N} is nested if the follo-wing chain of inclusions holds

I1 I2 In In+1 .2By taking nk := k for k N.


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Example 2.28 If for n N,In :=

0,

1

n

then it is clear that In

In+1 for each n

N so that this sequence of intervals is nested.

In this case, the element0 belongs to allIn and the Archimedean property (theorem 1.32)can be used to show that 0 is the only such common point. We denote this by writing

n=1

In = {0} .

Generally, a nested sequence of intervals need not have a common point. Let usconsider the following example.

Example 2.29 If for n N,Jn :=

0,

1

n

then this sequence of intervals is nested, but there is no common point because for everygiven x > 0, there exists m N such that

x >1

m

so that x / Jm. We denote this by writing

n=1

Jn = .

It is an important property of R that every nested sequence of closed, boundedintervals does have a common point (see example 2.28). Notice that the completenessofR plays an essential role in establishing this property.

Theorem 2.30 (Nested intervals property) If In = [an, bn], n N, is a nestedsequence of closed bounded intervals then there exists a number x

R such that x

In

for all n N.

Theorem 2.31 If In = [an, bn], n N, is a nested sequence of closed bounded intervalssuch that the lengths bn an of In satisfy

limn

(bn an) = 0

then the number x In for all n N is unique.

Proof. Since In In+1, it is clear that an an+1 bn+1 bn. Let us define

S = {an : n N}


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30 Sequences

then, since S = and an b1 for all n N, S is bounded above so that there exists asupremum of S. Let us denote this supremum as

x = sup S.

Moreover, {an} is an increasing sequence, by the monotone convergence theorem (theo-rem 2.20), we can say that

limn

an = x.

Let us notice that in fact, it is essential to show that x In for all n N if and onlyif an x bn (refer to [2]). In order to show the uniqueness of x, let y In thenan y bn for all n N. Since

0 y an bn anand lim(bn

an) = 0, by the squeeze theorem (theorem 2.15),

limn

(y an) = 0 implies x = limn

an = y.

Therefore, we can conclude that x = y is the only point that belongs to In for everyn N.

We will now use the above theorem 2.31 to prove an important Bolzano-Weierstrasstheorem, which states that every bonded sequence has a convergent subsequence.

Theorem 2.32 (Bolzano-Weierstrass theorem) A bounded sequence of real num-

bers has a convergent subsequence.

Proof. Let {xn} be a bounded sequence then for all n N, there exists a positive realnumber M such that

|xn| < M.1. For all n N, we define an interval I0 = [a0, b0] satisfying

xn [M, M] = I0.

We now bisect I0 into two equal subintervals

I0

= [M, 0] and I0

= [0, M].

2. One of these intervals must contain xn for infinitely many positive numbers n N.We denote this interval by I1 = [a1, b1].

3. We repeat this process with the interval I1, i.e., we bisect I1 into two equal subin-tervals I1 and I

1 . Notice that if I1 = I

0 then

I1 =

M, M

2

and I1 =

M

2, 0

.

4. Similarly with the previous case, one of these intervals I1 and I1 must contain xn

for infinitely many positive numbers n N. We denote this interval by I2 = [a2, b2].


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5. Continuing this process, we can obtain a nested sequence of interval {In} satisfyingI0 I1 I2

and a subsequence

{xnk

}of

{xn

}such that

{xnk

} Ik for k

N.

6. Since the length of interval In is

(bn an) = M2n1

,

we can obtain

limn

(length of interval In) = limn

M

2n1= lim

n(bn an) = 0.

Therefore, by theorem 2.31, there exists a unique common point x In for alln

N.

7. Moreover, since {xnk} and x both belongs to Ik, we have

|xnk x| 0 there exists a natural number N such that forall natural numbers n, m N, the terms xn and xm satisfy

|xn xm| < .

Example 2.34 The sequence

{xn} = 1n : n Nis a Cauchy sequence.

Proof. If > 0 is given, we choose a natural number N such that 1N

< . Then ifm, n N and n > m (m > n case is similar), we have

|xn xm| =

1

m 1

n

=n m

nm M

and write

limn

xn = +.


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36 Sequences

2. We say that{xn} diverges to minus infinity (or tends to minus infinity) iffor every M R, there exists a natural number N such that if n N then

xn < M

and writelimn

xn = .

3. We say that {xn} is properly divergent in case we have either

limn

xn = + or limn

xn = .

We should realize that we are using the symbols + and purely as a convenientnotation in the above expressions. Results that have been proved in earlier sections for

conventional limits lim xn = L (for L R

) may not remain true when lim xn = .Theorem 2.45 Let {xn} and{yn} be two sequences of real numbers such that

limn

xn = + and limn

yn > 0

thenlimn

xnyn = +.

Proof. Let M be a positive real number. Since, lim yn > 0, choose a positive number Lsuch that

0 < L < limn

yn.

Then there exists a natural number N1 such that, if n N1 then

yn > L.

Since lim xn = +, there exists a natural number N2 such that, if n N2 then

xn >M

L

.

Let N = max {N1, N2} then if n N then

xnyn >M

LL = M

Therefore, we infer that lim xnyn = +.Monotone sequences are particularly simple in regard to their convergence. We have

seen in the monotone convergence theorem that a monotone sequence is convergent ifand only if it is bounded. The next theorem is a reformulation of that result.

Theorem 2.46 A monotone sequence of real numbers is properly divergent if and onlyif it is unbounded.


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2.5 - Upper and Lower Limits of Bounded and Unbounded Sequences 37

1. If{xn} is an unbounded increasing sequence thenlimn

xn = +.

2. If{xn} is an unbounded decreasing sequence thenlimn

xn = .

Proof. Suppose that {xn} is an unbounded increasing sequence. Then for any M R,there exists N N such that

xN > M.

But since {xn} is increasing, for all n N, we have

xn > M.

Since M is arbitrary, it follows that lim xn = +. Remaining part can be proved in asimilar fashion.

The following comparison theorem is frequently used in showing that a sequence isproperly divergent.

Theorem 2.47 Let{xn} and {yn} be two sequences of real numbers and suppose thatfor all n N,

xn yn.

Then the followings are holds :

If limn

xn = + then limn

yn = +.If lim

nyn = then lim

nxn = .

Proof. Let lim xn = + and M R is given. Then there exists a natural number Nsuch that, if n N then

M < xn yn by hypothesis

implies M < yn.

Since M is arbitrary, it follows that lim yn = +. The proof of 2. is similar.Let us notice that since it is sometimes difficult to establish an inequality such as

xn yn, the following limit comparison theorem is often more convenient to use.

Theorem 2.48 (Limit comparison theorem) Let {xn} and {yn} be two sequencesof positive real numbers and suppose that for some positive real number L > 0, we have

limn

xnyn

= L.

Thenlimn

xn = + if and only if limn

yn = +.


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38 Sequences

The following theorem is also useful for obtaining the limit of sequences.

Theorem 2.49 Let {xn} be a sequence of real numbers such that x > 0 for all n N.Then

limnxn = + if and only if limn1

xn = 0.

Proof. Let lim xn = + and for given > 0, set M = 1 R. Then there exists anatural number N such that, if n N then

xn > M =1

implies

1

xn< .

Therefore

1

xn 0 < implies limn

1

xn= 0.

Conversely, let us assume that lim 1xn

= 0 and let = 1M

for a positive number M R.Then there exists a atural number N N such that, if n N then 1xn 0

< = 1M.Since xn > 0 for all n N,

0 M for all nN, we have

limn

xn = +.

Now, let us consider the limit superior and limit inferior of an arbitrary sequence.

Definition 2.50 Let{xn} be a sequence of real numbers1. Let Ak = sup {xk, xk+1, } = sup {xn : n k}. Then L is the limit superior of

{xn} ifL := lim

kAk = lim

ksup xk.

2. Let Bk = inf{xk, xk+1, } = inf{xn : n k}. Then L is the limit inferior of{xn} if

L := limk

Bk = limk

infxk.

The notations limxn and limxn are also used for lim sup xn and lim infxn, respectively.

Theorem 2.51 Let {xn} be a sequence of real numbers thenlimxn = inf

nsup

kn {xk

}and limxn = sup

n

infkn

{xk

}.

Proof. See the theorem 2.28 and exercise 3 of section 2.5 of main textbook.


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2.5 - Upper and Lower Limits of Bounded and Unbounded Sequences 39

Example 2.52 Compute the limit superior and limit inferior of sequence

{xn} =

(1)n + 1n

: n N

.

Proof. Let us define a set Ak = {xn : n k, n N}. If k is an even number then k + 1is an odd number and so on. Then

Ak =

1 +

1

k, 1 + 1

k + 1, 1 +

1

k + 2, 1 + 1

k + 3

implies

sup Ak = 1 +1

kand infAk = 1.

Similarly, if k is an odd number then

Ak =

1 + 1

k, 1 +

1

k + 1, 1 + 1

k + 2, 1 +

1

k + 3

implies

sup Ak = 1 +1

k + 1and infAk = 1.

Therefore,

limxn = 1 and limxn = 1.

Example 2.53 Compute the limit superior and limit inferior of sequence

{xn} =

1

n: n N

.

Proof. Let us define a set Ak = {xn : n k, n N}. Then

Ak =

1

k,

1

k + 1,

1

k + 2,

1

k + 3

implies

sup Ak =1

kand infAk = 0.

Therefore,

limxn = limxn = 0.

Above example says that if a sequence is convergent, its limit superior and limitinferior are same.

Theorem 2.54 Let{xn} be a bonded sequence of real numbers. Iflimn

xn = L if and only if L = limxn = limxn.


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40 Sequences

Proof. Let

L = limn

xn.

Then for > 0 given, there exists a natural number N such that, if n N then

|xn L| < 2

.

Therefore, ifn N then

L 2

< An = sup {xn, xn+1, } L + 2

implies

L 2

< limn

An L + 2

.

Since lim An = limxn,

L < L 2

< limxn L + 2

< L + implieslimxn L < .

Therefore, L = limxn. Similarly, one can show that L = limxn.

Conversely, let us assume that L = limxn = limxn. Then, since

L = limxn = limn

sup {xn, xn+1, } ,

for > 0 given, there exists a natural number N1 such that, if n

N1 then

|sup {xn, xn+1, } L| < implies xn < L + .

Similarly, since

L = limxn = limn

inf{xn, xn+1, } ,for > 0 given, there exists a natural number N2 such that, if n N2 then

|inf{xn, xn+1, } L| < implies L < xn.

Let N = max {N1, N2}. If n N thenL < xn < L + implies |xn L| < .

Therefore,

limn

xn = L.

Corollary 2.55 Let{xn} be a sequence of real numbers. If

limn

xn =

if and only if limxn = limxn =

.

Proof. See the theorem 2.30 of main textbook.


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1. Prove that

(a) Let {an} and {bn} be sequences such that {an} is bounded and lim bn = 0.Then lim

nanbn = 0.

(b) Give an example of sequences {an} and {bn} such that lim bn = 0 but

limn

anbn = 0.

2. Let {an} and {bn} be sequences of real numbers and x R. If for some k > 0 andevery natural number n,

|xn

x

|< k

|an

|and lim

n

an = 0

thenlimn

xn = x.

3. Establish the convergence or the divergence of the following sequences.

(a) xn =3 2n1 + n

.

(b) xn =(1)nn2n 1 .

(c) xn =

n2

2

n + 1 .

(d) xn =1 n

2n.

(e) xn =n!

2n.

(f) xn =n2

2n.

4. Prove that every contractive sequence (see definition 2.41) is a Cauchy sequence.

5. Prove the remaining part 2. of theorem 2.20 (theorem 2.13 of main textbook).

6. Prove that if the subsequences {x2n} and {x2n1} of {xn} converges to a realnumber x then {xn} converges.

7. Calculate the limit superior and limit inferior of following sequences.

(a) {xn} = {1 + (1)n : n N}.(b) {yn} =

1

2, 1,

1

4,

1

3,

1

6,

1

5,

1

8,

1

7,

.

(c) {zn} =

n2(1 + (1)n) : n N.(d) {tn} =

n sin

n

2: n N

.


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42 Limits of Functions


Mathematical analysis is generally understood to refer to that area of mathematicsin which systematic use is made of various limiting concepts : the limit of a sequence of

real numbers. In this chapter, we will encounter the notion of the limit of function.

3.1 Limits of Functions

The intuitive idea of the function f having a limit L at the point a is that the valuesf(x) are close to L when x is close to (but different from) a. But it is necessary to havea technical way of working with the idea ofclose to and this is accomplished in the definition given in this section.

In order for the idea of the limit of a function f at a point a to be meaningful, it is

necessary that f be defined at the point close to a. It is need not be defined at the pointa, but it should be defined at enough points close to a to make the study interesting.These are the reasons for the following definitions :

Definition 3.1 (Neighborhood (revisited)) Leta R and > 0.1. The neighborhood of a is the set

N(a) := {x R : |x a| < } = {x R : a < x < a + } .

2. D is called the neighborhood of a if there exists an neighborhood N(a) suchthat

N(a) D.3. The deleted neighborhoodof a is the set

N (a) := {x R : 0 < |x a| < } = {x R : a < x < a + } {a} .

Example 3.2 Let us consider the following examples of neighborhood :

1. Let I := {x : 0 < x < 1} and a I. Let = min {a, 1 a} then N(a) is anneighborhood of a. Moreover, for arbitrary x N(a),

0 a < x < a + 1 implies x I.It means that N(a) is contained in I. Thus I is neighborhood of a.

2. Let I := {x : 0 x 1} then for any > 0, N(0) contains points not in I, andso N(0) is not contained in I. For example, the number x = 2 is in N(0) butnot in I.

3. Let a be a real number. For given > 0, if x N(a) then x = a by theorem 1.23.

Definition 3.3 Let D R. A point x is an accumulation point or cluster point(orlimit point) ofD if for every

neighborhood N(x) ofx contains at least one point

of D distinct from a, i.e.,

(x , x + ) (D {x}) = .


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3.1 - Limits of Functions 43

Let us notice that the point x may or may not be a member of D, but even if it isin D, it is ignored when deciding whether it is an accumulation point of D or not, sincewe explicitly require that there be points in N(x) D distinct from x in order for x tobe an accumulation point of D. For the sake of simplicity, we set the domain D be a

non-empty subset ofR

.

Example 3.4 Let us consider the following examples of accumulation point :

1. The setS =

1

n: n N

has only the point 0 as an accumulation point. None of

the points in S is a cluster point of S.

2. The set S = {0}

1

n: n N

has only the point 0 as an accumulation point.

3. For intervals (0, 1), (0, 1], [0, 1) and [0, 1], every point of the closed interval [0, 1]is an accumulation point of them.

Theorem 3.5 A numbera R is an accumulation point of a subset D R if and onlyif there exists a sequence {an} in D such that for all n N

limn

an = a and an = a.

Proof. If a is an accumulation point of D then for any n N, the 1nneighborhood

N1n

(a) contains at least one point an in D distinct from a. Then

an A, an = a and |an a| < 1n

implies limn

an = a.

Conversely, if there exists a sequence {an} in D {a} with limn

an = a then for any

> 0, there exists N N such that if n N thenan N(a).

Therefore, for n N, N(a) contains the points an such thatan

D and an

= a.

It means that a is an accumulation point of D.

We now state the precise definition of the limit of a function f at a point a. It isimportant to note that in this definition, it is immaterial whether f is defined at a ornot. In any case, we exclude a from consideration in the determination of the limit.

Definition 3.6 (Limit of function) Let D R and a be an accumulation point ofD. A function f : D R, a real number L is said to be a limit of f at a if for given > 0, there exists a () > 0 such that if x D and 0 < |x a| < () then

|f(x) L| < .If the limit of f at a does not exists, we say that f diverges at a.


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1. Let us notice that the notation () is used to emphasize that the choice of depends on the value of . However, it is often convenient to write instead of(). For the sake of simplicity, we will use instead of ().

2. IfL is a limit off at a then we also say that f converges to L at a. We often write

limxa

f(x) = L.

3. Sometimes, the symbolism

f(x) L as x a

is used in order to indicate the intuitive idea that the f has limit L at a.

The following theorem indicates that the value ofL ofthe limit of function is uniquelydetermined. This uniqueness is not part of the definition of limit, but must be deduced.

Theorem 3.7 (Uniqueness of limits) Let f : D R be a function and if a is anaccumulation point of D then f can have only one limit at a.

Proof. Let L1 and L2 are both limits off at a. Then for given > 0, there exists 1 > 0such that if x D and 0 < |x a| < 1 then

|f(x) L1| < 2

.

Also there exists 2 > 0 such that if x D and 0 < |x a| < 2 then|f(x) L2| <

2.

Now, let = min {1, 2} then if a D and 0 < |x a| < ,

|L1 L2| |L1 f(x)| + |f(x) L2| < 2

+

2= .

Since > 0 is arbitrary, L1 = L2.

The definition of limit can be described in terms of neighborhoods, refer to Fig. 3.1.We observe that because

N(a) = {x : |x a| < } ,the inequality 0 < |xa| < is equivalent to saying that x = a and x N(a). Similarly,the inequality |f(x) L| < is equivalent to saying that f(x) N(L). In this way wecan obtain the following result. The proof is left to reader.

Theorem 3.8 Let f : D R be a function and a be an accumulation point of D.Then the following statements are equivalent :

1. limxa

f(x) = L.

2. Given anyN(L), there exists aN(a) such that if x = a is any point in N(a)Dthen f(x) N(L).


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L

N(L)

N(a)

a

given

there exists

x

y

Fig. 3.1 The limit of f at a.

Example 3.9 Show thatlimxa

b = b.

Proof. Let f(x) := b for all x R then if > 0 is given, we let = (in fact, anystrictly positive will serve the purpose, e.g., = 1). Then if 0 < |x a| < then

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

Since > 0 is arbitrary, we conclude that

limxa

b = b.

Example 3.10 Show that

limxa

x = a.

Proof. Let f(x) := x for all x R then if > 0 is given, we let = . Then if0 < |x a| < then

|f(x) a| = |x a| < = .Since > 0 is arbitrary, we deduce that

limxa

x = a.

Example 3.11 Show that

limxa

x2 = a2.

Proof. Let f(x) := x2 for all x R. We want to make the difference as

|f(x) a2| = |x2 a2| <


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for a preassigned > 0 by taking x sufficiently close to a. To do so, we note thatx2 a2 = (x + a)(x a). Moreover if |x a| < M then3

|x| |a| + M so that |x + a| |x| + |a| 2|a| + M.

Therefore, if |x a| < M, we have|f(x) a2| = |x2 a2| = |x + a||x a| (2|a| + M)|x a|. (3.1)

The last term of above inequality will be less than provided we take

|x a| < 2|a| + M.

Consequently, if we choose

:= minM,

2|a| + M,

then if 0 < |x a| < , it follow first that |x a| < M so that (3.1) is valid, andtherefore,

|f(x) a2| = |x2 a2| = |x + a||x a| (2|a| + M) 2|a| + M < .

Since we have a way of choosing > 0 for an arbitrary choice of > 0, we infer that

limxa

f(x) = limxa

x2 = a2.

Example 3.12 Show that if a > 0,

limxa

1

x=

1

a.

Proof. Let f(x) := 1x

for all x R. We want to make the difference asf(x) 1a =

1x 1

a

< for a preassigned > 0 by taking x sufficiently close to a. To do so, we note that for

x > 0, 1x 1

a

= 1ax(a x)

= 1ax |x a|.It is useful to get an upper bound for the term 1

axthat holds in some neighborhood of

a. In particular, if |x a| < 12

a then 12

a < x < 32

a, so that

0 0, we infer that

limxa

f(x) = limxa

1

x=

1

a.

There are times when a function f may not posses a limit at a point a, yet a limitdoes exist when the function is restricted to an interval on one side of the accumulationpoint a. For example, the following signum function sgn defined by (see Fig. 3.2)

sgn(x) :=

+1 for x > 00 for x = 0

1 for x < 0

has no limit at a = 0. However, if we restrict the sgn(x) to the interval (0, ), theresulting function has a limit of 1 at a = 0. Similarly, if we restrict the sgn(x) to theinterval (

, 0), the resulting function has a limit of

1 at a = 0. These are elementary

examples of right-hand and left-hand limits at a = 0.

x

y

0

1

-1

f(x)=sgn(x)

Fig. 3.2 The signum function f(x) = sgn(x).

Definition 3.13 (Right-hand and Left-hand limits) Let f : D R be a func-tion.

1. If a is an accumulation point of D (a, ), then we say that L R is a right-hand limit of f at a if given any > 0, there exists a > 0 such that for allx D with 0 < x a < ,

|f(x) L| < .

In this case, we write

limxa+

f(x) = L or f(a+) = L.


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2. If a is an accumulation point of D (, a), then we say that L R is a left-hand limit of f at a if given any > 0, there exists a > 0 such that for allx D with 0 < a x < ,

|f(x) L| < .In this case, we write

limxa

f(x) = L or f(a) = L.

Let us notice that the limits limxa+

f(x) and limxa

f(x) are called one-sided limits of f

at a. It is possible that neither one-sided limit may exists. Also, one of them may existwithout the other existing. Similarly, as is the case for f(x) = sgn(x) at x = 0, theymay both exist and be different.

The following result relates the notion of the limit of function to one-sided limits.

Theorem 3.14 Let f : D R be a function and a be an accumulation point ofD (a, ) and D (, a). Then

limxa

f(x) = L if and only if limxa+

f(x) = L = limxa

f(x).

Proof. Suppose thatlimxa

f(x) = L.

For given > 0, there exists > 0 such that if 0 0, there exists 1 > 0 such that if 0 < x a < 1 then|f(x) L| < .

Moreover, there exists 2 > 0 such that if 0 < a x < 2 then|f(x) L| < .

Let = min {1, 2} then if 0 < |x a| < either 0 < x a < 1 or 0 < a x < 2 sothat|f(x) L| < implies lim

xaf(x) = L.


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Example 3.15 For x = 0, let us consider the function

f(x) := |x| + x|x| .

Thenlimx0+

f(x) = 1 and limx0

f(x) = 1.

Example 3.16 Calculate the right-hand and left-hand limit of the function f(x) atx = 1 (see Fig. 3.3)

f(x) :=

2x + 1 for x > 1

x

2for x 1

Proof. For given > 0 there exists = 2

> 0 such that if1 < x < 1 + then |x 1| < so that

|f(x) 3| = |(2x + 1) 3| = 2|x 1| < 2 = .Therefore, we infer that

limx1+

f(x) = 3.

For given > 0 there exists = 2 > 0 such that if 0 < 1 x < then

|x 1| <

so that

|f(x) 3| =x2

1

2

= 12 |x 1| 0 there exists > 0 such that if 0 0, there exists N N such that if n N then

|xn a| < .

But for each xn, xn = a, we have 0 < |xn a| so that; ifn N then 0 < |xn a| < ,we have

|f(xn) a| < .

Therefore, {f(xn)} converges to L.Conversely, in order to apply the contrapositive argument, assume that

limxa

f(x) = L.

Then there exists N0(L) such that no matter what N(a) we pick, there exists at leastone number x with 0 < |x a| < and x = a such that

|f(x) L| 0.

Hence for every n N

, there exists N1

n (a) contains a number xn D such that0 < |xn a| < 1

nbut |f(xn) L| 0 for all n N.


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3.2 - Some Properties of Limits of Functions 51

So, we conclude that the sequence {xn} in D that converges to a such that xn = a, butthe sequence {f(xn)} does not converges to L. This is contradiction. So, we have

limxa

f(x) = L.

Sometimes, it is important to be able to show that a certain number is not the limitof a function at a point or that the function does not have a limit at a point. Thefollowing result is a consequence of theorem 3.17.

Theorem 3.18 (Divergence criterion) Let f : D R be a function and a be anaccumulation point of D. Then the following are equivalent.

1. limxa

f(x) = L.2. There exists a sequence{xn} in D with xn = a for all n N such that

limnxn = a but limn f(xn) = L.Example 3.19 Let

f(x) := sin1

x

then limx0

f(x) does not exist inR (See Fig. 3.4).

0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

xaxis

yaxis

1

1

3

1

2

Fig. 3.4 Graph of f(x) in example 3.19.

Proof. Now, we recall from Elementary Calculus that for integer n,

sin x =

0 if x = n1 if x =

2+ 2n.

For n N, let xn := 1n thenlimn

xn = 0 and f(xn) = sin n = 0 so that limn

f(xn) = 0.


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On the other hand, for n N, let yn :=

2+ 2n

1then

limn

yn = 0 and f(yn) = sin

2+ 2n

= 1 so that lim

nf(yn) = 1.

Therefore,limn

f(xn) = limn

f(yn).

This implies that

limx0

f(x) = limx0

sin1

x

does not exist.

Definition 3.20 (Bounded function) Letf : D R be a function and let a be anaccumulation point of D. We say that f is bounded on a neighborhood of a if there

exists a N(a) and a constant M > 0 such that for all a D N(a),|f(x)| M.

Theorem 3.21 Let f : D R be a function and a be an accumulation point of D. If

limxa

f(x) = L

then f is bounded on some neighborhood of a.

Proof. Sincelimxa

f(x) = L,

for = 1, there exists > 0 such that if 0 < |x a| < then

|f(x) L| < 1 implies |f(x)| |L| |f(x) L| < 1.

Therefore, ifx D N(a) and x = a, then |f(x)| < L + 1. Now let

M = |L| + 1 if a / D

sup {|f(a)|, |L| + 1} if a D.It follows that if x D N(a) then

|f(x)| M.

This shows that f is bounded on some neighborhood N(a) of a.

Theorem 3.22 Let f : D R be a function and a be an accumulation point of D. If

limxa

f(x) = L > 0

then there exists a neighborhood N(a) such that f(x) > 0 for all x D N(a) andx = a.


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Proof. Sincelimxa

f(x) = L > 0,

suppose that = L2

> 0. Then there exists > 0 such that if0 < |x a| < and x D

then |f(x) L| < L2

implies L2

< f(x) L < L2

.

Therefore it follows that if 0 < |x a| < and x D then

f(x) >L

2> 0.

The next definition is similar to the definition for sums, differences, products, andquotients of sequences given in section 2.2.

Definition 3.23 Let f : D R be a function and g : D R be functions. Wedefine

1. Sum f + g :(f + g)(x) = f(x) + g(x).

2. Difference f g :(f g)(x) = f(x) g(x).

3. Product f g :f g(x) = f(x)g(x).

4. Multiple kf for k R :(kf)(x) = kf(x).

5. Quotient fg

: f

g

(x) =

f(x)

g(x)

with g(x) = 0 for all x D.

Theorem 3.24 Letf : D R andg : D R be functions anda be an accumulationpoint of D. Further, let k

R. If

limxa

f(x) = L and limxa

g(x) = M

then :

1. limxa

(f + g)(x) = L + M.

2. limxa

kf(x) = kL.

3. limxa

(f g)(x) = L M.4. lim

xa

(f g)(x) = LM.

5. limxa

f

g

(x) =

L

Mwhere M = 0.


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Proof. The proof of this theorem is very similar to that of theorem 2.10. Notice thatone can prove this theorem by using Sequential criterion (theorem 3.17).

1. For given > 0, there exists a 1 > 0 such that ifx D and 0 < |x a| < 1 then

|f(x) L| 0 such that if x D and 0 < |x a| < 2 then

|g(x) M| < 2

.

Let us take = min {1, 2}. If x D and 0 < |x a| < then

|f(x) + g(x) (L + M)| |f(x) L| + |g(x) M| < 2

+

2= .

Therefore,limxa

(f + g)(x) = L + M.

2. Ifk = 0 then this property holds so let us assume that k = 0 case. For given > 0,there exists > 0 such that if x D and 0 < |x a| < then

|f(x) L| < |k| .

Analogously, if x D and 0 < |x a| < then

|kf(x) kL| = |k||f(x) L| < |k| |k| = .

Therefore,limxa

kf(x) = kL.

3. By combining 1. and 2., we can prove it.

4. In order to prove this property, we will consider the following estimation :

|f(x)g(x) LM| = |f(x)g(x) Lg(x) + Lg(x) LM|

|f(x)

L| |

g(x)|

+|L| |

g(x)

M|

.

Since g has limit M, according to theorem 3.21, g is bounded on some neighbo-rhood of a so that ; there exists 1 > 0 such that if x D and 0 < |x a| < 1then

|g(x)| < |M| + 1.By hypothesis, for given > 0 there exists 2 > 0 such that if x D and0 < |x a| < 2 then

|f(x) L| < 2(|M| + 1) .

And there exists 3 > 0 such that if x D and 0 < |x a| < 3 then|g(x) M| <

2|L| .


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Let us take = min {1, 2, 3}. If x D and 0 < |x a| < then|f(x)g(x) LM| |f(x) L| |g(x)| + |L| |g(x) M|

0 such that ifx

D and 0 0, there exists > 0 such that if x

D and 0 k2

> 0.

This is contradiction. So, we have

limxa

f(x) limxa

g(x).

We now state an analogue of the squeeze theorem 2.15. We leave its proof to thereader.

Theorem 3.31 (Squeeze theorem) Let f , g , h : D R be functions and a be anaccumulation point of D. If for x D and x = a,

f(x) g(x) h(x) and limxa

f(x) = L = limxa

h(x)

then

limxa g(x) = L.

Theorem 3.32 Let f, g : D R be functions and a be an accumulation point of D.If for x D and x = a,

g(x) is bounded and limxa

f(x) = 0

thenlimxa

f(x)g(x) = 0.

Example 3.33 Prove that

limx0

x sin1

x= 0.


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Proof. Let

f(x) = x sin1

xfor x = 0. Since for all x R, 1 sin x 1, we have the following inequality

|x| f(x) = x sin 1x

|x|for all x R, x = 0. Since lim

x0x = 0, it follows from the squeeze theorem 3.31 that

limx0

x sin1

x= 0.

For a graph, see Fig. 3.5.

0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.40.25

0.2

0.15

0.1

0.05

0

0.05

0.1

0.15

0.2

0.25

xaxis

yaxis

1

1

2

1

3

Fig. 3.5 Graph of f(x) in example 3.33.

Let us consider the function

f(x) =1

x2

for x

= 0, refer to Fig. 3.6. f(x) is not bounded on a neighborhood of 0, so it cannot

have a limit in the sense of definition 3.6. While the symbols (= +) and do notrepresent real numbers, it is sometimes useful to be able to say that f(x) approaches(or tends) to as x 0. This use of will not cause any difficulties, provided weexercise caution and never interpret or as being real numbers.

Definition 3.34 Letf : D R be a function and a be an accumulation point of D.1. We say that f approaches to infinity (or tends to infinity) as x a if for

everyM R there exists = (M) > 0 such that for allx D with0 < |xa| < then

f(x) > M

and writelimxa

f(x) = (or + ).


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2 1.5 1 0.5 0 0.5 1 1.5 22

0

2

4

6

8

10

x0

y

2 1.5 1 0.5 0 0.5 1 1.5 25

4

3

2

1

0

1

0

x

y

Fig. 3.6 Graph of f(x) = 1/x2 (left) and f(x) = log |x| (right) for x = 0.

2. We say that f approaches to minus infinity (or tends to minus infinity)as x a if for every M R there exists = (M) > 0 such that for all x Dwith 0 < |x a| < then

f(x) < M

and writelimxa

f(x) = .

Example 3.35 Prove that (see Fig. 3.6)

limx0

1

x2= .

Proof. For every M R, there exists = 1M

such that if 0 < |x 0| < then

x2 M.

Example 3.36 Prove that (see Fig. 3.6)

limx0

log |x| = .

Proof. For every M R, there exists = eM such that if 0 < |x 0| < then

|x| < eM so that log |x| < M.

Similarly with the definition 3.13, it will be useful to consider one-sided infinite limits

as

limxa+

f(x) = , limxa

f(x) = , limxa+

f(x) = and limxa

f(x) = .


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Example 3.37 For x = 0,f(x) =

1

xdoes not tend to either or as x 0. In fact

limx0+

f(x) = and limx0

f(x) = .

It is also desirable to define the notion of the limit of a function as x . Thedefinition as x is similar.

Definition 3.38 Letf : D R be a function.1. We say that L R is a limit of f as x if given any > 0 there exists M

such that for any x > M, then

|f(x)

L

|<

and writelimx

f(x) = L.

2. We say that f approaches to infinity (or tends to infinity) as x ifgiven any M R there exists K R such that for any x > K, then

f(x) > M

and writelimx

f(x) = .

3. Similarly, we can definelim

xf(x) = L, lim

xf(x) = , lim

xf(x) = and lim

xf(x) = .

Example 3.39 Compute

limx

1

2x + 3

Proof. For given > 0 there exists M =1

2> 0 such that if x > M then

12x + 3 0 = 12x + 3 < 12x < 12M < .

Therefore,

limx

1

2x + 3

= 0.

Example 3.40 Computelimx

x

Proof. For any M > 0, there exists K = M2 > 0 such that if x > K thenx >

K = M implies lim

x

x = .


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1. Show that the following limit does not exist

limx0

x

|x|.

2. Evaluate the following limits, or show that they do not exist.

(a) limx0

2|x|x

.

(b) limx2+

x2 4|x 2| .

(c) limx0+

sin

1

x

.

(d) limx0+x

1

x

.

(e) limx0+

4x[x]

2x + |x| .

(f) limx0

4x[x]

2x + |x| .3. A function f : R R satisfies

f(x + y) = f(x) + f(y)

for all x, y

R and there exists lim

x0

f(x). Show that the following holds.

(a) limx0

f(x) = 0.

(b) For all a R, there exists limxa

f(x).

4. Prove the following by using squeeze theorem 3.31 (or theorem 3.16 of main text-book).

(a) limx0

sin x = 0.

(b) limx0

cos x = 1.

(c) limx0

sin x

x= 1.

(d) limx0

cos x 1x

= 0.


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62 Continuous Functions


We now begin the study of the most important class of functions that arises inanalysis : the class of continuous functions.

4.1 Continuous Functions

In this section, we will define what it means to say that a function is continuous at apoint, or on a set. This notion of continuity is one of the central concepts of mathematicalanalysis, and it will be used in almost all of the following material. For convenience, weset the domain D be a non-empty subset ofR.

Definition 4.1 (Continuous function) Letf : D

R be a function and leta

D.We say that f is continuous at a if, given any number > 0 there exists > 0 suchthat if x D satisfying |x a| < then

|f(x) f(a)| < .

If f is continuous on every point of D, then we say that f is continuous on D. If ffails to be continuous at a, then we say that f is discontinuous at a.

As with the definition of limit, the definition of continuity at a point can be formu-

lated in terms of neighborhoods. This is done in the next result.

Theorem 4.2 A function f : D R is continuous at a point a D if and only ifgiven any > 0 there exists > 0 such that if x D N(a) then

f(x) N(f(a)).

Let us notice that ifa is an accumulation point ofD, then a comparison of definitions3.6 and 4.1 show that f is continuous at a if and only if

limxa

f(x) = f(a).

Thus, if a is an accumulation point of D, then three conditions must hold for f to becontinuous at a :

1. f must be defined at a (so that f(a) makes sense)

2. the limit of f at a must exist in R (so that limxa

f(x) makes sense)

3. these two values must be equal (so that limxa

f(x) = f(a)).

Example 4.3 Constant function, f(x) = x and f(x) = x2 are continuous functions,refer to examples 4.1, 4.2 and 4.3 of main textbook.


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4.1 - Continuous Functions 63

Example 4.4 Let f : R R be a function defined as

f(x) =

x sin1

xif x = 0

0 if x = 0

then f is continuous at x = 0.

Proof. For any > 0, there exists = > 0 such that if |x 0| < , x R then

|f(x) f(0)| =x sin 1x

|x| < = .Therefore, combining the result in example 3.33, f is continuous at x = 0.

A slight modification of the proof of theorem 3.17 for limits yields the followingsequential version of continuity at a point.

Theorem 4.5 (Sequential criterion for continuity) A function f : D R iscontinuous at the pointa D if and only if for every sequence{xn} inD that convergesto a, the sequence {f(xn)} converges to f(a).

The following discontinuity criterion is a consequence of the last theorem.

Theorem 4.6 (Discontinuity criterion) A functionf : D R is discontinuous atthe point a D if and only if for every sequence {xn} in D that converges to a, but thesequence {f(xn)} does not converges to f(a).

Example 4.7 Let f : R R be a function defined by

f(x) :=

0 if x is rationalx if x is irrational.

We claim that f is continuous at x = 0 and is discontinuous except x = 0.

Proof. For given > 0, there exists = > 0 such that if |x 0| < then

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

Therefore,limx0

f(x) = 0,

i.e., f is continuous at x = 0.

Now, assume that x = 0 then since |x| > 0, there exists n0 N such that|x| > 1

n0.


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For each n N, let us definexn = x 1

n0 + n.

First, let us assume that x is irrational number. Since xn is irrational number, there

exists a rational number yn such that

xn < yn < x.

Since limn

xn = x, by squeeze theorem

limn

yn = x.

Therefore by the sequential criterion for continuity (theorem 4.5),

limn

f(yn) = 0= x = f(x). (4.1)

Next, let us assume that x is rational number. Since xn is rational number, thereexists an irrational number yn such that

xn < yn < x.

Since limn

xn = x, by squeeze theorem

limn

yn = x.

Therefore by the sequential criterion for continuity (theorem 4.5),

limn

f(yn) = x = 0 = f(x). (4.2)

Since every real number if either rational or irrational, by (4.1) and (4.2), f is disconti-nuous at x = 0.

Example 4.8 Letf : R R be Dirichlets discontinuous function4 defined by

f(x) := 1 if x is rational

0 if x is irrational.

We claim that f is not continuous at any point ofR.

Proof. If a be a rational number, let {xn} be a sequence of irrational numbers suchthat5

limn

xn = a.

Since f(xn) = 0 for all n N, we have

limn

f(xn) = 0= 1 = f(a).

4This function was introduced in 1829 by P. G. L. Dirichlet.5The density theorem 1.35 assures us that such a sequence does exist.


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On the other hand, if a be an irrational number, let {xn} be a sequence of rationalnumbers such that6

limn

xn = a.

Since f(xn) = 0 for all nN, we have

limn

f(xn) = 1 = 0 = f(a).

Since every real number if either rational or irrational, we deuce that f is not continuousat any point in R.

Example 4.9 Let h : (0, 1) R be Thomaes function7 defined by

h(x) :=

1

nif x =

m

nfor m, n N and gcd(m, n) = 1

0 if x is irrational.

We claim that h is not continuous at every irrational number in (0, 1), and is disconti-nuous at every rational number in (0, 1), refer to Fig. 4.1.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

Fig. 4.1 Graph of h(x). Left : largest n = 10. Right : largest n = 20.

Proof. If c (0, 1) be a rational number, let {xn} be a sequence of irrational numbersin (0, 1) such that

limn

xn = c.

Since h(xn) = 0 for n N,limn

h(xn) = 0.

But c is rational, h(c) > 0. Therefore,

limn

h(xn) = h(c).6Similarly with the previous situation, the density theorem 1.35 assures us that such a sequence

does exist.7This function was introduced in 1875 by K. J. Thomae.


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Hence h is discontinuous at rational number c (0, 1).On the other hand, if d (0, 1) be an irrational number and > 0 then by the

Archimedean property (theorem 1.32), there exists n0 N such that

1n0

< .

Let us define a set of rational numbers

Qn0 =m

n: 1 n n0, 0 < m < n, gcd(m, n) = 1

=

1

2,

1

3,

2

3,

1

4,

3

4,

1

5,

2

5, , 1

n0, , n0 1

n0

.

Then there are only a finite number of rationals in Qn0. For d / Qn0, if we set > 0such that

= min {|d x| : x Qn0}then the neighborhood (d, d + ) contains no rational numbers with denominator lessthan n0. It means that n > n0 since, a rational number

m

n (0, 1) satisfies

mn

d < .

It follows that for |x d| < , x (0, 1), we have

|h(x) h(d)| = |h(x)| 1n0

< .

Thus h is continuous at the irrational number d (0, 1).The next result is similar to theorem 3.24, from which it follows.

Theorem 4.10 Let f, g : D R be functions and k R. Suppose that a D andthat f and g are continuous at a. Then

1. f + g, f g, f g and kf are continuous at a.2. If g(x) = 0 for all x D then the quotient f

gis continuous at a.

The next result is an immediate consequence of theorem 4.10, applied to every pointof D. This is an important result.

Theorem 4.11 Let f, g : D R be continuous on D and k R.1. f + g, f g, f g and kf are continuous on D.2. If g(x) = 0 for all x D then the quotient f

gis continuous on D.

Example 4.12 A polynomial function

p(x) = a0 + a1x + a2x2 + + an1xn1 + anxn

is continuous onR.


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Example 4.13 f(x) = sin x is continuous onR.

Proof. For all x,y,z R, we have

| sin z| |z|, | cos z| 1 and sin x sin y = 2 sin x y2

cos x + y2

.

Hence for a R, we have

0 | sin x sin a| =2sin x a2 cos

x + a

2

2x a2

= |x a|.Since lim

xa|x a| = 0, by squeeze theorem,

limxa

sin x = sin a.

Therefore f(x) = sin x is continuous at x = a. Since a R is arbitrary, it follows thatf is continuous on R.

Example 4.14 g(x) = x1 is continuous onR {0}.

In the next result, we will show that the composition of continuous functions is alsoa continuous function.

Theorem 4.15 Let A, b R and let f : A R and g : B R be functions suchthat f(A) B. If f and g are continuous at a A andb = f(a) B, respectively, thenthe composition

g f : A Ris continuous at a.

Proof. Since g is continuous at b = f(a) B, for given > 0, there exists 1 > 0 suchthat if y B and |y f(a)| < 1 then

|g(y)

g(f(a))

|< .

Since f is continuous at a A, with this 1 > 0, there exists > 0 such that if x Aand |x a| < then

|f(x) f(a)| < 1 implies |g(f(x)) g(f(a))| < .

It follows that g f is continuous at a.

Example 4.16 Since f(x) = x1 is continuous at every point onR {0} and g(x) =sin x is continuous onR,

(g f)(x) = sin 1x

is continuous at every point onR {0}.


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4.2 Properties of Continuous Functions

Functions that are continuous on intervals have a number of important propertiesthat are not possessed by general continuous functions. In this section, we will establish

some results that are of considerable importance and that will be applied later.

Definition 4.17 A function f : D R is said to be bounded on D if there exists aconstant M > 0 such that

|f(x)| Mfor all x D. On the other hand, F is said to be unbounded on D if given M > 0,there exists a point x D such that

|f(x)| > M.

Example 4.18 f(x) =

x is bounded on [0, 2].

Example 4.19 Let us consider the function f defined on the interval (0, 1) by

f(x) =1

x.

For any M > 0, we can take the point x0 =1

M + 1 (0, 1) such that

|f(x0)| = 1x0

= M + 1 > M.

Therefore, f is not bounded on (0, 1).

Example 4.19 shows that continuous functions need net be bounded however, wellshow that continuous functions on a certain type of interval are necessarily bounded.

Theorem 4.20 (Boundedness theorem) Letf : [a, b] R be continuous on [a, b].Then f is bounded on [a, b].

Proof. Suppose that f is unbounded on [a, b]. Then for any n N there exists a numberxn [a, b] such that

|f(xn)

|> n.

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