F18CE2 Multivariable Calculus & Real Analysis Bbryan/F18CE2/slides/8CE2-slides.pdf ·...

F18CE2 Multivariable Calculus& Real Analysis B

Bryan RynneDepartment of Mathematics

Heriot-Watt University

March 12, 2020

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Chapter 1

Sequences and Limits

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1.1 Some basic notation and results

The modulus of a number

Definition 1.1 For any x ∈ R, the modulus (or absolute value) ofx, denoted by |x|, is defined to be:

|x| :=

{x if x > 0,

−x if x < 0.

So, we obtain the modulus of a number x by simply throwing awaythe minus sign if it is negative

(and leaving it alone if it is positive).

For example,

|3| = 3, | − 2.6| = 2.6.

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The following properties of the modulus are almost obvious, butmake sure you understand them.

Lemma 1.2 For any x, y ∈ R, we have:

(a) |x| = | − x| > 0 and |xy| = |x||y|(b) if r > 0 then: |x| < r ⇐⇒ −r < x < r

|x| 6 r ⇐⇒ −r 6 x 6 r

(c) the distance between x and y is |x− y| = |y − x|(d) the triangle inequality: |x+ y| 6 |x|+ |y|.

I if x and y have the same sign then

|x+ y| = |x|+ |y|;I if x and y have opposite signs then

|x+ y| 6 max{|x|, |y|} 6 |x|+ |y|.Note. We will regularly use part (b) of Lemma 1.2 to turn a singlemodulus inequality into a pair of ‘normal’ inequalities – these areoften easier to use or rearrange than the modulus inequality.

We will also use the triangle inequality a lot – learn it!4/336

There are some example of common calculations using modulus inthe notes(you saw similar examples last semester).

Look at these and make sure you understand them, we will be doingthese sort of calculations a lot in this course.

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1.2 Sequences

A sequence is an ordered (infinite) list of numbers, such as:

1, 7, −12, 0, 23, . . . ,π, 101, −11, . . . .

A general sequence is often written as

(an) = (a1, a2, a3, . . . ),

where a1 denotes the first entry in the list, a2 denotes the secondentry in the list, and so on. So, in the first example above, we wouldhave

a1 = 1, a2 = 7, a3 = −12, a4 = 0, . . . .

The number in position n in the list is called the nth term of thesequence.

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Note.

I Some books use curly brackets {an} for sequences, but thisis the same notation as for a set of numbers, so we will useround brackets here.

I Often, the nth term, an, is given by a formula involving n,but no formula need exist, and the entries can be completelyrandom.

For instance, if (an) is 3, 1, 4, 1, 5, 9, · · · (the successive inte-gers in the decimal expansion of π) there is no formula.

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I Sets and sequences are different: a set does not have anyordering, whereas a sequence is an infinite set of numbersin a specific order; if you change the order you change thesequence.

For instance, the following two sequences are different:

1, 7, −12, 0, 23, . . . ,7, 1, −12, 0, 23, . . . ,

even though, collectively, they contain the same set of num-bers.

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Example 1.7 Examples of sequences:

(a) 1, 12 ,13 ,

14 , · · · which can be written as ( 1n);

(b) 1, 2, 3, 4, · · · or (n);

(c) 1,−1, 1,−1, · · · or ((−1)n+1);

(d) 0, 2, 0, 2, · · · or (1 + (−1)n);

(e) 12 , −

14 ,

18 , −

116 , · · · or (−1)n+1

2n ;

(f) 1, 21/2, 31/3, · · · or (n1n ).

(g) c, c, c, c, · · · , for some fixed number c;this is a perfectly legitimate sequence, although a bit dull.Such a sequence is called a constant sequence.

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We can visualise sequences by drawing graphs of them, as in Fig. 1.This can sometimes help with understanding what is going on, butnone of our proofs will depend on figures or graphs.

n1 2 3 4 5 6 7 8 9 10

an

•a1•a2

•a3

•a4

•

•

•

•

•

•

Figure 1: The graph of a random sequence.We have labelled the axes and the first 4 points, but we won’tusually do so.

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Definition 1.8 For a sequence (an) :

I (an) is bounded above if there exists Ku such that

an 6 Ku, for all n > 1;

we call Ku an upper bound for (an).

I (an) is bounded below if there exists Kl such that

Kl 6 an, for all n > 1;

We call Kl a lower bound for (an).

I (an) is bounded if there exists a number K > 0 such that

|an| 6 K (i.e., −K 6 an 6 k), for all n > 1.

Clearly, (an) is bounded ⇐⇒ (an) is bounded above andbelow(with, say, upper bound Ku = K, lower bound Kl = −K).

I If a sequence (an) is not bounded it is said to be unbounded.

Note. In Definition 1.8, Ku, Kl and K must not depend on n.

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••

•

•

•

•

•

•

•

•

Ku

Kl

Figure 2: Bounds on the sequence in Fig. 1.

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Example 1.7 (continued)

(a) ( 1n) is bounded (take K = 1, or anything bigger).

(b) (n) is unbounded.

(c)-(f) are all bounded (take K = 2, or anything bigger).

(g) is bounded (take K = |c| — we use modulus signs here in casec is negative, and K has to be positive).

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1.3 The limit of a sequence

We have looked at the idea of a bounded sequence.

Boundedness tells us a little bit about the behaviour of a sequencefor large n.

However, if possible, we would like to get more precise informationabout the behaviour of a sequence for large n.

This leads us to the idea of a ‘limit’ of a sequence.

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The intuitive idea is that a sequence (an) converges to a limit L ifthe entries in the sequence an get closer and closer to L as n getslarger and larger.

For instance, when n is ‘large’ the terms in the sequence

−1, 12, −1

3,1

4, · · · , (−1)n 1

n, · · ·

are ‘close’ to 0, while the terms in the sequence

1

2,2

3,3

4, · · · , n

n+ 1, · · ·

are ‘close’ to 1.

We would like to say that the above sequences ‘converge’ to the‘limits’ 0 and 1 respectively — whatever that means.

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Example 1.9 The graph of the sequence an = 5 − 4/n, n =1, 2, . . . , is shown in the following figure, which shows that thissequence gets ‘close’ to 5 as n gets ‘large’.

5

•

•• • • • • • • •

Figure 3: The sequence an = 5− 4/n, n = 1, 2, . . . .

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Example 1.10 On the other hand, it seems clear that the sequence

−1, 1, −1, · · · , (−1)n, · · ·does not converge to a limit (whatever that means) — it keeps‘jumping around’ when n is ‘large’ (draw a graph).

Clearly, this is all far too vague to make mathematical sense.

What do all those terms in quotes mean?

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For instance:

I what does an ‘close’ to L mean?

To try to make sense of this, recall that the distance betweenan and L is |an−L|, so we could try to make |an−L| ‘small’.

I what does n ‘large’ mean?

we could pick a ‘large’ number, e.g., N = 1, 000, 000 or N =1, 000, 000, 000, and say ‘large’ n means n > N .

But what N should we pick?

In fact, we could pick a tolerance, and look for a number N suchthat whenever n > N then the distance |an − L| is less than thespecified tolerance, i.e., |an − L| < tolerance.

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We seem to be making some progress, and now have some idea ofwhat ‘large’ means.

But what should the tolerance be?

Actually, we don’t know what it should be — one person might want0.01 and another person might want 0.001.

So, let’s not try to pin it down and instead deal with any conceivabletolerance in one go — we will call this unspecified tolerance ε > 0.

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1.4 Formal definition of the limit of a sequence

Definition 1.13 A sequence (an) converges to a limit L if, givenany ε > 0, there exists an integer N > 1 such that

|an − L| < ε, for all n > N.

If (an) converges to L then we write

limn→∞

an = L, or an → L as n→∞.

If (an) does not converge to any limit L, then (an) diverges, or is adivergent sequence.

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Note.

I In general, N depends on ε; if we make ε smaller we need tomake N bigger.

I the condition |an−L| < ε is equivalent to −ε < L− an < ε,which can be written as

L− ε < an < L+ ε.

So, when n is ‘large’, the numbers an in the sequence aresqueezed between L− ε and L+ ε

(thinking of ε as a ‘small’ tolerance).

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Example 1.9 (continued). Fig. 4 shows the quantities in Defini-tion 1.13 for the sequence an = 5 − 4/n, n = 1, 2, . . . , in Exam-ple 1.9.

L+ ε = 5 + ε

L = 5

L− ε = 5− ε

N = 7

•

•• • • • • • • • ε

ε

Figure 4: The sequence an = 5− 4/n, n = 1, 2, . . . , withε = 1 and N = 7.

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Taking ε = 1 and N = 7, we see that when n > N(i.e., when we are to the right of the red line at n = 7)the entries an in the sequence lie between the red lines at heights5− ε and 5 + ε.

We can also see that no matter how small we make ε, if we go farenough to the right along the sequence(i.e., if we make make N big enough)then all the terms in the sequence will lie between 5− ε and 5 + ε.

Note. We see here that N = 5 or N = 6 or N = 57 . . . , alsowork.

We don’t need to find the smallest (first) N , any N that does thejob will do.

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Example 1.14 Prove directly from the definition of convergencethat a constant sequence (c) converges to the constant c.

Solution. Here an = c and L = c. Let ε > 0.

Then

|an − L| = |c− c| = 0 < ε, for all n > 1,

so the definition of convergence is satisfied with N = 1.

Of course, this was a slightly trivial example!

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Example 1.15 Prove directly from the definition of convergencethat

limn→∞

1

n= 0.

Solution. Here an =1

nand L = 0.

Let ε > 0. Then

|an − 0| = 1

n< ε if n >

1

ε·

Let N be an integer with N >1

ε· Then

|an − 0| = 1

n< ε for all n > N.

Thus the definition is satisfied and so limn→∞

1

n= 0.

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Note. To prove that limn→∞

an = L, from the definition, we need to

do the following steps:

Line 1: Let ε > 0

Line 2: |an − L| < something

Line 3: |an − L| < something bigger (and simpler)

...

Last line: Then |an − L| < ε for all n > N .

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Example 1.16 Prove directly from the definition of convergencethat if 0 < y < 1 then

limn→∞

yn = 0.

Solution. Suppose 0 < ε < 1. We want to find N > 1 such that

n > N =⇒ |yn| = yn < ε.

Taking logs and rearranging the inequality turns this into

n log y < log ε ⇐⇒ n >log ε

log y> 0.

Note. Since 0 < y < 1 and 0 < ε < 1, we have log y < 0 and

log ε < 0, which is why the direction of the inequality switches here.

So, if we choose an integer N >log ε

log y, we get

n > N =⇒ |yn| = yn 6 yN <(elog y

)(log ε)/(log y)= elog ε = ε,

which is what we wanted.

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Example 1.19 Let limn→∞

an = L. Prove directly from the definition

of limit that

limn→∞

(4an+1 + 5an) = 9L.

Solution. Let ε > 0. Writing

4an+1 + 5an − 9L = 4(an+1 − L) + 5(an − L),we see that

|4an+1 + 5an − 9L| = |4(an+1 − L) + 5(an − L)|

6 4 |an+1 − L|+ 5 |an − L| ,by the triangle inequality.

We want to make the RHS < ε, so we will make both terms in theaddition on the RHS < ε/2.

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Since limn→∞

an = L , there exists N such that if n > N then

|an − L| <ε

10,

so

5 |an − L| <ε

2.

In addition, if n > N , then clearly n+ 1 > N , so we also have

|an+1 − L| <ε

10,

and so

|4an+1 + 5an − 9L| 6 4 |an+1 − L|+ 5 |an − L|

<4ε

10+

5ε

106 ε, for all n > N .

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Example 1.20 Suppose that (bn) is bounded and (an) is conver-gent, with lim

n→∞an = 0. Prove from the definitions that

limn→∞

anbn = 0.

Give an example with (bn) bounded and (an) convergent such that(an bn) is not convergent.

Solution. Let ε > 0. We want to show that |anbn| < ε, so weneed to use the given information about |an| and |bn|.

I (bn) is bdd, so there exists K > 0 st. |bn| 6 K for all n.

I Then |anbn| 6 K |an| for all n.

I Since limn→∞

an = 0, there exists N, such that

|an| <ε

Kfor all n > N.

I Then |an bn| < Kε

K= ε for all n > N .

I Hence limn→∞

anbn = 0, which proves convergence.

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As an example of nonconvergence, let an = 1 and bn = (−1)n .Then an bn = (−1)n so that (an bn) is not convergent.

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Example 1.21 If limn→∞

an = 1, prove that there exists an integer

N such that an >3

4for all n > N (see Fig. 5).

Solution. Taking ε =1

4in the definition of limit, there exists an

integer N such that

|an − 1| < 1

4for all n > N.

Rearranging this gives

3

4= 1− 1

4< an < 1 +

1

4=

5

4, for all n > N.

and the left inequality is what we wanted.

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Note. The result and the argument in Example 1.21 (for a generallimit L > 0) is often quite useful, so it is worth understanding this.

L

3

4L

N

•

•• • • • • • • • ε =

1

4L

Figure 5: Illustration of Example 1.21, for a general limit L.

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We can also define the idea of limn→∞

an =∞.

Intuitively, this means that an can be as large as we want for alllarge values of n.

Definition 1.22

I We write limn→∞

an = ∞ or an → ∞ as n → ∞ if, for any

M > 0, there exists N such that

an > M for all n > N.

I We write limn→∞

an = −∞ or an → −∞ as n→∞ if, for any

M < 0, there exists N such that

an < M for all n > N.

Note. If limn→∞

an =∞, we do not say that an converges;

an is divergent, and this is telling us something about how it di-verges.

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Example 1.23 Show that:

(a) limn→∞

−n2 = −∞;

(b) limn→∞

(n3 − n2) =∞;

(c) if y > 1 then limn→∞

yn =∞.

Solution. Let M > 0.

(a) We want −n2 < −M , that is, we need n >√M .

So, choosing an integer N >√M + 1, we clearly have

−n2 < −M , for all n > N

[For convenience here, instead of using M < 0 we are taking M > 0and then showing that an < −M < 0, which does the same job.]

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(b) We have

n3 − n2 = n2(n− 1) > n2, if n > 2,

so choosing N > 2 and N >√M , we see that

n3 − n2 > n2 > M, for all n > N .

(c) Choosing an integer N >logM

log y+ 1, we see that if n > N

then

yn > yN >(elog y

)(logM)/(log y)= elogM =M.

[this calculation is almost identical to Example 1.16.]

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Remark. It is clear that if limn→∞

an = ∞ or limn→∞

an = −∞ then

(an) is unbounded.

The converse does not hold:

I the sequence

((−1)nn2) = −12, 22,−32, 42,−52, . . . ,is unbounded, but it jumps between being large and positiveand large and negative, so does not →∞ or → −∞.

I the sequence

(an) = 12, 0, 32, 0, 52, . . . ,

is unbounded, but it does not → ∞; when n is odd an is‘large’, but when n is even an = 0 is ‘small’, so an is not largefor all large n.

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The following result is also almost obvious.

Theorem 1.24 If an > 0, n = 1, 2, . . . , then

limn→∞

an = 0 ⇐⇒ limn→∞

1

an=∞.

Proof. (⇒) Suppose that limn→∞

an = 0, and M > 0. We can

choose N such that

n > N =⇒ 0 < an <1

M=⇒ 1

an> M,

so by definition, limn→∞

1

an=∞.

(⇐) This proof is very similar to the (⇒) proof, so will be omitted.

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Note. Theorem 1.24 need not be true if we don’t suppose thatan > 0, n = 1, 2, . . . .

For example,

(−1)nn−1 → 0 but1

(−1)nn−1= (−1)nn 6→ ∞

(all the terms in the second sequence get big as n → ∞, but theykeep switching sign, so they don’t tend to either ∞ or −∞).

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1.5 Some general theorems about limits

Theorem 1.25 Any convergent sequence is bounded.

Proof. Let (an) be a sequence with limn→∞

an = L. The idea behind

the proof is simple:

I for large N , each of the numbers aN , aN+1, · · · is close toL, hence this set of numbers is bounded;

I the finite set of numbers a1, a2, · · · , aN−1 is also bounded;

I combining these two results shows that the whole sequence isbounded.

L

K

••

•

•

•••• • • •

Figure 6: Illustrating proof of Theorem 1.25.

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We need to make these steps more precise. Let ε = 1(any other ε would do but this is simple).Then, by definition, there is an N such that

|an − L| < 1, for all n > N . (1)

Hence for n > N

|an| = |an − L+ L|6 |an − L|+ |L| (by the triangle inequality)

6 1 + |L| (by (1))

Letting

K = max{|a1|, |a2|, · · · , |an−1|, 1 + |L|},we see that

|an| 6 K, for all n,

that is, the definition of ‘bounded’ is satisfied.

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Remark. The converse of Theorem 1.25 is not true. For example,((−1)n) is bounded but not convergent.

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The following theorem shows that you can do most standard arith-metic operations (add, multiply, etc.) on convergent sequences andthey remain convergent, and the limits are what they ‘ought to be’.

We have to be careful about dividing by zero though.

Theorem 1.26 Suppose that (an), (bn) are convergent sequenceswith limits A,B. Then:

(a) (an + bn) is convergent, with limit A+B;

(b) (anbn) is convergent, with limit AB;

(c) (can) is convergent, with limit cA for any constant c;

(d) if bn 6= 0 for all n and B 6= 0 then

(anbn

)is convergent, with

limitA

B·

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Proof. The proof of all the parts (a)-(d) is in the notes, and parts(a)-(b) were proved in MV Calc A, so we just look at part (d) here.

(d) Since limn→∞

bn = B 6= 0, there exists a positive integer N such

that

|bn| >1

2|B|, for all n > N

(see Example 1.21 for a similar result), and hence∣∣∣∣ 1bn − 1

B

∣∣∣∣ = ∣∣∣∣B − bnbnB

∣∣∣∣ 6 2

|B|2· |bn −B|, for all n > N .

By the convergence of (bn) to B we can now make the RHS of thisinequality smaller than ε so we get

limn→∞

1

bn=

1

B.

So, finally, combining this with part (b) proves part (d).

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We can use Theorem 1.26 to work out limits without having to goback to the definition every time.

Example 1.27 Using Theorem 1.26, show that

limn→∞

3n2 − 6n+ 2

5n2 − 2n+ 3=

3

5.

Solution. We rearrange the terms in the sequence, and then takethe limits, as follows.

Dividing the top and bottom lines of the fraction by n2 gives:

an =3n2 − 6n+ 2

5n2 − 2n+ 3=

3− 6n + 2

n2

5− 2n + 3

n2

→ 3− 0 + 0

5− 0 + 0=

3

5.

You have probably seen this sort of calculation before.

However, you won’t have seen a proper, rigorous, justification of thestep in the middle where we take the limit.

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To do this step rigorously relies on repeatedly using Theorem 1.26,together with

limn→∞

c = c for any constant c, limn→∞

1

n= 0,

which were proved in Example 1.14 and Example 1.15.

For instance, we deal with the top line of the fraction by:

I limn→∞

6

n= 6 lim

n→∞

1

n= 0 (by Theorem 1.26 (c);

I limn→∞

1

n2= lim

n→∞

1

nlimn→∞

1

n= 0 (by Theorem 1.26 (b);

I we then add the limits together using Theorem 1.26 (a).

We deal with the bottom line similarly, and then do the divisionusing Theorem 1.26 (d).

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The following two theorems deal with what happens to inequalitieswhen we take limits.

The first proof (and lots of other proofs below) uses proof by con-tradiction.

You have probably seen this before, but basically this works as fol-lows.

To prove something by contradiction:

I you assume the opposite of what you want to prove;

I you show that this assumption leads to a contradiction;

I this contradiction shows that your assumption must be false;

I so the opposite of your assumption (i.e., the thing you wantedto prove) must be true!

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Theorem 1.28 Suppose that (an), (bn), are convergent, and M isa positive integer. Then

an 6 bn for all n >M =⇒ limn→∞

an 6 limn→∞

bn.

Proof. To prove this by contradiction, let cn = an−bn 6 0, n > 1,and let’s suppose that

K := limn→∞

cn = limn→∞

an − limn→∞

bn > 0.

Now, by Example 1.21, there exists N such that

n > N =⇒ cn >1

2K > 0,

which contradicts the assumption that an 6 bn for all n >M . Thiscontradiction proves the theorem.

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Note. In Theorem 1.28,

an < bn for all n >M 6=⇒ limn→∞

an < limn→∞

bn.

For example,

an := 0 < bn :=1

nfor all n > 1,

but

limn→∞

an = 0 = limn→∞

bn.

So:

I non-strict inequalities are preserved when taking limits,

I strict inequalities need not be.

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Theorem 1.29 Let (an), (bn), (cn), be sequences such that,

an 6 bn 6 cn, for all n >M , for some M > 1,

and (an), (cn), are convergent with

limn→∞

an = limn→∞

cn = L.

Then (bn) is convergent with

limn→∞

bn = L.

Remark. Theorem 1.29 is sometimes called the Sandwich Theorem.

Intuitively, it says that if the sequence (bn) is trapped (or ‘sand-wiched’) between the sequences (an) and (cn) (at least for large n)and these sequences converge to a single limit, then (bn) must goalong with them and converge to the same limit.

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L

•

•

•

••

•

••

•

•••

••••••••••••••••••

Figure 8: An illustration of Theorem 1.29, with (an) in blue,(bn) in green, (cn) in red.

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Proof. If we knew that (bn) is convergent then it would followimmediately from Theorem 1.28 that lim

n→∞bn = L, but we don’t yet

know that.

Let ε > 0. Since limn→∞

an = L and limn→∞

cn = L, there exist Na, Nc

such that

|an − L| < ε, for n > Na, and |cn − L| < ε, for n > Nc.

Let Nb = max{Na, Nc, M}.Then rearranging these inequalities gives

n > Nb =⇒ L− ε < an 6 bn 6 cn < L+ ε =⇒ |bn − L| < ε,

which proves that limn→∞

bn = L.

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Examples of sequences

We will give some more example of limits here, but the followingtheorem is often very useful for showing that specific sequences tendto 0 or ∞.

Theorem 1.30 [Ratio Test for sequences] Suppose that an > 0,n > 1, and

limn→∞

an+1

an= L

(here, L > 0, and we allow L =∞).

Then

L < 1 =⇒ limn→∞

an = 0, L > 1 =⇒ limn→∞

an =∞.

If L = 1 then no conclusion can be drawn.

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Proof. Suppose that L < 1.

Choose y with L < y < 1.

Since limn→∞

an+1

an= L , there exists an integer N such that

an+1

an6 y, for all n > N .

Now, setting n = N, n = N + 1, n = N + 2, . . . , in this gives

aN+1 6 yaN ,

aN+2 6 yaN+1 6 y2aN ,

aN+3 6 yaN+2 6 y3aN .

Continuing in this way, we see that

aN+s 6 ysaN = yN+s

(aNyN

), for all s > 1.

Now, yn → 0, by Example 1.16, so an → 0 by the sandwich thm.

(setting M in the theorem to be the above N).

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Alternatively, we observe that if we put

C := max

{a1y1,a2y2, . . . ,

aNyN

}> 0,

then

0 < an 6 Cyn, for all n > 1.

That is, the above inequality holds for all n > 1 if we change theconstant. We will use this again below.

Now suppose that L > 1.

Choose y with 1 < y < L.

As in the proof of the case L < 1, we can show that for some C > 0,

an > Cyn, for all n > 1.

Now, yn → ∞, by Example 1.23, so an → ∞ by a version of thesandwich theorem for infinite limits, which we haven’t proved, buthopefully is now obvious. . . .

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Example 1.31 The following sequences all tend to ∞ as n→∞:

(a) nα, α > 0, (b) yn, y > 1, (c) n!, (d) nn.

Each one tends to∞ faster than the previous one, in the sense that

yn

nα→∞, n!

yn→∞, nn

n!→∞.

Solution. Each of these follows from the ratio test. For example,

yn+1

(n+ 1)αnα

yn= y

(1

1 + 1/n

)α→ y > 1

(the limit here follows immediately from Theorem 1.26 if α is apositive integer, but needs slightly more work if not).

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1.6 The supremum and infimum of a set

Definition 1.33 Let A be a non-empty subset of R.

I If there exists a number Mu ∈ A such that a 6 Mu for alla ∈ A then we say that Mu is the maximum of A, and wewrite maxA :=Mu.

I If there exists a number Ml ∈ A such that a > Ml for alla ∈ A then we say that Ml is the minimum of A, and wewrite minA :=Ml.

Note. If they exist, the maximum and minimum of a set A haveto belong to A.

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A non-empty, finite set A ⊂ R always has a maximum and a mini-mum.

For example, if A = {−2, 0, 3}, then maxA = 3 and minA = −2.

For infinite sets, it is not always the case that they possess maximumand minimum elements.

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Example 1.34

(a) The set

A = {x ∈ R : x > 1}clearly does not have a maximum — it contains arbitrarilylarge elements, say: 1,000, 1,000,000, 1,000,000,000. . . .

It is also clear that A has a minimum, minA = 1;

(b) Similarly, the set

B = {x ∈ R : x > 1}does not have a maximum.

However, it does not have a minimum either, although this isslightly more subtle.

You might be tempted to say that 1 is the minimum elementin B — except that 1 is not in B!

However, we will see below that in some sense 1 delimits thelowest point of B.

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To try to clarify this we make some more definitions.


I A is bounded above if there exists Ku ∈ R such that a 6 Ku

for all a ∈ A, and we say that Ku is an upper bound for A;

I A is bounded below if there exists Kl ∈ R such that Kl 6 afor all a ∈ A, and we say that Kl is a lower bound for A;

I if A is bounded above and below, then A is bounded. If A isnot bounded above and below then it is unbounded.

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Example 1.36

I A = {x ∈ R : x < 1} is bounded above by 1 (or any numberKu > 1), but it does not have a max.It is not bounded below.

I A = {x ∈ R : x > 0} is bounded below by 0; its min is 0.It is not bounded above.

I A = {1− 1/n : n ∈ N} is bounded above by 1, below by 0.

I A = {(−1)nn : n ∈ N} = {−1, 2, −3, 4, −5, · · · } is notbounded above or below.

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Remarks.

(a) Upper and lower bounds for a set A might not exist (Exam-ple 1.34 (a)), and even if they do they do not have to be inA (Example 1.36), although they might be.

In fact, from Definition 1.33 and Definition 1.35 we see

I A has a max =⇒ A is bounded above, but not con-versely;A is bounded above and has an upper bound Ku ∈ A=⇒ A has a max, and maxA = Ku.

I Similar remarks apply to lower bounds and minima.

(b) If Ku is an upper bound for A then any number K > Ku isalso an upper bound

(K automatically satisfies the definition of upper bound).

Similarly anything below a lower bound is also a lower bound.

In particular, upper and lower bounds are not unique.

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So, even if a set A is bounded it might not have a max or a min.However, the max and min of a set A are very useful concepts, so ifthey don’t exist we would like to find some replacements for them.

To find these we will drop the requirement that they should belongto the set A(which is what causes the problem with the max and min of A).

Upper and lower bounds of a set A do not have to be in A, but ingeneral they are not good replacements for the max and min of Asince they might be a ‘long way away’ from A(e.g., if Ku is an upper bound for A then so is Ku + 1, 000, 000).

What we want to do is to choose upper and lower bounds for A thatare ‘as close as possible’ to A(we don’t mind if they are in A or not).

What does that mean!?

The next definition tries to make this precise.

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(a) Let A be bounded above.

The supremum of A is defined to be the least real numberKu that is an upper bound of A.

It is denoted by supA.

More precisely, supA is defined by the properties:

I supA is an upper bound for A;

I if Ku is any upper bound for A then supA 6 Ku

(this is what ‘least’ means in the above definition).

(b) Let A be bounded below.The infimum of A is defined to be the largest real number Kl

that is a lower bound of A.

It is denoted by inf A.More precisely, inf A is defined by the properties:

I inf A is a lower bound for A;

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I if Kl is any lower bound for A then Kl 6 inf A(this is what ‘largest’ means in the above definition).

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We first need to know if the sup and inf exist.

The following ‘completeness property’ of the real numbers tells usthat if a non-empty set is bounded above or below then it must havea sup or an inf.

Theorem 1.38 [The Completeness Axiom]

I Every non-empty subset of R which is bounded above has asupremum in R.

I Every non-empty subset of R which is bounded below has aninfimum in R.

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Remark. Although it might seem like a minor technicality, thecompleteness axiom is a deep and crucial property of the set of realnumbers R.

Its proof relies on details of the logical construction of the set R.

Since we have not described this construction here (and are notgoing to!) we will not prove this theorem.

However, we observe that it is not ‘obvious’, and it may fail for othersets of numbers, for example, the set of rational numbers Q. To seethis, let

A = {x ∈ Q : x2 < 2}.By a similar proof to that in Example 1.42 below we can show thatsupA =

√2, but it is well-known that

√2 6∈ Q.

Colloquially:

I the set Q has ‘holes’ in it and is not ‘complete’;

I the set R does not have ‘holes’ in it and is ‘complete’.

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We have seen that general upper and lower bounds are not unique.The following theorem shows that the sup and inf are unique.

Lemma 1.39 Let A be a non-empty subset of R.

(a) If A is bounded above then its sup is unique.

(b) If A is bounded below then its inf is unique.

(c) If A has a max then maxA = supA;if A has a min then minA = inf A.Hence, if they exist, maxA and minA are unique.

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Proof.

(a) By Theorem 1.38, A has a supremum, and we want to provethat it has only one supremum.

We will again use proof by contradiction.

So, let’s assume the opposite:

suppose that A has more than one supremum.

To find a contradiction, let’s pick two different suprema andcall them s1 and s2, with s1 6= s2 — our assumption meansthat we can do this.

Now, by the definition of supremum:

I s1 is an upper bound, and s2 is a supremum, so s2 6 s1;

I s2 is an upper bound, and s1 is a supremum, so s1 6 s2.

Combining these shows that s1 = s2, which contradicts ourchoice of different s1, s2; this means that our assumptionmust be wrong — so A has at most one supremum.

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(b) This is almost identical to the proof of part (a).

(c) It is easy to see that maxA and minA (if they exist) satisfythe definitions of supA and inf A.

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Remarks 1.40 Generally, to show that a number S is supA wemust show that:

I S is an upper bound for A;

I any number less than S is not an upper bound for A;we often do this by showing that for any ε > 0, the numberS − ε is not an upper bound, that is, there exists a ∈ A suchthat a > S − ε.

We use a similar procedure to show that a number I is the infimumof a set A.

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Example 1.41 Let A = {1− 1/n : n ∈ N}. Show that supA = 1(see Fig. 9).

Solution. We follow the procedure in Remarks 1.40.

I By the definition of A, x 6 1 for all x ∈ A, so 1 is an upperbound for A;

I Now suppose that ε > 0. Choosing an integer N > 1/ε, wesee that 1− 1/N ∈ A, and 1− 1/N > 1− ε, so 1− ε is notan upper bound for A.

Hence, we have supA = 1.

• • • • •••••• |Ku

|1

|1− ε

Figure 9: The set in Example 1.41

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Remark. In this process, you could say that instead of looking‘inside’ the set A (which may be complicated) for a largest element(which may not exist), we look in the set of upper bounds (which issimple) for a smallest upper bound (which will exist).

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Example 1.42 Show that supA = supB = 3 for the sets

(a) A = {x ∈ R : x 6 3}, (b) B = {x ∈ R : x < 3}.

Solution. (a) We follow the above procedure:

I by the definition of A, x 6 3 for all x ∈ A, so 3 is an upperbound for A;

I if K < 3, then K is not an upper bound for A, since 3 ∈ Aand 3 > K.

Hence, by the definition of sup, we have supA = 3.

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(b) Again:

I As above, 3 is an upper bound for B.

I The second step is not quite so easy — but not hard.We must show that, for any ε > 0, the number 3 − ε is notan upper bound for B.

We do this by showing that there exists b ∈ B with b > 3− ε.We can’t use b = 3, since 3 6∈ B.

However, any number b satisfying 3− ε < b < 3 does work.

Obviously, there are lots of such numbers, but a simple ex-ample is b = 3− ε/2(there is nothing special about this particular number — it isjust simple and does the job).

We see that

3− ε < b = 3− ε

2∈ B =⇒ 3− ε is not an upper bound.

Hence supB = 3.

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We saw in the above examples that applying the second step inRemarks 1.40 for showing that a number S is supA was a bit fiddly.

It gets worse for more complicated sets!

The following theorem simplifies this step.

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Theorem 1.43 Let A be a non-empty set of real numbers that isbounded above, and let S be an upper bound for A. Then

S = supA ⇐⇒ there exists a sequence (an) in A such that

limn→∞

an = S.

A similar result holds for inf A.

Proof. (⇒) We want to construct a suitable sequence.

For each n > 1, put ε = 1/n in the second item in Remarks 1.40and let an be the corresponding element of A.

This gives us a sequence in A, and by construction,

S − 1

n< an 6 S =⇒ lim

n→∞an = S.

(⇐) Given any ε > 0, by the definition of convergence we canchoose N > 1 such that |aN − S| < ε, and hence

|aN − S| < ε =⇒ −ε < aN − S =⇒ aN > S − ε.Since aN ∈ A, the second step in Remarks 1.40 holds.

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Example 1.45 Let A be a non-empty set of real numbers that isbounded above, with a > 0 for all a ∈ A.

Let B be the set of numbers of the form b =1

a, for a ∈ A.

Show that B is bounded below and inf B =1

supA.

Solution. For any a ∈ A, we have 0 < a 6 supA, so

b =1

a>

1

supA, for all b ∈ B.

Hence, 1/ supA is a lower bound for B.

Next, by Theorem 1.43 we choose (an) in A with limn→∞

an = supA.

Setting bn = 1/an ∈ B, n > 1, gives

limn→∞

bn =1

supA(by Theorem 1.26)

=⇒ inf B =1

supA(by Theorem 1.43.)

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1.7 Monotone sequences

Up to now we have only been able to prove the convergence of asequence to a limit if we know the limit in advance.

This is quite restrictive — we will often use sequences to ‘construct’some number (say the solution to an equation) without knowingwhat the number is

(in fact, if we already knew the number we would not be botheringmessing around with the sequence!).

So, we would like to have some property that guarantees that asequence converges without knowing what the limit is.

We will show that we can do this for the types of sequences in thefollowing definition.

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Definition 1.47 For any sequence (an):

I (an) is increasing if an 6 an+1, for all n > 1, that is,

a1 6 a2 6 a3 6 · · · 6 an 6 an+1 6 · · · ;Note. we don’t need strict < inequalities here.

I (an) is decreasing if an > an+1, for all n > 1;

I (an) is monotone, or monotonic (not monotonous!), if it iseither increasing or decreasing.

Note. Since we don’t have strict inequalities in the definition, aconstant sequence is both increasing and decreasing!

• •• • • • •

• • •

Figure 10: An increasing sequence.

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Example 1.48 The following sequences are increasing:(1) , (n2) , (1− 1/n) and (an) with

an = 1 +1

2!+

1

3!+ · · ·+ 1

n!.

The following sequences are decreasing: (−n) and (1/n) .

The sequence ((−1)n) is not monotone.

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Remark. Before stating the main theorem here we note that, givena sequence (an) we have the corresponding set of values {an : n >1}(recall that these are different — the order matters in a sequence,but not in a set).

Comparing Definition 1.8 and Definition 1.35, we see that:

I the sequence (an) is bounded above ⇐⇒ the set {an : n >1} is bounded above;

I the sequence (an) is bounded below ⇐⇒ the set {an : n >1} is bounded below.

We also note that an increasing sequence is automatically boundedbelow, with lower bound a1, and a decreasing sequence is automat-ically bounded above, with upper bound a1.

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Theorem 1.49 [Monotone Convergence Theorem (MCT)]A bounded monotone sequence is convergent.

In more detail (see Fig. 11)

(a) an increasing sequence is convergent if and only if it is boundedabove:

if so, it converges to the supremum of its set of values.

(b) a decreasing sequence is convergent if and only if it is boundedbelow:

if so, it converges to the infimum of its set of values.

supS

•

• •• • • • • • •

Figure 11: Illustration of Theorem 1.49 (a).

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Proof. We only prove part (a); the proof of (b) is very similar.

Let (an) be an increasing sequence which is bounded above, so thatthe set A = {an : n > 1} is bounded above.

Let S = supA. We will show that an → S.

Let ε > 0.

Then, by the properties of sup:

I an 6 S for all n;

I S − ε is not an upper bound for A, so for some N ∈ N wehave aN > S − ε;

I since (an) is increasing, we must have

S − ε < aN 6 an 6 S < S + ε, for all n > N ;

I rearranging this gives |an − S| < ε , for all n > N , so bydefinition lim

n→∞an = S.

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Remark. If we know that a sequence is monotone, then:

I by Theorem 1.49, to show that it is convergent we only needto show that it is bounded;

I by Theorem 1.25, if it isn’t bounded then it can’t be conver-gent.

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Example 1.50 (a) The sequence an =(−1)n

nis convergent but

not monotone.

(b) The sequence an = n is monotone but not convergent.

Example 1.51 Let an = n+1

n, n > 1. By considering an+1−an,

show that (an) is increasing. Does (an) converge?

Solution. For n > 1,

an+1 − an = n+ 1 +1

n+ 1− n− 1

n= 1− 1

n(n+ 1)> 0,

so (an) is increasing. Also,

an = n+1

n> n,

so (an) is not bounded above and so cannot be convergent (byTheorem 1.25).

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Example 1.52 Let (an) be an increasing sequence with a1 > 0.Define a new sequence (bn) by

bn =3anan + 4

, n > 1.

By considering bn+1 − bn, show that (bn) is increasing.

Also, by considering 3 − bn, find a bound on (bn) and hence showthat (bn) is convergent

(even if the original sequence (an) is not convergent).

Solution. Since a1 > 0 and (an) increasing we have an > 0 forall n. Now,

bn+1 − bn =3an+1

an+1 + 4− 3anan + 4

=3an+1(an + 4)− 3an(an+1 + 4)

(an+1 + 4)(an + 4)

=12(an+1 − an)

(an+1 + 4)(an + 4)> 0,

so (bn) is increasing.

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Next,

3− bn = 3− 3anan + 4

=3an + 12− 3an

an + 4=

12

an + 4> 0

so that bn 6 3 for all n.

Thus (bn) is increasing and bounded above, and hence is convergent,by Theorem 1.49.

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Example 1.53 Let an = n2/4n, n > 1.

By considering the ratio an+1/an, show that (an) is decreasing.

By finding a suitable bound on (an), prove that (an) converges.

Solution. For n > 1,

an+1

an=

(n+ 1)2 4n

n2 4n+1=

(n+ 1)2

4n26

(n+ n)2

4n2= 1,

so an > an+1 and hence (an) is decreasing.

Also, since an is positive, the sequence (an) is bounded below (by0) and hence converges.

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Example 1.54

(a) Suppose that 0 < y < 1, and let an = yn, n > 1.

Show that (an) is decreasing, and that limn→∞

yn = limn→∞

an =

0.

(b) Deduce that if −1 < y < 0 then we also have limn→∞

yn = 0.

Solution. (a) For n > 1,

an+1 = yn+1 = yyn = yan < an

(since 0 < y < 1), so that (an) is decreasing.

Also, (an) is bounded below by 0, so by Theorem 1.49, (an) con-verges to some limit L.

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In general, Theorem 1.49 does not tell us what L is, but in this casewe can find out what it is by the following calculation:

L = limn→∞

an = limn→∞

an+1 = limn→∞

yn+1 = limn→∞

yyn

= y limn→∞

yn = yL,(2)

and combining the left and right hand sides gives (1 − y)L = 0,which implies that L = 0 (since 1− y 6= 0).

Note. We saw this result before in Example 1.54, but the methodhere is useful for calculating more complicated limits.

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(b) If −1 < y < 0 then the sequence (yn) is not monotone

(it alternates in sign),

but we have

−|y|n 6 yn 6 |y|n, n > 1,

and part (a) shows that |y|n → 0 as n→∞, so it follows from thesandwich theorem (Theorem 1.29) that yn → 0.

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Remark.

I The calculation in (2) is only valid once we know that thelimit exists — it might be tempting to jump straight to itat the beginning of the question, but if the limits don’t existthen the whole calculation might make no sense and give thewrong answer.

I In (2) we used

L = limn→∞

an = limn→∞

an+1.

This looks slightly strange at first, but for each n > 1 thenumber an+1 is just one step further along the sequence (an)than the number an;

so, given that the whole sequence (an) is converging to L, wemust have L = lim

n→∞an+1

(see Example 1.19, where we did something similar).

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1.8 Subsequences

I Bounded, monotone sequences definitely converge, even if wedon’t know what the limit is.

I This is very useful, but of course most sequences are neithermonotone nor convergent.

I However, convergence is so useful that it is worth seeing if wecan get convergence for more general sequences.

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One very useful trick is to take a divergent sequence and make itconvergent by throwing some of it away

(the ‘bad’ bits, whatever that means).

I For example, if we take the divergent sequence

1,−1, 1,−1, . . . ,we could throw away all the negative terms to leave the se-quence 1, 1, 1, . . . , which obviously converges to 1.

I We could also throw away the positive terms to leave thesequence −1,−1,−1, . . . .

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I More generally, for a sequence (an) we can form a new se-quence by throwing away some of the numbers of (an) andkeeping the rest

• we will call this a subsequence

(a bit like a subset is a part of a set).

I However, we must keep the terms in the subsequence in the

same order as in the original sequence.

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Example 1.56 The following are sequences and subsequences:

(an) = 1,−1, 1,−1, . . . 1, 1, 1, . . . ;

(an) = 1,1

2,1

3,1

4, · · · 1

2,1

4,1

8, · · · ;

(an) = 1,1

2,1

3,1

4, · · · 1

10,

1

100,

1

1000, · · ·

We need to describe this process more clearly.I Rather than thinking of throwing elements away, it is better

to compile a list of the elements that we will keep.

I For example, in the above example we keep the elements

a1, a3, a5, . . . ;

a2, a4, a8, . . . ;

a10, a100, a1000, . . . .

I In other words we need a ’list’ of the subscripts we keep – butthis will be a sequence of subscripts.

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We need some notation to describe this.

Definition 1.57 Let nk, k > 1, be a strictly increasing sequence ofpositive integers, that is

n1 < n2 < n3 < · · · .Then, given a sequence (an), we can define a new sequence (bk) by

bk = ank, k = 1, 2, . . . ,

that is

b1 = an1 , b2 = an2 , b2 = an3 · · · .Then (bk) is a subsequence of (an).Note.

I It is clear that all the terms in the sequence (bk) are in thesequence (an).

I The assumption that the sequence (nk) is strictly increasingensures that the terms in (bk) are in the same order as in (an),so it is legitimate to call (bk) a subsequence of (an).

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I For example, for a sequence (an), taking

a1 a2︸︷︷︸b1

a3 · · · a57︸︷︷︸b2

· · · a131︸︷︷︸b3

· · · ,

we get the subsequence

(bn) = (a2, a57, a131, . . . ).

Here, (nk) is the strictly increasing sequence of integers

(nk) = (2, 57, 131, . . . ).

I As another example, the sequence

(a2k) = (a2, a4, a6, . . . )

is a subsequence of (an)

Here, (nk) is the strictly increasing sequence of integers

(nk) = (2, 4, 6, . . . ), i.e., nk = 2k, k > 1.

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More examples:

sequence subsequence (nk)

1,1

2,1

3,1

4, · · · 1

10

1

100,

1

1000, · · · 10, 100, 1000, . . .

1,−1, 1,−1, . . . 1, 1, 1, 1, . . . 1, 3, 5, . . .

a1, a2, a3, . . . a2, a4, a6, . . . 2, 4, 6, . . .

Note. The sequence (nk) in Definition 1.57 is strictly increasing,so

n1 > 1,

n2 > n1 =⇒ n2 > 2,

n3 > n2 =⇒ n3 > 3,

...

=⇒ nk > k, k > 1.

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We can now start to prove the main theorems about subsequences.The first one is almost obvious.

Theorem 1.58 Suppose that (an) is a sequence with limn→∞

an = L,

and (ank) is a subsequence of (an).

Then limk→∞

ank= L.

Proof. Let ε > 0. Since limn→∞

an = L, there exists N such that

|an − L| < ε, for all n > N .

Also, since nk > k for all k > 1, we have

k > N =⇒ nk > N =⇒ |ank− L| < ε,

which shows that limk→∞

ank= L.

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Although Theorem 1.58 is almost obvious, it can provide a usefulway of showing that a subsequence does not converge.

Example 1.59 Show that the sequence

((−1)n) = (−1, 1,−1, 1, . . . )does not converge.

Solution. This is obvious, but would be a bit fiddly to prove fromthe definition.

However, the sequence has two convergent subsequences

1, 1, 1, · · · → 1, and − 1,−1,−1, · · · → −1,converging to different limits, so the original sequence cannot con-verge

(if it converged to some limit L, then every subsequence would haveto converge to the same limit L).

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Theorem 1.60 Every sequence has a monotone subsequence.

Proof. Let (an) be a sequence. We define an integer k to be apeak point of (an) if ak > an for all n > k, (see Fig. 12).

1 2 3 4 5 6 7 8 9 10

•

•

•

•

••

•• • •

Figure 12: The numbers 2, 4, 7, 9, are peak points of thissequence(at least, as far as we can see on this figure).

The peak point values a2, a4, a7, a9 (in red) are decreasing.103/336

We now split the proof into two cases.

Case 1. The sequence (an) has infinitely many peak points.

In this case, let us list the peak points as

n1 < n2 < n3 < · · · .Then, by the definition of peak points,

an1 > an2 > an3 > · · · ,which is a decreasing subsequence of (an).

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Case 2. The sequence (an) has only finitely many peak points.

In this case, let m be the last peak point, and let n1 = m+ 1.

Then:

I n1 is not a peak point so there exists an integer n2 > n1 suchthat an2 > an1 ;

I similarly, n2 is not a peak point so there exists n3 > n2, suchthat an3 > an2 ;

I continuing this gives an increasing subsequence of (an).

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Theorem 1.61 [Bolzano-Weierstrass theorem (BWT)] Every boundedsequence has a convergent subsequence.

Proof. Let (an) be a bounded sequence.

Any subsequence of (an) is also bounded so, by Theorem 1.60, (an)has a bounded, monotone subsequence, and this must be convergent(by Theorem 1.49).

Example 1.62 The sequence ((−1)n) = (−1, 1,−1, 1, . . . ) has thefollowing convergent subsequences

1, 1, 1, · · · , and − 1,−1,−1, · · · .These are simple and obvious, but there are lots of other convergentsubsequences, for example

1,−1, 1,−1, 1, 1, 1, . . . .

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Example 1.63 Consider the sequence (an) where

an =3n cos(n2 − 5)

n+ 2, n > 1.

Does (an) have a convergent subsequence?

Solution. Since | cos(n2 − 5)| 6 1,

|an| =3n | cos(n2 − 5)|

n+ 26

3n

n= 3, n > 1,

so the sequence (an) is bounded and hence has a convergent sub-sequence, by Theorem 1.61.

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Example 1.64 Give an example of a sequence (an) with

2 6 an 6 3, n = 1, 2, . . . ,

and (an) not convergent.

Write down a convergent subsequence of (an).

Solution. If (an) is the sequence 2, 3, 2, 3, 2, · · · then (an) is notconvergent.

A simple example of a convergent subsequence is 2, 2, 2, 2, 2, · · · .

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Chapter 2

Functions:Limits and Continuity

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2.1 Notation

We first introduce some notation for subsets of the real line R.

Definition 2.1 Suppose that a, b ∈ R, with a < b. Then we definethe following sets.

I (a, b) := {x ∈ R : a < x < b}. This set is an open interval.

I [a, b] := {x ∈ R : a 6 x 6 b}. This set is a closed interval.

I (a, b] := {x ∈ R : a < x 6 b}, [a, b) = {x ∈ R : a 6 x <b}(these sets are intervals, but are neither open nor closed).

I (a,∞) := {x ∈ R : a < x}, (−∞, b) = {x ∈ R : x < b}(these sets are also called open intervals).

I [a,∞) := {x ∈ R : a 6 x}, (−∞, b] = {x ∈ R : x 6 b}(these sets are also called closed intervals).

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Note. Basically:

I in the notation for an interval, we use a square bracket if theinterval contains the corresponding end point and a roundbracket if it does not;

I an interval is closed if it contains each of its end points andopen if it contains neither of its end points;

I you might ask why the final intervals in Definition 2.1 areclosed when they only seem to contain one end point?

It is because they only have one end point — so they contain‘each of their end points’ !.

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Definition 2.2 Let A be a subset of R.

A function f from A to R (often called a ‘real-valued’ function),written f : A→ R is a rule which assigns to each element a ∈ A aunique number f(a) ∈ R.

The set A is called the domain of f .

Example 2.3

(a) The functions defined by,

f(x) = sin

(1

x

), g(x) =

sinx

x, x 6= 0,

are defined for all non-zero x (see Fig. 15), so their domainis R \ {0} (the set R with the point 0 removed).

(b) A more complicated example is h : R→ R defined by

h(x) =

{sinx, if x > 0,

cosx, if x < 0.

The function h is defined for all x ∈ R.

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We can do arithmetic operations on functions, such as add themand multiply them, at points where they are both defined.

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Definition 2.4 Suppose that f : Af → R, g : Ag → R, arefunctions with domains Af , Ag, and let A := Af ∩Ag.

Then we can define the functions f + g, fg, f/g, on A as follows.

For any x ∈ A,

(f + g)(x) := f(x) + g(x),

(fg)(x) := f(x)g(x),

(f/g)(x) := f(x)/g(x) (assuming that g(x) 6= 0).

We also define |f | : A→ R by

|f |(x) := |f(x)|, x ∈ Af .

Remark. You might think ‘what is the difference between the leftand right hand sides of the formulae in Definition 2.4?’.

What we mean is that, say, the symbol ‘f + g’ stands for a newfunction, defined on the domain A, whose value at any point x ∈ Ais defined to be f(x) + g(x), that is, the sum of the the values ofthe original functions f and g at the point x.

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Clearly, since we need the values of both f and g to be defined atx, the new function only makes sense on the domain

A := Af ∩Ag.

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We also define ‘functions of functions’ or the ‘composition’ of twofunctions.

Definition 2.5 Suppose that f : A → B ⊂ R, g : B → R. Thenwe can form the function h : A→ R defined by

h(x) = g(f(x)), x ∈ A.The notation g ◦ f is often used for this function, and it is oftencalled the ‘composition of f and g.

Remark. The definition of the function h in Definition 2.5 makessense:

f is defined at the point x ∈ A, and, by the assumption that f :A→ B, gives a value f(x) in the domain B of g,

so we can apply g to f(x) to give g(f(x)).

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We will now investigate various properties of functions, similar tothose of sequences, although we will see that they can be morecomplicated.

The first obvious property is boundedness.

The following definition is almost identical to Definition 1.8 for se-quences.

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Definition 2.6 For a function f : A→ R:

I f is bounded above if there exists Ku such that

f(x) 6 Ku, for all x ∈ A;

we call Ku an upper bound for f .

I f is bounded below if there exists Kl such that

Kl 6 f(x), for all x ∈ A;

We call Kl a lower bound for f .

I f is bounded if there exists a number K > 0 such that

|f(x)| 6 K, for all x ∈ A.

Clearly, f is bounded ⇐⇒ f is bounded above and below.

I If a function f is not bounded it is said to be unbounded.

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Whether or not a function f is bounded depends on the set A aswell as f .

Example 2.7

I The function f(x) = 1/x is bounded on [1,∞) but is notbounded on (0, 1].

I The function f(x) = x2 is bounded on [0, 3] (by 9) but isnot bounded on the set [0,∞).

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2.2 Limits of functions as x→ ±∞We now want to look at the idea of the limit of a function.

Initially, this will imitate the idea of the limit of a sequence as n→∞, but for functions there will be several alternative types of limits.

Firstly, suppose that f : (a,∞)→ R.

We want to define the limit L of f(x) as x gets large, in a similarway that we defined the limit L of a sequence (an) as n gets large.

The idea is that the values f(x) should be close to L for all largevalues of x.

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Definition 2.8 Suppose that f : (a,∞)→ R.

Then f converges to a limit L, as x→∞, if, given any ε > 0, thereexists X > a such that

|f(x)− L| < ε for all x > X,

and we write limx→∞

f(x) = L, or f(x)→ L as x→∞.

Definition 2.8 says that

L− ε < f(x) < L+ ε for all x > X,

see Fig. 13.

You should compare Definition 2.8 with Definition 1.13, and Fig. 13with Fig. 4.

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x

f(x)

L+ ε

L

L− ε

X

Figure 13: Illustration of Definition 2.8.

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Example 2.9 Show that limx→∞

x

1 + x= 1.

Solution. Let ε > 0.∣∣∣∣ x

1 + x− 1

∣∣∣∣ = ∣∣∣∣ 1

1 + x

∣∣∣∣ < 1

x(if x > 0)

< ε (if x >1

ε).

Fix X > 1ε · Then for all x > X,∣∣∣∣ x

1 + x− 1

∣∣∣∣ < ε.

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When taking the limit of a sequence (an), the only thing that n cando is n→∞.

However, for a function there are other possibilities for what x cando, for example x→ −∞.

Definition 2.10 Suppose that f : (−∞, b)→ R.

Then f converges to a limit L, as x → −∞, if, given any ε > 0,there exists X < b such that

|f(x)− L| < ε for all x 6 X,

and we write limx→−∞

f(x) = L, or f(x)→ L as x→ −∞.

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x

f(x)

L+ ε

L

L− ε

X


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We can also define the idea of limx→∞

f(x) =∞.

Intuitively, this means that f(x) can be as large as we want for alllarge values of x.

Definition 2.11 Suppose that f : (a,∞)→ R.

We will write limx→∞

f(x) =∞, or f(x)→∞ as x→∞, if for each

M > 0 there exists X such that

f(x) > M for all x > X.

Note. If limx→∞

f(x) = ∞, we do not say that f converges; f is

divergent, and this is telling us something about how it diverges.

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Example 2.12 Show that limx→∞

(x3 − x2) =∞.

Let M > 0. We have

x3 − x2 = x2(x− 1) > x2, if x > 2,

so choosing X such that X > 2 and X >√M , we see that

x3 − x2 > x2 > M, for all x > X.

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Remark. We can also define the following limits, but these aresimple variants on the above definitions so we won’t actually writeout these definitions.

I limx→∞

f(x) = −∞

I limx→−∞

f(x) =∞

I limx→−∞

f(x) = −∞

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2.3 Limits of functions as x→ a

Next, we want to understand the behaviour of a function f near tosome point a.

Specifically, we wish to define limx→a

f(x) = L.

The function need not be defined at a, and it may behave in acomplicated manner near to a.

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Example 2.13 Some functions f, g, h : R \ {0} → R, defined by:

f(x) = sin

(1

x

), g(x) = x sin

(1

x

), h(x) =

sinx

x, see Fig. 15.

x

f(x)

f(x) = sin

(1

x

)x

g(x)

g(x) = x sin

(1

x

) x

h(x)

h(x) =sinx

x

Figure 15: Some ‘complicated’ functions(for simplicity, they are only drawn for x > 0).

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The intuitive idea of limx→a

f(x) = L is that we can make |f(x)− L|small by making |x− a| small (and x 6= a).

The formal definition is as follows.

From now on, A will denote an open interval in R(we won’t always state this).

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Definition 2.14 Suppose that f is defined on an open interval A,except possibly at some point a ∈ A. Then we will write

limx→a

f(x) = L

if, for every ε > 0, there exists δ > 0 such that

|f(x)− L| < ε for all x ∈ A with 0 < |x− a| < δ.

Remark. The condition in Definition 2.14 says that

L− ε < f(x) < L+ ε,

for all x 6= a satisfying

a− δ < x < a+ δ,

see Fig. 16.

Also in this condition, since A is an open interval, if δ is small enoughand 0 < |x− a| < δ then x ∈ A.

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x

f(x)

L+ ε

L

L− ε

a− δ a a+ δ


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Note. In Definition 2.14 we only look at values of f(x) with0 < |x − a|, that is, with x 6= a, so this definition does not makeany use of the value f(a) of f at a.

So, the limit L (if it exists):

I does not need f to be defined at a;

I does not depend on f(a), if f is defined at a.

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Example 2.15 Prove that limx→2

10x = 20 .

Solution. Let ε > 0. Then

|10x− 20| = 10|x− 2| < ε if |x− 2| < ε

10·

Now set δ =ε

10·. Then

|10x− 20| < ε if 0 < |x− 2| < δ.

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Example 2.16 Prove that

limx→3

x2 = 9.

Solution. Let ε > 0. Now |x2 − 9| = |x− 3||x+ 3| .We want to make this small by making |x− 3| small.

But what about the |x+ 3| term?

We would like to show that |x+3| 6 C, for some constant C, thencancel off the C by making |x− 3| small enough.

To do this, we note that |x+ 3| = |x− 3 + 6| 6 |x− 3|+ 6, so

|x− 3| < 1 =⇒ |x+ 3| < 7.

Hence

|x2 − 9| < 7|x− 3| if |x− 3| < 1

< ε if |x− 3| < ε

7.

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Thus, putting

δ = min{1,ε

7

},

we see that

|x− 3| < δ =⇒ the above calculations are valid

=⇒ |x2 − 9| < ε if 0 < |x− 3| < δ.

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Example 2.17 Prove that

limx→0

x sin1

x= 0

(f is not defined at x = 0 here). See Fig. 15 for a graph of thisfunction.

Solution. Let ε > 0. For x 6= 0,

|x sin 1

x− 0| = |x sin 1

x| 6 |x| < ε, if 0 < |x− 0| < ε.

Thus, putting δ = ε, we see that

|x sin 1

x− 0| < ε, if 0 < |x− 0| < δ.

[obviously, we don’t really need to write −0 in these calculations,but it is in to match up with the definition].

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We can do the usual arithmetic operations on limits(as long as we avoid dividing by zero).

Theorem 2.18 Let f, g : A \ {a} → R be functions with

limx→a

f(x) = L, limx→a

g(x) =M.

Then:

I limx→a

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

I limx→a

f(x)g(x) = LM.

I If g(x) 6= 0 for all x ∈ A and M 6= 0, then limx→a

f(x)

g(x)=

L

M.

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Proof. We prove the first result; the other proofs are similar.

Let ε > 0. Now,

|(f(x) + g(x))− (L+M)| = |(f(x)− L) + (g(x)−M)|

6 |f(x)− L|+ |g(x)−M |.We want the RHS < ε, so we make each separate term < ε/2.

By definition, there exists numbers δf > 0 and δg > 0 such that

|f(x)− L| < ε

2, for all x with 0 < |x− a| < δf ;

|g(x)−M | < ε

2, for all x with 0 < |x− a| < δg.

Hence, putting δ = min{δf , δg} , we see that if 0 < |x − a| < δthen

|(f(x) + g(x))− (L+M)| 6 |f(x)− L|+ |g(x)−M |

<ε

2+ε

2= ε.

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Example 2.19 [polynomials] Show that for any polynomial

p(x) = p0 + p1x+ p2x2 + · · ·+ pkx

k, x ∈ R

(where p0, p1, . . . , pk are constants), and any a ∈ R, we have

limx→a

p(x) = p(a).

Solution. It can easily be shown that for any constant c,

limx→a

c = c, limx→a

x = a

(see Example 2.15); here we have regarded c as a constant function.

Now, combining this (repeatedly) with Theorem 2.18 proves theresult.

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Theorem 2.20 Suppose that f : A \ {a} → R is a function withlimx→a

f(x) = L, and c, d, γ > 0 are constants.

Then

(a) c 6 f(x) for all |x− a| < γ =⇒ c 6 L;

(b) f(x) 6 d for all |x− a| < γ =⇒ L 6 d.

Proof. The theorem, and proof, are similar to Theorem 1.28 forsequences, so the proof will be omitted here.

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The next result will give us a useful way of showing that a functionf does not have a limit at a point a

Theorem 2.21 Suppose that f : A \ {a} → R has limx→a

f(x) = L,

and (an) is a sequence in A \ {a} with limn→∞

an = a.

Then

limn→∞

f(an) = L.

x

f(x)

•

a1

•

a2

•

a3

•

a

L

Figure 17: Illustration of Theorem 2.21: function valuesf(an) converging to L:

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Remark. By Theorem 2.21, we can show that a function f doesnot have a limit at a if we can find two sequences (an), (bn) inA \ {a} such that

limn→∞

an = limn→∞

bn = a but limn→∞

f(an) 6= limn→∞

f(bn).

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Example 2.22 Suppose that f : R→ R is defined by

f(x) = sin1

x, if x 6= 0.

Prove that f does not have a limit at 0.

Solution. Looking at the graph of f in Fig. 15 we see that f hasvalues ±1 arbitrarily close to x = 0, so we will try to find sequences(an), (bn) which converge to 0 with:

f(an) = 1, f(bn) = −1 for all n.In fact the following works:

I an =1

(2n+ 1/2)π=⇒ f(an) = sin(2n+ 1/2)π = 1;

I bn =1

(2n− 1/2)π=⇒ f(bn) = sin(2n− 1/2)π = −1;

So we conclude that f(x) = sin1

xdoes not have a limit at x = 0.

This seems reasonable, looking at the graph in Fig. 15.

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2.4 One-sided limits The limit limx→a

f(x) is concerned with the

values of f(x) for values of x near a and on either side of a.

We also need to consider limits when either:

x > a (limit

or

x < a (limit from the left).

We call these one-sided limits.

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Definition 2.23 [limit from the right] Suppose that f is defined onan open interval (a, b). Then we will write

limx→a+

f(x) = L

if, for each ε > 0, there exists δ > 0 such that

|f(x)− L| < ε, for all x with a < x < a+ δ.

Definition 2.24 [limit from the left] Suppose that f is defined onan open interval (a, b). Then we will write

limx→b−

f(x) = L

if, for each ε > 0, there exists δ > 0 such that

|f(x)− L| < ε for all x with b− δ < x < b.

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Example 2.25 Define f : [0,∞)→ R by f(x) =√x.

Show that limx→0+

f(x) = 0 .

Note. In this example, f(x) is not defined for x < 0 so neitherlimx→0

f(x) nor limx→0−

f(x) can be defined here.

Solution. Let ε > 0. Now |√x−0| =

√x , so putting δ = ε2 gives

|√x− 0| < ε if 0 < x < δ.

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The following theorem relating one-sided and normal limits is clearby combining Definition 2.14, Definition 2.23 and Definition 2.24, .

Theorem 2.26 The limit of f at a exists and equals L ⇐⇒ boththe left and right-sided limits of f at a exist and equal L.

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Example 2.27 Define f : R→ R by

f(x) =

−1, if x < 0,

0, if x = 0,

1, if x > 0.

It is clear that limx→0−

f(x) = −1, limx→0+

f(x) = 1, that is, the one-

sided limits exist at 0 but are not equal to each other.

So the limit limx→0

f(x) does not exist, see Fig. 18.

x

f(x)

•

Figure 18: The function f in Example 2.27 ‘jumps’ between−1 = lim

x→0−f(x) on the left, and 1 = lim

x→0+f(x) on the right.

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2.5 Continuous functions

For the rest of this chapter we suppose that A is an open intervaland a ∈ A (we won’t keep writing this).

Definition 2.28 [Limit definition of continuity] A function f : A→R is continuous at a if both the following conditions hold:

I limx→a

f(x) exists (and is finite);

I limx→a

f(x) = f(a)

(these are separate conditions and both must hold).

Looking at the definition of limx→a

f(x) in Definition 2.14, we see that

the following is an equivalent definition of continuity.

Definition 2.29 [ε−δ definition of continuity] A function f : A→R is continuous at a if for every ε > 0 there exists δ > 0 such that

|f(x)− f(a)| < ε, for all x ∈ A with |x− a| < δ.

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We also have the terminology:

I if f is not continuous at a, then f is discontinuous at a.

I if f is continuous at every point a ∈ A then it is continuouson A (or just ‘continuous’).

Remark. A function f can be discontinuous at a in various ways:

I it might be highly oscillatory, so that the one-sided limits donot exist (see Example 2.22);

I the one-sided limits may exist but not be equal to each other(so the overall limit does not exist)(see Example 2.27 and Fig. 19 (b)).In this case f is said to have a jump discontinuity at a;

I the overall limit exists but is not equal to the value f(a)(see Fig. 19 (c)).In this case the value can simply be changed to the limit togive us a continuous function!

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Intuitively, f is continuous if the graph of f does not have any ‘gaps’,that is, we can draw it without taking our pen off the paper.

x

f(x)

(a)x

f(x)

(b)x

f(x) •

(c)

Figure 19: Some functions: (a) continuous; (b) jumpdiscontinuity; (c) wrong value.

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We can use our results about limits to show that many standardfunctions are continuous.

Example 2.30 By Example 2.19, any polynomial is continuous onR.

In addition, with a bit more effort, it can be shown that the standardfunctions sin, cos, exp, are continuous on R.

Functions like ln and tan are also continuous, except where they‘blow up’.

We will not prove all this here.

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Example 2.31 Let f : R→ R be defined by

f(x) =

{3x, if x > 0,3, if x < 0.

By considering left and right limits at 0, prove that f is not contin-uous at 0.

Solution. It is easy to see (using Example 2.19) that

limx→0+

f(x) = limx→0+

3x = 0,

limx→0−

f(x) = limx→0−

3 = 3,

so limx→0

f(x) does not exist, so f is not continuous at 0.

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Example 2.32 Let f : R→ R be defined by

f(x) =

0, if x < −1,ax2 + bx, if − 1 6 x 6 1,cx, if x > 1.

where a, b and c are constants.Find conditions on a, b , c which ensure that f is continuous on R.

Solution. It is clear that f is continuous at all points of R, exceptpossibly at −1 and 1.

To ensure continuity at each of these points we first find conditionswhich ensure that the limits exist at these points.

For this, we need the left and right limits to be equal.

To check this:

0 = limx→−1−

f(x) = limx→−1+

f(x) = a− b =⇒ a = b;

a+ b = limx→1−

f(x) = limx→1+

f(x) = c =⇒ c = 2a,

so, if these conditions hold then the limits of f exist at ±1.

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It can now also be seen that these conditions ensure that the functionvalues f(±1) are the right ones, that is,

f(±1) = limx→±1

f(x).

So the required conditions are: a = b and c = 2a.

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Example 2.33 Suppose that f : R→ R is defined by

f(x) =

sin

(1

x

), if x 6= 0,

42, if x = 0,

and g : R→ R is defined by g(x) = xf(x).

Prove that f is discontinuous at 0 but g is continuous at 0.

Solution. It was shown in Example 2.22 that f does not have alimit at 0, so it cannot be continuous there(irrespective of what value we give f(0)).

On the other hand, it was shown in Example 2.17 that

limx→0

g(x) = 0 = g(0),

so g is continuous at 0.

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The next result follows immediately from Theorem 2.21 and Defini-tion 2.28.

Theorem 2.34 Suppose that f : A→ R is continuous at a,and (an) is a sequence in A with lim

n→∞an = a.

Then

limn→∞

f(an) = f(a).

Remark. By Theorem 2.34, we can show that a function f is notcontinuous at a if we can find a sequence (an) such that

limn→∞

an = a and limn→∞

f(an) 6= f(a).

Recall Example 2.22, and the preceding remark, where we showedthat a limit did not exist by finding two sequences of function valuesthat converged to different limits.

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2.6 Combinations of continuous functions

As with limits, we can do the usual arithmetic operations on contin-uous functions and they remain continuous(as long as we avoid dividing by zero).

Theorem 2.35 Let f, g : A→ R and a ∈ A.

Suppose that f, g are continuous at a.

Then:

I f + g is continuous at a;

I fg is continuous at a;

I if g(a) 6= 0 thenf

gis continuous at a.

Proof. The proof of these rules follow from Theorem 2.18 on limitsof functions and the definition of continuity in Definition 2.28.

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Theorem 2.36 [function of a function] Let f : A→ B, g : B → R(where B is an open interval),so we can form the function h : A→ R defined by

h(x) = g(f(x)), x ∈ A.Suppose: f is continuous at a ∈ A, g is continuous at f(a) ∈ B.

Then h is continuous at a.

Proof. Let ε > 0. We need to show there exists δ > 0 such that

|g(f(x))− g(f(a))| < ε, for all x with |x− a| < δ. (4)

By the continuity of g at f(a), there exists δ1 > 0 such that

|g(y)− g(f(a))| < ε, for all y with |y − f(a)| < δ1. (5)

Also, by the continuity of f at a, there exists δ2 > 0 such that

|f(x)− f(a)| < δ1, for all x with |x− a| < δ2. (6)

Thus, putting y = f(x) in (5) and combining (5) and (6) gives (4),which proves the theorem.

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It can be shown that the standard functions sin, cos, exp, are con-tinuous.

It now follows from Theorem 2.35 and Theorem 2.36 that manyother functions are continuous.

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2.7 Continuous functions on closed, bounded intervals

Throughout this section a, b will be finite numbers with a < b, andwe consider continuous functions f : [a, b] → R, defined on theclosed, bounded interval [a, b].

We will prove two important results about such functions.

First we must define what we mean by a continuous function definedon such an interval, since Definition 2.28 and Definition 2.29 onlydefined continuity on open intervals.

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Definition 2.37 A function f : [a, b] → R is continuous at anend-point a, or b, if

limx→a+

f(x) = f(a), or limx→b−

f(x) = f(b);

if f is continuous at every point c ∈ [a, b](in the usual sense that lim

x→cf(x) = f(c) if c ∈ (a, b))

then it is continuous on [a, b](or just ‘continuous’).

In other words, we just extend the definition of continuity at pointsc ∈ (a, b), in terms of normal limits, to continuity at the end pointsa and b by simply using one-sided limits.

We can also define continuity in a similar way on intervals that areneither open nor closed — we simply use one-sided limits at anyend-point the interval contains, and normal limits at points that arenot end-points.

There are lots of possibilities, so we won’t write any of them out.

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Theorem 2.38 [The Boundedness Theorem] Let f be continuouson the closed interval [a, b]. Then:

(a) f is bounded on [a, b];

(b) f attains its bounds on [a, b].

More specifically, if we let

Kl(f) = inf{f(x) : x ∈ [a, b]} , Ku(f) = sup{f(x) : x ∈ [a, b]}(these exist by (a)) then there exist c, d ∈ [a, b] such that

f(c) = Kl(f), f(d) = Ku(f).

x

f(x)

a b

Ku(f)

Kl(f)

d c

Figure 20: Illustration of Theorem 2.38.

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Note. Theorem 2.38 is not true on open intervals.

For example,1

xis continuous on (0, 1), but is not bounded.

Proof. (a) We prove this by contradiction.

Suppose that f is not bounded — say f is not bounded above(the case where f is not bounded below is similar).

Then we have the following results:

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I for each n > 1, there exists xn ∈ [a, b] with f(xn) > n(if xn didn’t exist then we would have f(x) 6 n for all x ∈[a, b], that is, f would be bounded above);

I since the sequence (xn) lies in the interval [a, b] it is bounded,so by Theorem 1.61 it has a subsequence (xnk

) which con-verges to some limit L ∈ [a, b];

I since f is continuous, we have f(xnk)→ f(L)

(by Theorem 2.34), so the sequence (f(xnk)) is bounded

(by Theorem 1.25);

I however, by our construction of the sequence (xn), we havef(xnk

) > nk > k, for all k > 1, so the sequence (f(xnk)) is

not bounded.

The final two bullet points are contradictory,

so our initial supposition that f is not bounded must be false,

that is, f must in fact be bounded, which proves the result.

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(b) The following results show there exists d ∈ [a, b] such thatf(d) = Ku(f) (the other case is similar):

I for each n > 1, Ku(f)− 1n is not an upper bound for the set

{f(x) : x ∈ [a, b]}, so there exists xn ∈ [a, b] with

Ku(f)−1

n< f(xn) 6 Ku(f);

so, by the sandwich thm, f(xn)→ Ku(f) as n→∞;

I as in the proof of part (a), the sequence (xn) has a convergentsubsequence (xnk

) with some limit, say d ∈ [a, b];

I (f(xnk)) is a subsequence of (f(xn)) so f(xnk

)→ Ku(f) ask →∞ (by Theorem 1.58);

I by continuity, f(d) = limk→∞

f(xnk) = Ku(f), which is what

we wanted.

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The next result is a very useful method of showing that an equationf(x) = 0 has a solution, even when we don’t know very much aboutthe function f .

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Theorem 2.39 [The Intermediate Value theorem (IVT)] Let f becontinuous on [a, b], and suppose that a number γ satisfies

f(a) 6 γ 6 f(b) or f(a) > γ > f(b).

Then there exists c ∈ [a, b] such that f(c) = γ.

x

f(x)

γ

a bc

f(a)

f(b)


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Proof. Suppose that f(a) 6 γ 6 f(b)(the proof is similar if f(a) > γ > f(b)).

Let

B = {x ∈ [a, b] : f(x) 6 γ},c = supB

(supB exists since B is non-empty (a ∈ B) and is bounded aboveby b).

We will show that f(c) = γ.

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Case (a) Suppose that a < c < b.

By the definition of c = supB, for each n > 1:

I set bn := c+ 1n , then bn 6∈ B;

I there exists an ∈ B such that c− 1n 6 an 6 c.

Hence,

c− 1

n6 an 6 c 6 bn = c+

1

n=⇒ lim

n→∞an = lim

n→∞bn = c,

and so, by continuity of f ,

f(c) = limn→∞

f(an) 6 γ, (since an ∈ B, f(an) 6 γ for all n),

f(c) = limn→∞

f(bn) > γ, (bn 6∈ B, f(bn) > γ for all n).

Combining these results implies that f(c) = γ.

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Case (b) Suppose that c = b.

By a similar argument to Case (a) we can construct a sequence (an)in [a, b] with, now,

an ∈ B, limn→∞

an = b, f(b) = limn→∞

f(an) 6 γ 6 f(b)

(the final inequality is the hypothesis in the theorem).

Combining these inequalities gives f(b) = γ.

Case (c) Suppose that c = a.

The proof in this case is very similar to the proof of Case (b), so wewon’t write it out.

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Remark. Fig. 21 makes it fairly clear why the IVT holds.

The following figure shows that the IVT need not hold if f is notcontinuous.

Here, f has a jump discontinuity, and the graph of f just ‘jumpsover’ the value γ at this jump.

x

f(x)

a b

γ

Figure 22: It is necessary that f be continuous in Theo-rem 2.39.

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We can use the IVT to show that complicated equations have solu-tions (although we usually won’t be able to find them).

Note. In the following examples we want to show that some equa-tion, say

H(x) = G(x),

has a solution.

This is not quite in the right form for the IVT(the equation in Theorem 2.39 has a constant γ on the RHS).

To fix this we define a new function by

f(x) = H(x)−G(x)and then use Theorem 2.39 to show that the equation

f(x) = 0

has a solution and this is then a solution of the original equation.

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Example 2.40 Show that the following equations have solutionson the given intervals:

(a) ex = 10x, in (0, 1).

(b) 17x7 − 19x5 − 1 = 0, in (−1, 0).

Solution. (a) Put f(x) = ex − 10x. This is continuous on [0, 1]and

f(0) = 1, f(1) = e− 10 < 0.

Hence, by Theorem 2.39, there exists c ∈ [0, 1] such that f(c) = 0,that is, ec = 10c, but it is clear that c 6= 0 and c 6= 1, so c ∈ (0, 1).

(b) Put f(x) = 17x7 − 19x5 − 1. Then f is continuous on [−1, 0]and

f(−1) = −17 + 19− 1 = 1 , f(0) = 0 + 0− 1 = −1.By the IVT, f(c) = 0 for some c ∈ (−1, 0).

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Example 2.41 (a) Let f : [0,∞) → R be continuous on [0,∞),and suppose that there exists K > 0 such that

0 < f(x) 6 K, for all x > 0 .

Prove that there exists c > 0 so that f(c) = c.Give an example of a continuous function f : [0,∞) → R withf(x) > 0 and f(x) 6= x for all x > 0.

Solution. (a) Put g(x) = f(x)− x. Then g is cts on [0,∞).

Since g(0) = f(0) − 0 > 0 , if we can find b st. g(b) < 0, it wouldfollow from the IVT that g(c) = 0 for some c, that is, f(c) = c.

Since

g(K + 1) = f(K + 1)−K − 1 6 K −K − 1 < 0,

b = K + 1 works.

An example is: f(x) = x+ 1.

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(b) Let f : [0, 2] → R be continuous on [0, 2] with f(0) = f(2).Prove that there exists c ∈ [0, 1] with f(c) = f(c+ 1).

Solution. (b) Define g : [0, 1]→ R by

g(x) = f(x+ 1)− f(x).

Then g is continuous on [0, 1] and

g(0) = f(1)− f(0), g(1) = f(2)− f(1) = f(0)− f(1) = −g(0)(since f(0) = f(2)).

Now:

I if g(0) 6= 0 then g(0) and g(1) have opposite signs so by theIVT, g(c) = 0 for some c ∈ (0, 1), that is,

f(c+ 1) = f(c);

I if g(0) = 0 then f(1) = f(0), so c = 0 does the job.

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2.8 Uniform continuity

Let f be a continuous function defined on an arbitrary interval I(I may be open, closed or neither, and bounded or unbounded).Then, f is continuous at each a ∈ I, so:

for every ε > 0 there exists δ > 0 such that

|f(x)− f(a)| < ε for all x ∈ I with |x− a| < δ.

We will say that f is uniformly continuous on I if δ can be chosento be independent of a ∈ I .

More precisely:

Definition 2.42 f is uniformly continuous on I if for every ε > 0there exists δ > 0 such that

|f(x)− f(y)| < ε, for all x, y ∈ I with |x− y| < δ.

Remark. The definitions of continuity and uniform continuity arevery similar.

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Clearly, a function which is uniformly continuous on I is also con-tinuous on I.

The following example shows that in general the converse is false.

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Example 2.43

(a) Show that f(x) = x2 is not uniformly continuous on R.

(b) Show that f(x) =1

xis not uniformly continuous on (0, 1).

(c) Show that f(x) = sin(1x

)is not uniformly continuous on

(0, 1).

Solution. (a) Suppose that f is uniformly continuous on R.

Putting ε = 1 in the definition, there must exist δ > 0 such that

|f(x)− f(y)| < 1, for all points x, y ∈ R with |x− y| < δ.

Now choose an integer n > 1/δ, and set y = n and x = n+ δ/2.

Then

|x− y| = δ/2 < δ but |f(x)− f(y)| = nδ +δ2

4> 1.

This contradiction shows that f is not uniformly continuous on R.

(b) and (c) Similar calculations proves these.182/336

In Example 2.43 (a) and (b), uniform continuity fails due to theunboundedness of the functions.

In (c) the function is bounded, but has infinitely many steeper andsteeper oscillations near to x = 0, and is not defined at x = 0.

However, we have seen in Theorem 2.38 that continuous functionson closed, bounded intervals must be bounded, and they generallyseem to behave better on closed, bounded intervals.

This motivates the following theorem.

Theorem 2.44 Let f be continuous on the closed bounded interval[a, b].

Then f is uniformly continuous on [a, b].

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Proof. We will prove this by contradiction, so we suppose, on thecontrary, that f is not uniformly continuous on [a, b].

Looking at Definition 2.42 and ‘reversing’ it, we see that this meansthat:

there exists at least one ε > 0 such that:

for any δ > 0 there exists x, y ∈ [a, b] such that

|x− y| < δ and |f(x)− f(y)| > ε.

(NUC)

The following results will derive a contradiction from this.

I Choose ε as in (NUC).

I Now, for each integer n > 1, setting δ = 1/n in (NUC) showsthat there exists xn, yn ∈ [a, b] such that

|xn − yn| <1

nand |f(xn)− f(yn)| > ε. (7)

I The sequences (xn) and (yn) are in [a, b], so are bounded.

Hence, by the BWT, (xn) has a convergent subsequence (xnk)

with limit L.

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I Letting k →∞ and using the first inequality in (7) gives,

xnk︸︷︷︸→L

− 1/nk︸︷︷︸→0

< ynk< xnk︸︷︷︸→L

+1/nk︸︷︷︸→0

,

and since nk > k, we have1

nk6

1

k→ 0 as k → ∞, so by

the sandwich theorem,

limk→∞

ynk= L.

I By continuity,

limk→∞

(f(xnk

)−f(ynk))= lim

k→∞f(xnk

)− limk→∞

f(ynk) = L−L = 0.

I This contradicts the fact that

|f(xnk)− f(ynk

)| > ε > 0, for all k

(as shown in (7)).

Hence the initial assumption must be wrong and the theorem mustbe true.

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Note. You might wonder why I didn’t just apply BWT to bothsequences (xn), (yn) to get the convergent subsequences.

You could, but the subsequences you get might be different, so theymight not satisfy (7) any more, and so they might have differentlimits.

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Chapter 3

Functions: Differentiability

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3.1 Introduction

In this section we are going to discuss differentiation.

You already know how to differentiate certain standard functions,and you know various rules for differentiating combinations of func-tions, such as the product rule and the chain rule.

However, you may not know what exactly a derivative actually is, orwhy these are the rules for working out derivatives.

Here, we will start from the very beginning, with a precise definitionof what a derivative is, and then prove various well-known resultsabout differentiation.

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You might think: ‘why do this when we know it already?’, but thepoint is that we need a precise definition in order to derive theserules, and to obtain further results in both calculus and other areasof mathematics in the future.

For example, in 3rd year you may encounter derivatives of complex-valued functions of complex-numbers, which at first you won’t knowhow to differentiate, so you will need to know what a derivativeactually is.

An engineer might be happy to just be told what to do (don’t quoteme on that!), but mathematicians should know why they are doingthings and how they work.

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Throughout this section we will suppose that A is an open intervaland f : A→ R.

Definition 3.1 The function f is differentiable at a point a ∈ A ifthe limit

f ′(a) = limx→a

f(x)− f(a)x− a

(8)

exists, in which case the number f ′(a) is said to be the derivativeof f at a.

Remark. The definition (8) has a geometric interpretation. Thefraction

f(x)− f(a)x− a

is the gradient of the straight line segment joining the points (a, f(a))and (x, f(x)), see Fig. 23, and the idea is that as we let x→ a thegradient of the straight line tends to the gradient of the function fat the point (a, f(a)), and we call this gradient f ′(a).

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x

•

•

a x

−f(x)

−f(a)

f(x)− f(a)

x− a


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Note. An alternative way of writing the limit in (8) (which oftenmakes calculating it easier) is to set x = a+ h , which gives

f ′(a) = limh→0

f(a+ h)− f(a)h

. (9)

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Example 3.2 Define the functions

f0(x) := 1, f1(x) := x, f2(x) := x2, x ∈ R.

Then, for any a ∈ R,

f0(a+ h)− f0(a)h

=1− 1

h= 0→ 0,

f1(a+ h)− f1(a)h

=a+ h− a

h= 1→ 1,

f2(a+ h)− f2(a)h

=(a+ h)2 − a2

h=

2ah+ h2

h= 2a+ h→ 2a,

so each of these functions are differentiable at any a ∈ R, withderivatives

f ′0(a) = 0, f ′1(a) = 1, f ′2(a) = 2a

(which you already knew).

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Theorem 3.3 If f is differentiable at a ∈ A then f is continuousat a.

Proof. We must show that limx→a

f(x) = f(a).

Now, by Theorem 2.18, the limit of a product is equal to the productof the limits, so

limx→a

(f(x)− f(a)) = limx→a

(f(x)− f(a))(x− a)x− a

=

(limx→a

f(x)− f(a)x− a

)limx→a

(x− a)

= f ′(a)× 0 = 0.

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Theorem 3.3 showed that differentiability at a point a implies con-tinuity at a.

The next example shows that the converse need not be true, thatis, a function may be continuous at a but not differentiable there.

Example 3.4 Show that the function f(x) = |x| , x ∈ R, is notdifferentiable at 0 (see Fig. 24).

Solution. We calculate the following one-sided limits:

limh→0−

f(0 + h)− f(0)h

= limh→0−

|h|h

= limh→0−

−hh

= −1;

limh→0+

f(0 + h)− f(0)h

= limh→0+

|h|h

= limh→0+

h

h= 1.

These one-sided limits are different so, by Theorem 2.26, the overall

limit limh→0

|h|h

does not exist.

Hence, by definition, f is not differentiable at x = 0.

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x

|x|

Figure 24: Graph of the function f(x) = |x|.

Remark. The function f(x) = |x| is continuous at x = 0, but it isnot differentiable there.

In view of our above interpretation of f ′(a) as the gradient of fat a, this is reasonable — clearly, the function |x| does not have agradient at (0, 0).

In fact, its graph has a ‘corner’ there: on the left (x < 0) thegradient is −1, while on the right (x > 0) the gradient is 1(this is reflected in the one-sided limits calculated in Example 3.4).So, it should not be differentiable there.

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Example 3.5 Define f : R→ R by

f(x) :=

{x3, if x < 0,

x2, if x > 0.

Prove that f differentiable at 0.

Solution. Again we calculate the one-sided limits

limh→ 0+

[f(h)− f(0)

h

]= lim

h→ 0+

h2

h= 0,

limh→ 0−

[f(h)− f(0)

h

]= lim

h→ 0−

h3

h= 0,

so f is differentiable at 0, with f ′(0) = 0.

Draw a graph of f to see why this works.

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Example 3.7 Suppose that f is differentiable at a.

Calculate the following limits in terms of the derivative f ′(a).

(a) limh→0

f(a+ 2h)− f(a)h

;

(b) limh→0

f(a+ h)− f(a− h)h

.

Solution. (a)

limh→0

f(a+ 2h)− f(a)h

= limh→0

2f(a+ 2h)− f(a)

2h,

and writing H = 2h now gives

limh→0

2f(a+ 2h)− f(a)

2h= 2 lim

H→0

f(a+H)− f(a)H

= 2f ′(a).

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(b) Using

f(a+ h)− f(a− h) = (f(a+ h)− f(a)) + (f(a)− f(a− h))gives

limh→0


= limh→0

f(a+ h)− f(a)h

+ limh→0

f(a)− f(a− h)h

,

and writing H = −h in the second limit on the right gives

limh→0

f(a)− f(a− h)h

= limH→0

f(a+H)− f(a)H

= f ′(a),

so that

limh→0


= 2f ′(a).

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3.2 Combinations of differentiable functions

The next two theorems prove some well-known, standard rules fordifferentiating various combinations of functions.

Theorem 3.8 Suppose that f, g : A → R are differentiable at apoint a ∈ A.

Then f + g and fg are differentiable at a, with derivatives

(f + g)′(a) = f ′(a) + g′(a) (the sum rule)

(fg)′(a) = f ′(a)g(a) + f(a)g′(a) (the product rule)

If g(x) 6= 0 on some open interval A containing a then f/g is definedon A and is differentiable at a, with derivative(

f

g

)′(a) =

f ′(a)g(a)− f(a)g′(a)g2(a)

(the quotient rule)

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Proof. (of the quotient rule) We will prove that(1

g

)′(a) = − g

′(a)

g(a)2,

and the quotient rule then follows from this and the product rule.

limx→a

1/g(x)− 1/g(a)

x− a= lim

x→a

g(a)− g(x)g(x)g(a)(x− a)

= − limx→a

1

g(x)g(a)limx→a

g(x)− g(a)x− a

=1

g(a)2(−g′(a)).

In the above calculation we used the following results:

I we suppose that x is close enough to a that g(x) 6= 0;

I the limit of a product equals the product of the limits;

I g is continuous at a, by Theorem 3.3, so limx→a

g(x) = g(a) =⇒

limx→a

1

g(x)=

1

g(a)

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The following result is known as the chain rule.

Theorem 3.9 [The chain rule] Suppose that the composite function

h(x) := g(f(x))

is defined on an interval containing a, and that f is differentiable ata, g is differentiable at f(a).

Then h is differentiable at a, with

h′(a) = g′(f(a))f ′(a).

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Proof. Let us first suppose that f ′(a) > 0(the case when f ′(a) < 0 is almost identical).

We need to work out the limit

limx→a

h(x)− h(a)x− a

= limx→a

g(f(x))− g(f(a))x− a

.

To do this it will be convenient to know that

x 6= a =⇒ f(x)− f(a) 6= 0.

To see this we note that if we put ε = 12f′(a) in the limit definition

of f ′(a), then there is a δ > 0 such that

0 < |x− a| < δ =⇒∣∣∣∣f(x)− f(a)x− a

− f ′(a)∣∣∣∣ < 1

2f ′(a)

=⇒ f(x)− f(a)x− a

>1

2f ′(a)

=⇒ |f(x)− f(a)| > 1

2f ′(a)|x− a| > 0.

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Hence, if 0 < |x− a| < δ we can write


=g(f(x))− g(f(a))f(x)− f(a)

f(x)− f(a)x− a

(the previous calculation shows we are not dividing by zero here),and letting x→ a in this gives

limx→a


= limx→a

g(f(x))− g(f(a))f(x)− f(a)

limx→a

f(x)− f(a)x− a

= limH→0

g(f(a) +H)− g(f(a))H

f ′(a) (putting H = f(x)− f(a))

= g′(f(a))f ′(a)

which is what we wanted to prove.

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The case when f ′(a) = 0 is similar, but we have to work harder todeal with the possibility of dividing by zero in the final calculation.

This is not too hard, but is a bit long-winded, and does not doanything that will be useful later, so will be omitted here.

If you want to see the proof, look in any ‘real analysis’ book.

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Example 3.10 [monomials] For any integer k > 0, define mk(x) =xk, x ∈ R.

Then mk is differentiable at any a ∈ R, with derivative

m′k(a) = kak−1. (10)

Proof. We will prove this by induction on k.

I By Example 3.2, (10) is true for k = 0, 1, 2.

I Now suppose that (10) is true for some k > 3.

Then by the product rule

m′k+1(a) = (m1mk)′(a) = m′1(a)mk(a) +m1(a)m

′k(a)

= ak + akak−1 = (k + 1)ak.Hence, by induction, (10) is true for all k > 0.

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Example 3.11 [polynomials] Any polynomial of the form

p(x) = p0 + p1x+ p2x2 + · · ·+ pkx

k, x ∈ R,

for any integer k > 0, is differentiable at any a ∈ R, with derivative

p′(a) = p1 + 2p2a+ 3p3a2 · · ·+ pkka

k−1. (11)

Proof. This follows immediately from Example 3.10 and repeatedapplications of Theorem 3.8.

Remark. We ought to now obtain the derivative of other standardfunctions such as sinx, cosx, ex, etc., but we will skip that here.

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3.3 Local maxima and minima

Suppose that A = (a, b) is an open interval and f : A→ R.

Definition 3.12

I c ∈ A is a local maximum of f if there exists δ > 0 such that

|x− c| < δ =⇒ f(x) 6 f(c).

I c ∈ A is a local minimum of f if there exists δ > 0 such that

|x− c| < δ =⇒ f(x) > f(c).

I A local extremum is either a local max or a local min.

x

f(x)

a b

•

••

Figure 25: Illustration of Definition 3.12 and Theorem 3.13.

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Theorem 3.13 [Local extremum theorem] If f is differentiable atc ∈ A and c is a local extremum then f ′(c) = 0.

Proof. Suppose that c is a local maximum.

Then, by definition, there exists δ > 0 such that

|x− c| < δ =⇒ f(x) 6 f(c).

Hence

δ > x− c > 0 =⇒ f(x)− f(c)x− c

6 0 =⇒ limx→c+

f(x)− f(c)x− c

6 0,

−δ < x− c < 0 =⇒ f(x)− f(c)x− c

> 0 =⇒ limx→c−

f(x)− f(c)x− c

> 0.

(the final inequalities in each line follow from Theorem 2.20).

But f is differentiable at c, so these one-sided limits must equaleach other;

combining the inequalities implies that they must each be zero, sowe must have f ′(c) = 0.

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Note. The converse of Theorem 3.13 is not true. That is

f ′(c) = 0 6=⇒ c is a local extremum.

For example, the function f(x) = x3 has f ′(0) = 0, but 0 is not alocal extremum.

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Remark.

I Theorem 3.13 is the basis for a standard way of searching forlocal maxima and minima of a differentiable function f : A→R inside an open interval A:

you look for the zeros of f ′.

I This test finds local maxima and minima of f in an openinterval.

However, the overall max or min of the function might not beone of these, it might occur at an end-point of the interval.

The values at the end-points might be bigger or smaller thanthe maxima and minima inside the interval, but this test won’tnotice these.

In Fig. 25 we see that the overall max is the leftmost localmax, but the overall min is at the end-point b.

Actually, of course, we have only defined derivatives and localextrema at points inside an open interval

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3.4 Rolle’s theorem

Theorem 3.14 [Rolle’s Theorem] Suppose that f is continuous on[a, b] and differentiable on (a, b).

Suppose also that f(a) = f(b).

Then there exists c ∈ (a, b) such that f ′(c) = 0.

Proof. If f is constant then f ′(c) = 0 for all c ∈ (a, b).

So, let’s suppose that f is not constant.

By the boundedness theorem, f has an overall max and min on [a, b],and at least one of these must be different from f(a) = f(b)(otherwise f would be constant).

Suppose that

f(c) = sup {f(x) : x ∈ [a, b]} > f(a) = f(b),

so that c ∈ (a, b).

Clearly, c is a local max and f is differentiable at c, so by the localextremum theorem f ′(c) = 0. See Fig. 26.

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x

f(x)

• •

a bc

•

Figure 26: Illustration of Rolle’s Theorem, Theorem 3.14.

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Example 3.15 Prove that the equation

x3 − 3x2 + 4x+ 1 = 0

cannot have more than one solution on R.

Solution. Define f : R→ R by

f(x) = x3 − 3x2 + 4x+ 1, x ∈ R

(so the equation becomes f(x) = 0).

Now suppose that the equation has two solutions a, b, with a < b.

Then f(a) = f(b) (= 0), so by Rolle’s Theorem applied on [a, b],f ′(d) = 0 for some d ∈ (a, b).

But

f ′(x) = 3x2 − 6x+ 4 = 3(x− 1)2 + 1 > 0, for all x ∈ R,

which is a contradiction, so two solutions cannot exist.

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Example 3.16 Prove that the equation

x3 + x+ 1 = 0

has exactly 1 solution on R.

Solution. We have to do two things:

(a) Show that there is at least one solution (we will use the IVT(Theorem 2.39) for this);

(b) Show that there cannot be more than one solution (we willuse Rolle’s theorem (Theorem 3.14) for this).

(a) Define f(x) = x3 + x+ 1.

This is a polynomial, so it is continuous and differentiable on R(see Example 2.30 and Example 3.11).

Now,

f(0) = 1, f(−2) = −8− 2 + 1 = −9 < 0,

so by the IVT we have f(c) = 0 for some c ∈ (−2, 0).

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(b) Suppose there are two solutions a, b with a < b.

Then f(a) = f(b), so by Rolle’s Theorem, applied on [a, b], f ′(d) =0 for some d ∈ (a, b).

But f ′(x) = 3x2 + 1 > 0 for all x, which is a contradiction so twosolutions cannot exist.

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Example 3.17 Let f : R → R be twice differentiable on R withf(a) = f(b) = f(c) where a < b < c.

Prove that there exists α ∈ R such that f ′′(α) = 0.

Solution.

I Applying Rolle’s Theorem to f on [a, b] gives f ′(d1) = 0, forsome d1 ∈ (a, b).

I Applying Rolle’s Theorem to f on [b, c] gives f ′(d2) = 0, forsome d2 ∈ (b, c).

I From the previous results, f ′(d1) = f ′(d2), and d1 < d2, soapplying Rolle’s Theorem to the function f ′ on [d1, d2], givesf ′′(α) = 0, for some α ∈ (d1, d2).

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3.5 The first mean value theorem

Theorem 3.18 [The First Mean Value Theorem (MVT)] Supposethat f is continuous on [a, b] and differentiable on (a, b).

Then there exists a point c ∈ (a, b) such that

f(b)− f(a) = (b− a)f ′(c). (12)

Remark. We can rewrite (12) as

f ′(c) =f(b)− f(a)

b− a,

which says that the gradient of the function f at the point c is thesame as the gradient of the straight line joining the points (a, f(a)),(b, f(b)) (see Fig. 27).

In a sense, this is the average gradient of f between these points,and this is why Theorem 3.18 is called a ‘mean value theorem’.

Since it is called the first mean value theorem, you can guess thatthere are others.

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x

•

•

a bc

•

f(a)

f(b)

b− a

f(b)− f(a)


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Proof. Define F : [a, b]→ R by

F (x) = f(x)− f(a)− (x− a)f(b)− f(a)b− a

, x ∈ [a, b].

Then F is continuous on [a, b] and differentiable on (a, b). Also,

F (a) = F (b) = 0,

so, by Rolle’s Theorem, there exists c ∈ (a, b) such that F ′(c) = 0.Now, differentiating F at c directly gives

0 = F ′(c) = f ′(c)− f(b)− f(a)b− a

,

and rearranging this gives (12).

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One of the main uses of the MVT is in proving inequalities.

To do this it will be helpful to first define monotone functions.

Recall Definition 1.47 which defined monotone sequences.

Definition 3.19 If f : B → R (for some set B ⊂ R):

I f is increasing on B if x 6 y =⇒ f(x) 6 f(y) for allx, y ∈ B;

I f is decreasing on B if x 6 y =⇒ f(x) > f(y) for allx, y ∈ B;

I f is monotone on B if it is either increasing or decreasing.

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Example 3.20 The function f(x) = x3 is increasing on R.

The function g(x) = x2 is neither increasing nor decreasing on R;

it is decreasing on (−∞, 0] and increasing on [0,∞).

x

x2

x

x3

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Monotone functions need not be differentiable, but if a function isdifferentiable we can use the MVT to test whether it is monotone.

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Theorem 3.21 Suppose that f is continuous on [a, b] and differ-entiable on (a, b), for some a, b ∈ R. Then:

(a) f is increasing on [a, b] ⇐⇒ f ′(c) > 0 for all c ∈ (a, b);

(b) f is decreasing on [a, b] ⇐⇒ f ′(c) 6 0 for all c ∈ (a, b);

(c) f is constant on [a, b] ⇐⇒ f ′(c) = 0 for all c ∈ (a, b).

Proof. We only prove part (a), the proofs of (b) and (c) are similar.

(⇒) If f is increasing then, for any c ∈ (a, b),

f ′(c) = limx→c

f(x)− f(c)x− c

> 0,

by the sign condition in the definition of ‘increasing’.

(⇐) Suppose that f ′(c) > 0, for all c ∈ (a, b), and x, y ∈ (a, b)with x < y. Then, by the MVT (Theorem 3.18) on [x, y], thereexists c ∈ (x, y) such that

f(y)− f(x) = (y − x)f ′(c) > 0,

by the sign condition on f ′, which shows that f is increasing.224/336

x

f(x)

a b

•

Figure 28: Illustration of Theorem 3.21, with an increasingfunction f , and f ′ > 0.

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Remarks.

I We also say that f is strictly increasing on a set B if

x < y =⇒ f(x) < f(y), for all x, y ∈ B.

A similar proof to that of Theorem 3.21 shows that

f ′(c) > 0 for all c ∈ (a, b) =⇒ f is strictly increasing on [a, b].

The reverse implication is not true.

For example, the function f(x) = x3 is strictly increasing onR, but it has f ′(0) = 0.

I A similar definition and remarks hold for strictly decreasingfunctions.

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We can now derive inequalities for the values of functions on aninterval.

Example 3.22

(a) Show that ln(1 + x) >x

2for x ∈ [0, 1].

(b) Show that ex > 1 + x for all x > 0.

Solution. We will do both of these by moving everything to theleft and then showing that what we get is > 0.

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(a) Define f : [0, 1]→ R by

f(x) = ln(1 + x)− x

2, x ∈ [0, 1].

We show that f(x) > 0 for x ∈ [0, 1].

To do this we note that f is continuous on [0, 1] and differ-entiable on (0, 1), with

f(0) = 0, f ′(x) =1

1 + x−1

2=

1− x2(1 + x)

> 0, x ∈ (0, 1).

So, by the above remarks, f is strictly increasing on [0, 1], so

f(x) > f(0) = 0, for all x ∈ (0, 1],

which proves the result.

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(b) Define f : [0,∞)→ R by

f(x) = ex − 1− x, x ∈ [0,∞).

We will show that f(x) > 0 for arbitrary x > 0.

We first note that the above MVT results were on finite inter-vals [a, b], and here we have an infinite interval [0,∞), withno right hand end-point.

To deal with this, let’s just consider an arbitrary x > 0 andconsider the problem on the interval [0, x].

Now, f is continuous on [0, x] and differentiable on (0, x),with

f(0) = 0, f ′ > 0 on [0, x],

so f is strictly increasing on [0, x] and f(x) > f(0) = 0,which proves the result.

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In the above proof of Theorem 3.21, and in Example 3.22, we hadf ′ > 0, so we could get rid of the unknown term f ′(c) that camefrom the MVT by replacing it with 0.

We can sometimes make more use of this term by using some otherinequalities for it, for example in terms of the values of f ′(a) orf ′(b), which we can work out.

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Example 3.23 Let 0 < p < 1. By applying the first mean valuetheorem to xp, prove that for any integer n > 1,

p (n+ 1)p−1 6 (n+ 1)p − np 6 p np−1 . (13)

Solution. At first sight this might look a bit baffling, but thetrick is to notice that the term in the middle of (13), between theinequalities, is the difference between two values of the function xp,evaluated at the points x = n and x = n + 1, and the terms onthe left and right of the inequalities looks like the derivative of thisfunction.

So, (13) looks a bit like the MVT (with inequalities), for the functionxp.

But we need to do this properly.

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Define f(x) = xp, x > 0.

Then f is continuous on [n, n+ 1] and differentiable on (n, n+ 1).

Applying the MVT to f on the interval [n, n+ 1] gives

f(n+ 1)− f(n) = f ′(c), for some c ∈ (n, n+ 1),

that is

(n+ 1)p − np = pcp−1, for some c ∈ (n, n+ 1). (14)

So, we have shown that the term in the middle of (13) is equal tosomething, but we actually want to get the inequalities in (13).

We will do this by moving the point c in (14) to the end-pointsn and n + 1, which will break the equalities, but will give us theinequalities that we want.

Since n < c < n+ 1 and 0 < p < 1, we have

p(n+ 1)p−1 6 pcp−1 = (n+ 1)p − np = pcp−1 6 pnp−1,

which is (13).

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3.6 The nth mean value theorem

There is a version of Theorem 3.18 involving higher order derivatives.

To describe this we first need to define higher order derivatives!

Definition 3.25 Suppose that f is continuous on [a, b] and differ-entiable on (a, b).

I If the derivative f ′ is itself differentiable on (a, b) then we willwrite the derivative of f ′ as (f ′)′ = f ′′, and call it the secondderivative of f .

I If the second derivative is differentiable on (a, b) then we willwrite the derivative of f ′′ as (f ′′)′ = f ′′′, and call it the thirdderivative.

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The dash notation for derivatives starts to become a bit unwieldy ifwe need more derivatives!

Alternatively we write:

f (1) = f ′,

f (2) = f ′′ = (f ′)′,

f (3) = f ′′′ = (f ′′)′,

...

f (n) = (f (n−1))′,

for any integer n > 1.

The round brackets in the exponent are used to denote derivatives,to distinguish them from powers, or iterates, of f .

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Theorem 3.26 [The nth Value Theorem] Suppose that f is con-tinuous on [a, b] and n > 2 times differentiable on (a, b).

Then there exists a point c ∈ (a, b) such that

f(b) = f(a) + f (1)(a)(b− a) + 1

2!f (2)(a)(b− a)2 + . . .

+1

(n− 1)!f (n−1)(a)(b− a)n−1 + 1

n!f (n)(c)(b− a)n.

(15)

Proof. The proof is similar to that of Theorem 3.18, just a bitlonger, so it will be omitted here – it is in the notes.Remark. Theorem 3.26 is sometimes called Taylor’s theorem withremainder(the final term, involving f (n)(c) is the ‘remainder’).

The theorem is often useful in its own right, and is a step towardsderiving infinite Taylor expansions of functions(or MacLaurin expansions if a = 0).

However, we won’t go any further with this here.

The third year complex analysis course will do a lot more of this.235/336

Chapter 4

Series

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4.1 Convergence of series

A series is an infinite summation of the form∞∑r=1

ar = a1 + a2 + a3 + · · ·

(a series is sometimes called an ‘infinite series’ to emphasize thatthere are infinitely many terms in the summation, but this is a bitredundant since no one would call a finite summation a ‘series’, sowe will omit the word ‘infinite’).

You will have seen this before.

However, what does this mean?

How do you add up infinitely many terms?

Yet again, we need a precise definition of something that we haveseen before but didn’t know precisely what it meant.

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Definition 4.1 Suppose that we have a series∞∑r=1

ar.

For each integer n > 1, let

Sn = a1 + a2 + a3 + · · ·+ an =

n∑r=1

ar, n > 1,

so Sn is just the (finite) sum of the first n terms in the series.

The numbers Sn are the partial sums of the series∑∞

r=1 ar.

I If the sequence (Sn) converges, with S = limn→∞

Sn, then:

I the series∞∑r=1

ar converges;

I the sum of the series is S, and we write∞∑r=1

ar = S.

I If the sequence (Sn) diverges then the series diverges.

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Note. In general we will suppose that the summation in a seriesstarts at r = 1, but there are times when it is more convenient tostart at r = 0(for example, for power series – see below).

Clearly, whether we start at r = 0 or at r = 1 does not affect theconvergence of the series.

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Theorem 4.2 If the series∞∑r=1

ar converges, then limr→∞

ar = 0.

Proof. If∞∑r=1

ar converges, then limn→∞

Sn = S.

Since an = Sn − Sn−1 ,limn→∞

an = limn→∞

Sn − limn→∞

Sn−1 = S − S = 0.

Theorem 4.2 gives the following simple, but useful, criterion fordivergence of a series:

limr→∞

ar 6= 0 =⇒∞∑r=1

ar diverges. (16)

Note. Theorem 4.2 does not say

limr→∞

ar = 0 =⇒∞∑r=1

ar converges.

We will see below that, in general, this is false.

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Example 4.3 Discuss the convergence of∞∑r=1

ar for the cases:

(a) ar = (−1)r , (b) ar =r

r + 1, (c) ar =

r

r2 + 1.

Solution.

(a) For ar = (−1)r , the limit limr→∞

ar does not exist so the series

diverges.

(b) For ar =r

r + 1, limr→∞

ar = 1 so the series diverges.

(c) For ar =r

r2 + 1, limr→∞

ar = 0 so we cannot, yet, determine

if the series converges.

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Definition 4.1 is OK as a definition, but unfortunately there are veryfew series whose partial sums we can work out, so we need somemore useful tests for whether series converge or not.

We first give two very useful examples of series, whose convergenceor divergence we can work out, and then in the following sectionwe give some convergence tests that we can apply to more generalseries.

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For our first example we look at a series whose partial sums we canwork out.

Theorem 4.4 [The geometric series] The series∞∑r=0

xr = 1 + x+ x2 + . . .

with constant x, diverges if |x| > 1 and converges if |x| < 1, inwhich case its sum is

∞∑r=0

xr = 1 + x+ x2 + · · · = 1

1− x

Proof. If |x| > 1 then ar = xr 6→ 0 (see Example 1.23), so theseries diverges, by (16).

If |x| < 1 then, by the formula for the sum of a geometric series,

Sn = 1+x+x2+ · · ·+xn =1− xn+1

1− x=

1

1− x− xn+1

1− x→ 1

1− x(by Example 1.54) which proves the result.

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The next example will be the basis for many of our tests for conver-gence of series below.

This series is about as simple you could imagine, but the proof ofconvergence is surprisingly long — see the notes for this!

Theorem 4.5 The series∞∑r=1

r−α =1

1α+

1

2α+

1

3α+ . . . (17)

with constant α, converges if α > 1 and diverges if α 6 1.

Remark. As you make α > 0 bigger, the terms in the series (17)shrink to zero more quickly as r → ∞, so the series is more likelyto converge, which agrees with what the theorem says:

I when α is below, or equal to, 1 the series diverges;

I when α is above 1 the series converges.

Remember this!

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As usual, we have a theorem on combining series, whose proof comesfrom our results on combining limits.

Theorem 4.6 Suppose that∞∑r=1

ar,

∞∑r=1

br are convergent series

with sums A,B, and c is a constant.

Then:

(a) the series∞∑r=1

(ar + br) is convergent, with sum A+B;

(b) the series∞∑r=1

c ar is convergent, with sum cA.

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Proof. (a) Let (Sn) , (Tn) be the sequences of partial sums of∑∞r=1 ar,

∑∞r=1 br.

Then, for any n > 1, the partial sumn∑r=1

(ar + br) =

n∑r=1

ar +

n∑r=1

br = Sn + Tn,

that is, the sequence of partial sums of the series∑∞

r=1 (ar + br) isthe sequence (Sn + Tn).

By definition, the separate sequences (Sn) , (Tn) converge, so byTheorem 1.26 (a), the sequence (Sn + Tn) also converges, with

limn→∞

(Sn + Tn) = A+B,

which proves part (a).

(b) The proof is similar to (a), using Theorem 1.26 (c).

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4.2 Series with positive terms

In this section we consider series∞∑r=1

ar whose terms are all positive,

that is,

ar > 0, for all r > 1. (18)

If (18) holds then for any n > 1 the partial sums satisfy

Sn+1 = a1 + a2 + a3 + · · ·+ an + an+1 = Sn + an+1 > Sn.

That is, the sequence (Sn) is increasing, and so the series∑∞

r=1 arconverges if and only if the sequence (Sn) is bounded above(by Theorem 1.49).

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We can now prove the most basic test for convergence of series.

Theorem 4.7 [Comparison Test] Suppose that∑∞

r=1 ar,∑∞

r=1 brare series with

0 6 ar 6 br, for all r > 1. (19)

Then:

(a)∞∑r=1

br converges =⇒∞∑r=1

ar converges;

(b)∞∑r=1

ar diverges =⇒∞∑r=1

br diverges.

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r

•

•

b1

a1 ••

•••••••• •• •• •• ••

Figure 29: Two series as in Theorem 4.7.

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Proof. Let (Sn) , (Tn) be the partial sums of∞∑r=1

ar,∞∑r=1

br.

(a) Suppose that∞∑r=1

br converges.

Then the sequence (Tn) is bounded, say Tn 6 K for all n.

Hence, by (19),

Sn 6 Tn 6 K, for all n > 1,

so that the sequence (Sn) is bounded and so∞∑r=1

ar converges.

(b) Suppose that∞∑r=1

ar diverges.

Then the sequence (Sn) is unbounded.

Since Tn > Sn, for all n > 1, the sequence (Tn) is also unbounded

so∞∑r=1

br diverges.

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Remark. Intuitively, Theorem 4.7 says that:

I if∞∑r=1

br converges, and∞∑r=1

ar is smaller, then it must also

converge;

I if∞∑r=1

ar diverges, and∞∑r=1

br is bigger then it must also di-

verge.

To use Theorem 4.7 to decide whether some given series convergesor diverges, we need to ‘compare’ it with some other series whoseconvergence properties are known.

In fact, we will usually use the series in Theorem 4.4 and Theorem 4.5as the ‘comparison’ series.

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Example 4.8 Discuss the convergence or divergence of∞∑r=1

ar for

the cases:

(a) ar =r + 4

r3 + 1; (b) ar =

r2

r3 + 1; (c) ar =

(r + 1) sin2 r

r3 + 3r + 1.

Solution.

(a) For all r > 1,

r + 4

r3 + 16r + 4r

r3=

5

r2.

Since∞∑r=1

1/r2 converges (by Theorem 4.5),∞∑r=1

ar converges.

Note. Strictly, the comparison series should be∞∑r=1

5/r2 here, but

the constant 5 does not affect the convergence or divergence so weignore it — similarly below.

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(b) For all r > 1 ,

r2

r3 + 1>

r2

r3 + r3=

1

2r.

Since∞∑r=1

1

rdiverges, the original series diverges.

(c) For all r > 1 , ar > 0 and

(r + 1) sin2 r

r3 + 3r + 16

(r + r)

r3=

2

r2.

Since∞∑r=1

1

r2converges, the original series converges.

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The next theorem gives another, equally useful, convergence test.

Theorem 4.9 [Ratio Test] Suppose that for all r > 1, ar > 0 and

limr→∞

ar+1

ar= L > 0.

Then the series∞∑r=1

ar converges if L < 1 and diverges if L > 1 .

If L = 1 then no conclusion can be drawn.

Proof. Suppose that 0 6 L < 1.

Choose y with L < y < 1. By the argument in the proof of Theo-rem 1.30, there exists C > 0 such that

0 < ar 6 Cyr for all r > 1.

Now, the geometric series∞∑r=1

yr converges, by Theorem 4.4, so

∞∑r=1

ar converges, by the comparison test (Theorem 4.7).

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Suppose that L > 1.

Choose y with 1 < y < L. Now, by the argument in the proof ofTheorem 1.30, for some C > 0,

ar > Cyr, for all r > 1,

so ar 6→ 0, and hence the series∞∑r=1

ar must diverge, by Theo-

rem 4.2.

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Example 4.10 Discuss the convergence or divergence of∞∑r=1

ar for

the cases:

(a) ar =r

5r; (b) ar =

3r

r!; (c) ar =

(r!)2

(2r)!.

Solution. For each r > 1,

(a)ar+1

ar=r + 1

5r+1

5r

r=r + 1

5r→ 1

5=⇒ converges.

(b)ar+1

ar=

3r+1

(r + 1)!

r!

3r=

3

r + 1→ 0 =⇒ converges.

(c)ar+1

ar=

((r + 1)!)2

(2r + 2)!

(2r)!

(r!)2=

(r + 1)2

(2r + 2)(2r + 1)→ 1

4

=⇒ converges.

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4.3 General series

We now consider the general case of series with positive and negativeterms and obtain some convergence tests for these.

The first result shows that if we convert a general series into a seriesof positive terms by taking the modulus (or absolute value) of all theterms then, if this new series is convergent then the original seriesis convergent.

This is a useful test since we can use the results of the previoussection to consider the convergence of the series of absolute values.

Definition 4.11 A series∞∑r=1

ar is absolutely convergent if the

series of absolute values∞∑r=1

|ar| converges.

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Theorem 4.12 [Absolute convergence test] If a series∞∑r=1

ar is

absolutely convergent then it is convergent.

Proof. For all r > 1, let

br =|ar|+ ar

2, cr =

|ar| − ar2

,

and hence,

ar = br − cr , 0 6 br 6 |ar| , 0 6 cr 6 |ar|.

Now, we have assumed that∞∑r=1

|ar| converges,

so by the comparison test,∞∑r=1

br and∞∑r=1

cr also converge, and

hence∞∑r=1

ar =

∞∑r=1

(br − cr) converges.

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Remark. We will see below that the converse to Theorem 4.12need not be true.

That is, there exist sequences which converge but whose series ofabsolute values diverge.

Using Theorem 4.12 we can now extend the convergence parts ofthe comparison and ratio tests to series which need not be positive(by the previous remark, we cannot extend the divergence cases).

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Corollary 4.13

(a) (comparison) |ar| 6 br, for all r > 1, and∞∑r=1

br converges

=⇒∞∑r=1

ar converges;

(b) (ratio) ar 6= 0, for all r > 1, and limr→∞

|ar+1||ar|

= L < 1

=⇒∞∑r=1

ar converges.

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Example 4.14 Consider the series∞∑r=1

(−1)r+1 1

r2= 1− 1

4+

1

9− 1

16+ · · · .

Since the series∞∑r=1

∣∣∣∣(−1)r+1 1

r2

∣∣∣∣ = ∞∑r=1

1

r2

converges, the series∞∑r=1

(−1)r+1 1

r2converges.

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Example 4.15 Discuss the convergence of∞∑r=1

ar for the cases:

(a) ar =r sin r

r3 + 5r2 + 3; (b) ar =

r (−1)r

3r.

Solution.

(a) In this case we will use the comparison test to show that∞∑r=1

|ar| converges.

For all r > 1 ,

|ar| =∣∣∣∣ r sin r

r3 + 5r2 + 3

∣∣∣∣ 6 ∣∣∣∣ r

r3 + 5r2 + 3

∣∣∣∣ 6 r

r3=

1

r2.

Now,∞∑r=1

1

r2converges =⇒

∞∑r=1

|ar| converges =⇒∞∑r=1

ar converges

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(b) In this case we will use the ratio test to show that∞∑r=1

|ar|

converges.For all r > 1,

|ar+1||ar|

=r + 1

3r+1

3r

r=r + 1

3r→ 1

3.

By the ratio test,∞∑r=1

|ar| converges, so by the absolute con-

vergence test the original series converges.

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Theorem 4.16 [Alternating series test] Suppose that (ar) is adecreasing sequence, with ar > 0, r > 1, and lim

r→∞ar = 0.

Then the series∞∑r=1

(−1)r+1ar = a1 − a2 + a3 − a4 + . . .

converges.

Proof. The trick in this proof is to notice that:

I because the signs of the terms in the series are alternating,the partial sums jump up and down at each step;

I because the sequence (ar) is decreasing, the size of thesejumps gets smaller at each step and tends to zero;

I hence the sequence of partial sums gets ‘squeezed’ to a limit(see Fig. 30).

But this needs to be made more precise!

It isn’t hard, but see the notes for this.264/336

n

Sn

•

•

•

•

•••• • •

Figure 30: The partial sums of an alternating series as inTheorem 4.16.

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Example 4.17 Consider the series∞∑r=1

(−1)r+1 1

r= 1− 1

2+

1

3− 1

4+ · · · .

Here, ar =1

r, r > 1, and this sequence is decreasing with ar → 0

as r →∞.

Hence, by the alternating series test, the series converges.

Remark. We know from Theorem 4.5 that the series∞∑r=1

1

rdoes

not converge.

Thus, Example 4.17 shows that a series may converge even thoughits series of absolute values does not converge.

Hence, the converse to the absolute convergence test, Theorem 4.12,is not true.

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4.4 Summary of convergence tests

We consider the series∞∑r=1

ar.

I If ar 6→ 0 then the series diverges.

I

∞∑r=1

1

rαconverges if α > 1 and diverges if α 6 1;

∞∑r=1

xr converges if |x| < 1 and diverges if |x| > 1.

I The comparison test applies when all terms are positive andcan be used to show convergence and divergence.

Useful when (ar) is the ratio of polynomials such as

ar =r + 6

r3 + 7r2 + 2

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I The ratio test applies when all terms are positive and can beused to show convergence and divergence.Useful when (ar) contains powers and factorials, for example,

ar =r2

3rand ar =

7r

r!

I The absolute convergence test can only prove convergence. Itapplies to series with both positive and negative terms.

I The alternating series test can only prove convergence.It applies when the terms alternate in sign and the size of theterms decreases to zero.

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4.5 Power series

Definition 4.18 A series of the form∞∑r=0

prxr, (20)

where pr, r = 0, 1, 2, . . . , are constants, and we regard x as avariable, is said to be a power series.

Remarks 4.19 Power series are important because many functionsare actually defined in terms of them. For example,

ex = 1 + x+1

2!x2 +

1

3!x3 + . . . ,

cosx = 1− 1

2!x2 +

1

4!x4 − . . . ,

sinx = x− 1

3!x3 +

1

5!x5 − . . . .

So, we would like to know for what values of x the series (20)converges, and some of the properties of the function defined by(20) when it does converge.

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We start with convergence.

Clearly, (20) converges when x = 0, but does it converge for othervalues of x, and if so, which values?

To answer this question, let

S := {|x| : (20) converges}.The set S is non-empty (since 0 ∈ S) and bounded below (sinceany element is positive) so we can make the following definition.

Definition 4.20

I if S is bounded, let R := supS;

I if S is unbounded, let R :=∞.

R is called the radius of convergence of the power series (20).

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Theorem 4.21 For the power series (20), there are three possibili-ties (see Fig. 31):

(a) if R = 0 then (20) converges only for x = 0;

(b) if 0 < R <∞ then (20) converges for all |x| < R and divergesfor all |x| > R;

(c) if R =∞ then (20) converges for all x ∈ R.

0 R−R

Figure 31: The radius of convergence of a power series asin Theorem 4.21.The series:

I converges for any red x with |x| < R;

I diverges for any black x with |x| > R;

I can converge or diverge when |x| = R.

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Proof.

(a) This is obvious from the definition of R.

(b) If |x| > R then the divergence is again obvious.

If |x| < R then the convergence can be proved by comparisonwith a suitable geometric series.

The proof is in the notes – it isn’t hard but is moderately longand detailed, so will be omitted here.

(c) In this case any x ∈ R satisfies |x| < R, and so the proof ofpart (b) shows that (20) converges for all x ∈ R.

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Corollary 4.22 For the power series (20), if the limit

L = limr→∞

|pr+1||pr|

exists then the radius of convergence of (20) is R =1

L.

Note. We allow L =∞ here, and we set: R = 0 if L =∞ andR =∞ if L = 0.

Proof. Since L exists, we have

limr→∞

|pr+1xr+1|

|prxr|= |x|L

{< 1 if |x| < 1/L,

> 1 if |x| > 1/L,

so by the ratio test,∞∑r=0

|prxr| converges if |x| < 1/L and diverges

if |x| > 1/L.

Hence, by Theorem 4.21, R must coincide with 1/L.

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Remark.

I Corollary 4.22 is a useful result for finding the radius of con-vergence of a power series.

Unfortunately, in general, the limit L might not exist, so thisresult does not apply to every power series.

I The proof of Corollary 4.22 shows that when the limit L existsthen the series (20) converges for all |x| < R, which is part (b)of Theorem 4.21.

This is not a complete proof of the result in Theorem 4.21since it only works when L exists.

I Neither Theorem 4.21 nor Corollary 4.22 tell us anythingabout convergence or divergence of (20) when |x| = R.

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Example 4.23 Examples of the three possible cases in Theo-rem 4.21:

(a)∞∑r=0

r!xr, by Example 1.31, for any x 6= 0, limr→∞

r!xr =∞,

so the series diverges;

(b)∞∑r=0

xr by Theorem 4.4, the geometric series has R = 1;

(c) choose m > 1 and let pr = 0 for all r > m, so that the seriesonly has finitely many terms — such a series clearly convergesfor all x;if that seems a bit contrived, a proper infinite series whichconverges for all x is

∞∑r=0

xr

r!

(by the ratio test).

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Example 4.24 Use Corollary 4.22 to find the radius of convergenceR of the following power series.

Also, if R <∞, determine whether the series converge for x = −Rand x = R.

(a)

∞∑r=0

xr

r4r(b)

∞∑r=0

r2xr

r!(c)

∞∑r=0

xr

7r(r3 + 2r)

Solution. (a)pr+1

pr=

r4r

(r + 1)4r+1=

r

4(r + 1)→ 1

4=⇒ R = 4.

x = R = 4: the series is∞∑r=0

1

r, which diverges (Thm 4.5).

x = −4: the series is∞∑r=0

(−1)r

rwhich converges (Thm 4.16).

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(b)pr+1

pr=

(r + 1)2

(r + 1)!

r!

r2=

(r + 1)2

r2(r + 1)→ 0 =⇒ R =∞.

(c)pr+1

pr=

(r3 + 2r)

7 ((r + 1)3 + 2r + 2)→ 1

7=⇒ R = 7.

When x = ±R = ±7, the series is∞∑r=0

(±1)r

r3 + 2r, and

0 61

r3 + 2r6

1

r3, r > 1,

and∞∑r=0

1

r3converges, so by comparison and absolute convergence

the power series converges when x = ±R.

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Properties of functions defined by power series.

We now define a function

f(x) :=

∞∑r=0

prxr, x ∈ (−R,R), (21)

where R is the radius of convergence of the power series (21), sothat the domain of f is D(f) = (−R,R).Theorem 4.25 If R > 0 then the function f defined by (21) iscontinuous and differentiable on (−R,R), and the derivative f ′ isgiven by the power series

f ′(x) =

∞∑r=1

prrxr−1, x ∈ (−R,R), (22)

and this power series also has radius of convergence R.

Remarks 4.26 Clearly, we can obtain the formula (22) for f ′ bydifferentiating each of the terms in the formula (21) for f .

We know that this would be valid for a finite summation, but wemust prove that this process is valid for an infinite summation.

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Proof.

By the definition of the derivative, we need to show that

f(x+ h)− f(x)h

−∞∑r=1

prrxr−1 =

1

h

∞∑r=1

pr[(x+ h)r − xr − hrxr−1

],

tends to zero as h→ 0.

To do this we use the the second MVT(Theorem 3.26 with n = 2),to show that

(x+ h)r − xr − hrxr−1 = 1

2r(r − 1)h2|cr|r−2

for suitable cr, and then the h2 term more than kills off the h termon the bottom line of the preceding formula.

The details are in the notes.

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Theorem 4.25 shows that a function f defined by a power series on(−R,R) is differentiable, and its derivative f ′ is itself given by apower series on (−R,R).Hence, we can now apply Theorem 4.25 to f ′ to show that this isalso differentiable on (−R,R).In fact, we can keep on differentiating f arbitrarily often.

Corollary 4.27 The function f defined by (21) is infinitely differ-entiable on (−R,R), that is, every derivative f (n), n = 1, 2, . . . ,exists on (−R,R).

There will be much more about this in the 3rd year Complex Analysiscourse.

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Chapter 5

The Riemann Integral

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5.1 Introduction

Let f : [a, b]→ R be a bounded function.

The aim of this chapter is to define the integral of f over the interval[a, b].

We begin by considering this as the area under the graph of f(more precisely, the area between the graph of f and the x-axis).

However, integrals are much more general than just an area — wewill say more about this at the end.

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Looking at Fig. 32, it is clear that the area under the graph of f liesbetween the areas of the upper and lower sets of rectangles drawnthere, and we can work out what these areas are(the area of a rectangle is width×height).

x

f(x)

a b x

f(x)

a b x

f(x)

a b

Figure 32: The area under the blue graph lies between theareas under the upper and lower sets of red rectangles.

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Now, to obtain the area under the graph, we imagine splitting thisarea into narrower and narrower rectangles and taking the limit ofthe areas of the upper and lower sets of rectangles as the widthtends to zero.

We would guess that these upper and lower areas should tend to thesame value, and this should be the area under the graph.

In the following sections we try to make this precise.

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5.2 Riemann sums

Definition 5.1 A partition P of [a, b] is a finite set of points suchthat {a, b} ⊂ P ⊂ [a, b].

We can write P as P = {x0, x1, x2, . . . , xk} where

a = x0 < x1 < x2 < · · ·xk = b.

For a partition P we also define

δr := xr+1 − xr, for each r = 0, . . . , k − 1,

δmax(P ) := max{δ0, . . . , δk−1}.

• • • • • •x0 = a xk = bxr xr+1

δr

Figure 33: A partition, as in Definition 5.1.

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Example 5.2 P = {0, 0.4, 0.75, 1} is a partition of [0, 1], withδ0 = 0.4, δ1 = 0.35, δ2 = 0.25.

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Remarks.

I Partitions will do the splitting process in Fig. 32.

A partition P splits the interval [a, b] into k subintervals[xr, xr+1], r = 0, 1, . . . , k − 1, of width δr, and these subin-tervals will be the bases of the rectangles, as in Fig. 32..

This is the first step in the process of calculating the area.

I The subintervals need not all have the same width

(although they do in each of the figures in Fig. 32).

I δmax(P ) is the maximum width of all the subintervals.

I Clearly, the sum of the widths of the subintervals equals thetotal width b− a of the interval [a, b], that isk−1∑r=0

δr =

k−1∑r=0

(xr+1 − xr)

= (x1 − x0) + (x2 − x1) + · · ·+ (xk − xk−1)= b− a.

(23)

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We now introduce a notation for the areas of the lower and uppersets of rectangles.

Definition 5.3 The lower Riemann sum L(P, f) and the upperRiemann sum U(P, f) of f with respect to the partition P are:

L(P, f) =

k−1∑r=0

Ir(f)δr, U(P, f) =

k−1∑r=0

Sr(f)δr, (24)

where, for r = 0, . . . , k − 1,

Ir(f) = inf{f(x) : x ∈ [xr, xr+1]},Sr(f) = sup{f(x) : x ∈ [xr, xr+1]}.

(25)

Remark. For a function f and partition P (as in Fig. 32):

I L(P, f) is the area under the set of lower rectangles;

I U(P, f) is the area under the set of upper rectangles.

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Remark.

I Clearly, the quantities Ir, Sr and δr depend on the partitionP , so we should really write Ir(P ), Sr(P ) and δr(P ).

However, this would make all the formulae fairly unwieldy andhard to read (and write).

So we have omitted the dependence on P in these quantities,although we have kept it in the notation L(P, f) and U(P, f).

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Example 5.4 Let f(x) = x2 and let P = {0, 1/2, 3/4, 1}.

S0(f) = sup{x2 : x ∈ [0, 1/2]} =(1

2

)2

=1

4, δ0=

1

2,

S1(f) = sup{x2 : x ∈ [1/2, 3/4]} =(3

4

)2

=9

16, δ1=

1

4,

S2(f) = sup{x2 : x ∈ [3/4, 1]} = 12 = 1 , δ2=1

4,

U(P, f) =1

4× 1

2+

9

16× 1

4+ 1× 1

4=

33

64.

A similar calculation shows that

I0(f) = 0, I1(f) =1

4, I2(f) =

9

16.

L(P, f) = 0× 1

2+

1

4× 1

4+

9

16× 1

4=

13

64.

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We now define the ‘trivial’ partition

T0 := {a, b},that is T0 just contains the end points of the interval [a, b], and nointerior points, and let

I(f) := inf{f(x) : x ∈ [a, b]},S(f) := sup{f(x) : x ∈ [a, b]}.

(26)

It is clear from the definitions that

L(T0, f) = I(f)(b− a),U(T0, f) = S(f)(b− a).

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Lemma 5.5 For any partition P of [a, b],

L(T0, f) 6 L(P, f) 6 U(P, f) 6 U(T0, f). (27)

Proof. It follows immediately from (25) and (26) that, for anypartition P ,

I(f) 6 Ir(f) 6 Sr(f) 6 S(f), r = 0, . . . , k − 1,

so substituting this into (24), and using (23) gives

I(f)(b− a) = I(f)

k−1∑r=0

δr 6k−1∑r=0

Ir(f)δr = L(P, f)

6k−1∑r=0

Sr(f)δr = U(P, f) 6 S(f)

k−1∑r=0

δr = S(f)(b− a).

Remark. Lemma 5.5 shows that for any single partition P thelower sum L(P, f) is less than the upper sum U(P, f)(which you would expect!).

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Next, suppose that we have two partitions P and R, with P ⊂ R,that is, all the points in P are also in R.

In this case we say that R is a refinement of P .

In effect, we refine a partition P by adding points to it to give thenew partition R, so that all the gaps have either stayed the same,or got smaller.

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Lemma 5.6 Let P and R be partitions of [a, b] such that R is arefinement of P . Then

L(P, f) 6 L(R, f), U(P, f) > U(R, f). (28)

Proof. We first suppose that R contains just one more point thanP , say y ∈ (xt, xt+1), for some integer t with 0 6 t 6 k − 1.

• • • • • •x0 = a xk = bxt xt+1

•y

Nothing else changes, so we only need to look at [xt, xt+1], with thetrivial partition T0 = {xt, xt+1} and the refinement R = {xt, y, xt+1}.Now, by Lemma 5.5,

L(T0, f) 6 L(R, f) 6 U(R, f) 6 U(T0, f),and so

L(P, f) 6 L(R, f) 6 U(R, f) 6 U(P, f)

(since the contributions to each of these summations outside theinterval [xt, xt+1] are all the same).

That is, (28) holds for a single point refinement.294/336

We can now prove the general case as follows:

I we can obtain the partition R from the partition P in a stepby step manner by adding a single point at a time;

I the required inequalities (28) hold at each step in this process,so they hold at the end of the process.

Remark. Lemma 5.6 shows that if we refine a partition then thelower sum goes up and the upper sum goes down

(again, we would probably expect this to happen).

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The next theorem deals with any two partitions P and Q, neitherof which need be a refinement of the other one.

Theorem 5.7 If P and Q are any partitions of [a, b] then

L(P, f) 6 U(Q, f).

Proof. We first note that the set R = P ∪Q is a partition, and isa refinement of both the partitions P and Q

(since P ⊂ P ∪Q and Q ⊂ P ∪Q).

Hence, by Lemma 5.5 and Lemma 5.6,

L(P, f) 6 L(R, f) 6 U(R, f) 6 U(Q, f).

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Remark.

I Theorem 5.7 extends the result of Lemma 5.5 and shows thatany lower Riemann sum is less than any upper Riemann sum.

This is not immediately obvious since, in the general casewhere Q is not a refinement of P , there does not seem tobe any link between the sups and infs that we work out forthe two separate partitions P and Q, and the widths of thesubintervals in these partitions.

I In the proof of Theorem 5.7 we took two separate partitionsP and Q and combined them into a single partition P ∪ Q,which is a refinement of both P and Q.

This trick is very useful and will be used again below.

It often means that if we have one partition P that does somejob, and another partition Q that does another job, we canget a single partition that does both jobs by using P ∪Q.

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5.3 The Riemann integral of f

For any given partition P we now have the idea of the upper andlower Riemann sums of f , with respect to P , which give us the totalareas of the upper and lower rectangles in Fig. 32.

What we now want to do is take some sort of limit as the widths ofthe gaps in the partitions shrinks to zero — and we hope that theseupper and lower areas will tend to the same limit!

Unfortunately, it is not easy to define what we mean by the limit ofthe set of partitions, so instead we will first look at the ‘smallest’upper sum, and the ‘largest’ lower sum.

That is, we will look at the inf and the sup of these sums.

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It follows from Lemma 5.5 that the sets of values of lower and upperRiemann sums

L(f) :={L(P, f) : P is a partition of [a, b]

},

U(f) :={U(P, f) : P is a partition of [a, b]

},

are bounded, so we can define

L(f) := supL(f), U(f) := inf U(f).

Theorem 5.8 For any partitions P , Q,

L(P, f) 6 L(f) 6 U(f) 6 U(Q, f). (29)

Proof. The outer inequalities in (29) follow immediately from thedefinitions of L(f) and U(f).

The middle one follows from Theorem 5.7, which shows that everyelement of L(f) is 6 every element of U(f), so

L(f) = supL(f) 6 U(f) = inf U(f).

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Definition 5.9 A bounded function f : [a, b] → R is (Riemann)integrable on [a, b] if

L(f) = U(f).

If f is integrable, then this common value is defined to be the (Rie-mann) integral of f over [a, b] and is denoted by∫ b

af or

∫ b

af(x) dx .

The latter notation is useful if we need to emphasize which variableis being integrated, or we want to put in a specific function, e.g.∫ 1

0x2 dx.

The former notation is shorter and simpler if we are doing an ab-stract integration and haven’t defined the letter we are using for thevariable of integration.

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It is also convenient (notationally) to define the integral of f whenthe lower limit is greater than, or even equal to, the upper limit.

So, we define ∫ a

bf := −

∫ b

af,

∫ a

af := 0, a < b. (30)

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Remark. The above constructions have been motivated by findingthe area between the graph of f and the x-axis, and worked out theareas of rectangles by multiplying ‘heights’ by ‘widths’, where the‘heights’ are given by function values.

So, if any function values are negative the corresponding ‘areas’ arenegative.

That is, the Riemann integral counts ‘areas’ below the x-axis asnegative.

x

f(x)

a

b

Figure 34: Positive (blue) and negative (red) areas ‘under’the graph of f .

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We first observe that not every bounded function is integrable.

Example 5.10 Consider the function f : [0, 1]→ R defined by

f(x) =

{0, if x ∈ [0, 1] is irrational,

1, if x ∈ [0, 1] is rational.

For any partition P , every interval [xr, xr+1] contains both rationaland irrational numbers, so we see that

Ir(f) = inf{f(x) : x ∈ [xr, xr+1]} = 0 ,

Sr(f) = sup{f(x) : x ∈ [xr, xr+1]} = 1,

so

L(P, f) = 0, U(P, f) = 1 =⇒ L(f) = 0, U(f) = 1

=⇒ f is not integrable.

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Remark. We would now like to answer the following questions.

I In view of Example 5.10, are there any simple criteria for whena function is integrable?

I How do we calculate the integral of an integrable function?

To answer these questions, we first give a simpler characterisationof an integrable function, in terms of sequences of partitions, ratherthan the inf/sup definition in Definition 5.9.

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Theorem 5.11 For a bounded function f : [a, b]→ R, the followingconditions are equivalent:

(a) f is integrable on [a, b];

(b) there exists a sequence of partitions (Pn) such that

limn→∞

U(Pn, f) = limn→∞

L(Pn, f); (31)

(c) there exists a sequence of partitions (Pn) such that

limn→∞

(U(Pn, f)− L(Pn, f)

)= 0. (32)

If any of these conditions hold then the integral

∫ b

af exists and is

given by the limits in (31).

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Proof. ((a) ⇒ (b)) By Theorem 1.43 (on finding sups and infs)there exists sequences of partitions (Qun), (Q

ln) such that

limn→∞

U(Qun, f) = U(f) = L(f) = limn→∞

L(Qln, f) (33)

(we have U(f) = L(f) here, since f is integrable).

The sequences (Qun), (Qln) need not be the same, but if we set

Rn = Qun ∪Qln, n = 1, 2, . . .

(so that Rn is a refinement of each of Qun and Qln), then

U(f) 6 U(Rn, f) (by (29), with Q = Rn)

6 U(Qun, f) (by (28), since Rn is a refinement of Qun)

→ U(f) (by (33))

so by (33) and the sandwich theorem: U(Rn, f)→ U(f).

Similarly, L(Rn, f)→ L(f).

Now, setting Pn = Rn, n = 1, 2, . . . , and using U(f) = L(f), from(33), gives (31).

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((b)⇒ (c)) This is obvious.

((c)⇒ (a)) Now,

0 6 U(f)− L(f) 6 U(Pn, f)− L(Pn, f) (by (29))

→ 0 (by (32))

=⇒ U(f)− L(f) = 0 (by the sandwich theorem)

=⇒ f is integrable (by Definition 5.9).

Combining these results proves the equivalence of the conditions(a)-(c).

The final result is now easy – see the notes.

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Remarks 5.12

I The implication (b) ⇒ (c) in Theorem 5.11 is obvious, butthe opposite implication (b) ⇐ (c) is not. At first sight (32)does not imply that the separate limits in (31) exist.

I Condition (c) in Theorem 5.11 is a useful test for integrabilityof a function f .

It says that a function f is integrable ⇐⇒ the areas betweenthe upper and lower sets of rectangles in Fig. 32 shrinks tozero, see Fig. 35.

x

f(x)

a b x

f(x)

a b x

f(x)

a b

Figure 35: The areas between the upper and lower sets ofrectangles in Fig. 32.These are shrinking to zero as the partitions get finer.

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Example 5.13 Define f : [0, 5]→ R by

f(x) =

4, if 0 6 x < 3,

2, if x = 3,

1, if 3 < x 6 5,

x

f(x)

0 3 5

1−2−

4−

•

The area under this graph looks like two large rectangles of area4×3 = 12 and 1×2 = 2, together with an ‘infinitely thin’ rectangleover the point x = 3, with zero area.

So, the area ‘ought to be’ 12 + 2 = 14, but we need to do thisproperly (what does ‘infinitely thin’ mean??).

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To deal with the jump at x = 3, for each n > 1, let Pn be thepartition given by

Pn ={0, 3− 1

n, 3 +

1

n, 5}.

Find U(Pn, f), L(Pn, f), and show that f is integrable on [0, 5] and

find

∫ 5

0f .

x

f(x)

0 3 5

1−2−

4−

•

3− 1n 3 + 1

n

Figure 36:

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Solution. We see that

sup

{f(x) : x ∈

[0, 3− 1

n

]}= 4,

sup

{f(x) : x ∈

[3− 1

n, 3 +

1

n

]}= 4,

sup

{f(x) : x ∈

[3 +

1

n, 5

]}= 1,

so that

U(Pn, f) = 4

(3− 1

n

)+ 4

(3 +

1

n−(3− 1

n

))+ 1

(5−

(3 +

1

n

))= 14 +

3

n·

Similarly,

L(Pn, f) = 4

(3− 1

n

)+ 1

(3 +

1

n−(3− 1

n

))+ 1

(5−

(3 +

1

n

))= 14− 3

n·

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Thus

limn→∞

(U(Pn, f)− L(Pn, f)) = limn→∞

6

n= 0.

Hence f is integrable on [0, 5] and∫ 5

0f = lim

n→∞U(Pn, f) = lim

n→∞

(14 +

3

n

)= 14.

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5.4 Properties of the integral

We can now prove some of the standard properties of integrals.

Theorem 5.14 Suppose that f, g : [a, b] → R are both integrableand C is a constant.

(a) f + g is integrable, with

∫ b

a(f + g) =

∫ b

af +

∫ b

ag.

(b) Cf is integrable, with

∫ b

aCf = C

∫ b

af .

(c) fg is integrable, with(∫ b

afg)2

6(∫ b

af2)(∫ b

ag2)

(Cauchy’s inequality).

(d) f(x) 6 g(x) for all x ∈ [a, b] =⇒∫ b

af 6

∫ b

ag.

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(e) if f(x) = k, for some constant k, for all x ∈ [a, b], then∫ b

af = k(b− a).

(f) |f | is integrable, with∣∣∣ ∫ b

af∣∣∣ 6 ∫ b

a|f | .

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Proof. (a) Since f and g are integrable separately, by Theo-rem 5.11 we can choose sequences of partitions, say (Pn) and (Qn),such that

0 6 U(Pn, f)− L(Pn, f)→ 0,

0 6 U(Qn, g)− L(Qn, g)→ 0.

Now, for each n > 1, let Rn = Pn ∪Qn(so Rn is a refinement of both Pn and Qn).

Then, by Lemma 5.7 (and the sandwich theorem, Theorem 1.29),

0 6 U(Rn, f)− L(Rn, f) 6 U(Pn, f)− L(Pn, f)→ 0,

0 6 U(Rn, g)− L(Rn, g) 6 U(Qn, g)− L(Qn, g)→ 0

In other words, the sequences of partitions (Pn) and (Qn) do whatwe want, separately, for each of the function f and g, but thesequence of refined partitions (Rn) does the job for both f and g.

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Next, for any one of the partitions Rn, for any r and any x ∈[xr, xr+1],

f(x) + g(x) > Ir(f) + Ir(g) =⇒ Ir(f + g) > Ir(f) + Ir(g)

f(x) + g(x) 6 Sr(f) + Sr(g) =⇒ Sr(f + g) 6 Sr(f) + Sr(g)

so that

0 6 U(Rn, f + g)− L(Rn, f + g)

6[U(Rn, f)− L(Rn, f)

]+[U(Rn, g)− L(Rn, g)

]→ 0

=⇒ U(Rn, f + g)− L(Rn, f + g)→ 0

(again using Theorem 1.29),

which shows that f + g is integrable (by Theorem 5.11).

The rest of the proof is now straightforward, and similar to this.

There are more details in the notes.

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Next, we show that if f is integrable on [a, b] then it is integrableon any subinterval of [a, b].

Also, we can split the integration interval [a, b] into two subintervals

[a, c] ∪ [c, b], a < c < b,

then integrate separately over the two parts and add the results toget the same answer as integrating over [a, b] (see Fig. 37).

Theorem 5.15 Suppose that f : [a, b]→ R is integrable.

(a) If [c, d] ⊂ [a, b] then f is integrable on [c, d].

(b) If c ∈ (a, b) then ∫ b

af =

∫ c

af +

∫ b

cf.

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x

f(x)

a bc d

(a)

x

f(x)

a bc

(b)

Figure 37: (a):

∫ d

cf (blue).

(b):

∫ c

af (red) +

∫ b

cf (blue) =

∫ b

af.

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Proof. Let (Pn) be a sequence of partitions such that (32) holds.

(a) By adding c, d to each partition Pn (if necessary) we mayassume that c, d ∈ Pn for all n > 1, so that

Qn := Pn ∩ [c, d]

are partitions of [c, d].

Now,

0 6 U(Qn, f)− L(Qn, f) 6 U(Pn, f)− L(Pn, f)→ 0

=⇒ U(Qn, f)− L(Qn, f)→ 0,(34)

so f is integrable on [c, d].

(b) This is similar to part (a) – see the notes for details.

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5.5 Some classes of integrable functions

We now have two characterisations of integrable functions, in termsof partitions (Definition 5.9 and Theorem 5.11).

However, we really don’t want to have to go back to these everytime we want to know if a function is integrable.

The following two theorems show that two broad classes of functionsare integrable, and these results show that most functions that wenormally meet are integrable.

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Theorem 5.16 If f is bounded and monotone on [a, b] then it isintegrable.

Proof. Suppose f is increasing (the decreasing proof is similar).

Then, by definition, for any partition P ,

U(P, f)−L(P, f) =k−1∑r=0

(Sr(f)− Ir(f))δr

6 δmax(P )

k−1∑r=0

(f(xr+1)− f(xr)

)(see Fig. 38)

= δmax

[(f(x1)− f(x0)) + (f(x2)− f(x1)) + . . .

+ (f(xxk)− f(xk−1))]

= δmax(P )(f(b)− f(a)

).

Now, let (Pn) be a sequence of partitions with δmax(Pn)→ 0. Then

0 6 U(Pn, f)− L(Pn, f) < δmax(Pn)(f(b)− f(a))→ 0,

so f is integrable, by Theorem 5.11.321/336

x

f(x)

xr xr+1

Ir(f) = f(xr)

Sr(f) = f(xr+1)

Figure 38: Sr(f) and Ir(f) for increasing f .

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Theorem 5.17 If f is continuous on [a, b] then it is integrable.

Proof. By Theorem 2.44, f is uniformly continuous on [a, b], sofor each integer n > 1, there exists δn > 0 such that

|x− y| < δn =⇒ |f(x)− f(y)| < 1

n.

Now, for any partition Pn with δmax(Pn) < δn, by the BoundednessTheorem, (Theorem 2.38), for each r = 0, . . . , k − 1,

f(yr) =Ir(f), f(zr) = Sr(f), for some yr, zr ∈ [xr, xr+1],

=⇒ Sr(f)− Ir(f) = f(zr)− f(yr) <1

n.

Hence,

0 6 U(Pn, f)− L(Pn, f)

=

k−1∑r=0

(Sr(f)− Ir(f))δr 61

n

k−1∑r=0

δr =1

n(b− a)→ 0

(using (23)), so f is integrable, by Theorem 5.11.

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The proofs of Theorem 5.16 and Theorem 5.17 also prove the fol-lowing result.

Corollary 5.18 Suppose that f is either monotone or continuouson [a, b].

Let (Pn) be a sequence of partitions such that δmax(Pn)→ 0.

Denote these partitions by

Pn = {xn,0, . . . , xn,kn}, n > 1,

and for each n > 1 and r = 0, 1, . . . , kn−1, suppose that

ξn,r denotes an arbitrary point in [xn,r, xn,r+1],

δn,r = xn,r+1 − xn,r.Then ∫ b

af =

∫ b

af(x) dx = lim

n→∞

k−1∑r=0

f(ξn,r)δn,r. (35)

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Remarks.

I Corollary 5.18 holds for general integrable functions. However,this is a bit harder to prove, so we have put it in an Appendixin the notes.

The most important cases are those considered in Corollary 5.18.

I The formula (35) motivates the notation for the integral.

The idea is that, when we take the limit:

I the summation∑k−1

r=0 on the RHS of (35) becomes the

integral sign∫ ba

(this is an elongated ‘S’, for ‘summation’);

I the small gaps δn,r tend to zero and become dx(this is an ‘infinitely small’ gap);

that is,

k−1∑r=0

→∫ b

aand δn,r → dx so we get

∫ b

af(x) dx.

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5.6 The Fundamental Theorem of Calculus

Theorem 5.16 and Theorem 5.17 give us a large class of integrablefunctions, but they don’t tell us how to calculate integrals.

We can now (finally) relate the above idea of an integral to thederivative and show that, in a sense, integration is the reverse ofdifferentiation (in 1-dimension).

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Theorem 5.19 [The Fundamental Theorem of Calculus]

(a) Suppose that g : [a, b] → R is continuous, and define G :[a, b]→ R by

G(x) :=

∫ x

ag(t) dt, x ∈ [a, b].

Then G is differentiable on (a, b), and G′(x) = g(x) for allx ∈ (a, b).

(b) Suppose that I is an open interval and F : I → R is differ-entiable on I.

Suppose also that [a, b] ⊂ I and F ′ is continuous on [a, b].

Then ∫ x

aF ′(t) dt = F (x)− F (a), x ∈ [a, b].

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Remarks.

(a) Theorem 5.19 (a) says that if you integrate a continuous func-tion g then differentiate, you get back to where you startedfrom.

(b) Theorem 5.19 (b) says that if you differentiate a function F(and the derivative F ′ is continuous)and then integrate, you get back to where you started from,

apart from the ‘constant of integration’ −F (a).

Or you can rewrite the result as

F (x) = F (a) +

∫ x

aF ′(t) dt,

which you can regard as saying:

starting with the value F (a) and integrating F ′ from a to xgives you F (x).

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Proof. (a) Let x ∈ (a, b) and let ε > 0.

Since g is continuous at x, there exists δ > 0 such that

|t− x| < δ =⇒ |g(t)− g(x)| < ε. (36)

Then, if 0 < h < δ,∣∣∣G(x+ h)−G(x)h

− g(x)∣∣∣ = ∣∣∣1

h

(∫ x+h

0g(t) dt−

∫ x

0g(t) dt

)− g(x)

∣∣∣=∣∣∣1h

∫ x+h

xg(t) dt− g(x)

∣∣∣ (by Theorem 5.15)

=∣∣∣1h

∫ x+h

x(g(t)− g(x)) dt

∣∣∣ (by Theorem 5.14 (e))

61

h

∫ x+h

xε dt (by (36) and Theorem 5.14 (f))

= ε (by Theorem 5.14) (e))

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so that

limh→0+

G(x+ h)−G(x)h

= g(x) .

We can also prove the same result for the left-sided limit h→ 0−

(using the definition of the integral when the lower limit of integra-tion is larger than the upper limit).

Hence, G is differentiable at x and G′(x) = g(x).

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(b) Define

H(x) =

∫ x

aF ′(t) dt, x ∈ [a, b].

Then by part (a), H is differentiable on (a, b), and H ′(x) = F ′(x),x ∈ (a, b).

Hence, by Theorem 3.21 (c), H − F is constant on [a, b], so

H(a)− F (a) = H(x)− F (x).Thus

F (x)− F (a) = H(x)−H(a) =

∫ x

aF ′(t) dt .

See Tutorial 9 for some simple integrations using both Riemann sumsand the fundamental theorem of calculus – the latter is much easier!

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5.7 Finally, some general remarks about integration.

We have defined the integral as the area under a curve, but actuallywhat we have done is added up an infinite number of infinitely smallcontributions to obtain some limiting quantity.

Such a process occurs in many other contexts.

For example, consider a metal rod with varying density ρ(x)(the density is the mass per unit length)lying along the x-axis, between 0 and 1, see Fig. 39.

What is the total mass of the rod?

x0 1xr xr+1

mr = ρ(xr)δr

Figure 39: Variable density rod.

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Let’s partition the rod(i.e., choose a partition P of the interval [0, 1]),and look at the small interval on the rod in Fig. 39.

The mass in this interval is, approximately, mr = ρ(xr)δr.

To get an approximate value for the total mass we add up the massesin each of the subintervals to get

approximate mass of the rod =∑r

ρ(xr)δr.

This won’t be exact since the density is varying in the subintervals.

To get the exact answer we then let the lengths of the subintervalstend to zero (i.e., δmax(P )→ 0) and take the limiting value of theapproximate masses.

In other words (by all the above setup)

mass of the rod =

∫ 1

0ρ.

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That is, we obtain the total mass of the rod, with variable density,by integrating the density function.

There was nothing special about mass/density here, and there wasnothing special about the fact that the rod lay along a straight linealong the x-axis.

Integrals arise whenever we want to know how much there is of‘something’ in some ‘region’.

The region might be:

I a curve (e.g., the mass of a curved, variable density wire);

I a surface (e.g., the mass of a variable density shell);

I or a solid region (e.g., the mass of a lump of metal).

These would lead to integrals along curves, or surfaces, or over 3D-regions — you have seen some of these types integrals in the firstsemester Multivariable Calculus course, and this is why there are somany types of integrals.

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Finally, we note that in 1D we have the fundamental theorem ofcalculus, which says that integration is the reverse of differentiation.

However, this is just a ‘fluke’ of 1D-integration, and this is not truefor the more general types of integrals mentioned above.

These have nothing in particular to do with differentiation.

So, we are lucky to have this result in 1D(it is what we usually use to calculate integrals, even (ultimately)multi-dimensional ones),but it is just luck — it is not the definition of the integral.

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THE END

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