Limits and Continuity - Hong Kong University of Science ...majhu/Math1023/Section 1.1.pdf · 4...

Chapter 1Limits and Continuity

1.1 Limits of Sequences

1.1.1 The Concept of Limit and Its Properties

A sequence {xn} is an ordered infinite list

x1, x2, . . . , xn, . . . .

The n-th term of the sequence is xn, and n is the index of the term.In this course, we will always assume that the terms xn are real numbers. Here are

some examples of sequences of real numbers

xn = n ! 1,2,3, . . . , n, . . . ;yn = 2 ! 2,2,2, . . . ,2, . . . ;zn = 1

n! 1, 1

2, . . . ,

1

n, . . . ;

un = ("1)n ! 1,"1,1, . . . , ("1)n, . . . ;vn = sinn ! sin 1, sin 2, sin 3, . . . , sinn, . . . .

Note that the index does not have to start from 1. For example, the sequence {un}actually starts at n = 0 (or any even integer). Moreover, a sequence does not have to begiven by a formula. For example, the decimal expansions of ! give a sequence {wn}:

wn ! 3,3.1,3.14,3.141,3.1415,3.14159,3.141592, . . . .If n is the number of digits after the decimal point in wn, then the sequence {wn} startsfrom n = 0.

We can examine the behaviors of these sequences as n becomes larger and larger.Apparently, in this case, the term xn gets larger. In fact, xn can be larger than any fixednumber if the index n is su!ciently large. On the other hand, yn remains constant, andzn can be smaller than any fixed positive number if n is su!ciently large. Our intuitiontells that yn approaches 2 and zn approaches 0. Moreover, the terms un and vn jumparound and do not approach any number. Finally, wn gets closer and closer to !. Infact, wn is equal to ! up to the n-th decimal point.

3

4 Chapter 1. Limits and Continuity

n

xn

yn

zn

unvn

wn

Figure 1.1: Examples of several sequences.

Therefore, as n gets larger, the sequences {yn},{zn},{wn} approach finite numbers2, 0, and !, while the sequences {xn},{un},{vn} do not approach any specific finitenumbers. These sequences are illustrated in Figure 1.1.

Definition 1.1.1 (Non-rigorous) If xn approaches a finite number l when n getslarger and larger (or goes to infinity), then we say that the sequence {xn} convergesto the limit l, or simply that lim

n!!xn exists, and write

limn!!xn = l.

A sequence diverges if it does not converge to any specific finite number when n goesto infinity.

Since the limit describes the behavior when n gets very large, the modification offinitely many terms in a sequence changes neither the convergence nor the limit value.

Proposition 1.1.2 If limn!!xn exists, and if {yn} is obtained from {xn} by adding,

deleting, or changing finitely many terms, then limn!!yn exists, and

limn!!yn = lim

n!!xn.

Example 1.1.1 The sequence ! 1"n + 2# is obtained from ! 1"

n# by deleting the first

two terms. By limn!!

1"n= 0 and Proposition 1.1.2, we get lim

n!!1"n= lim

n!!1"n + 2 = 0.

In general, the same reason tells us that if limn!!xn exists, then lim

n!!xn+k = limn!!xn for

any integer k.

Intuitively, we know that if x is close to 3 and y is close to 5, then the arithmeticcombinations x+y and xy are close to 3+5 = 8 and 3 #5 = 15. These lead to the followingproperties.

1.1. Limits of Sequences 5

Proposition 1.1.3 (Arithmetic Rule) Suppose limn!!xn = l and lim

n!!yn = k. Thenthe arithmetic combinations of the sequences converge, and

limn!!(xn + yn) = l + k, lim

n!!(cxn) = c l, limn!!(xn yn) = k l, lim

n!!xnyn= l

k,

where c is a constant and k ! 0 in the last equality.

The proposition tells us that limn!!(xn + yn) = lim

n!!xn + limn!!yn under the condition

that both limn!!xn and lim

n!!yn exist. However, if this condition fails, then in general we

cannot conclude that limn!!(xn + yn) = lim

n!!xn + limn!!yn. For instance, if xn = ("1)n and

yn = ("1)n+1, then limn!!(xn +yn) = 0, and both lim

n!!xn and limn!!yn diverge. We therefore

havelimn!!(xn + yn) ! lim

n!!xn + limn!!yn.

Example 1.1.2 By limn!! c = c, lim

n!!1

n= 0, and the arithmetic rule, we have

limn!!

2n2 + nn2 " n + 1 = lim

n!!2 + 1

n

1 " 1

n+ 1

n2

= limn!!$2 + 1

n%

limn!!$1 " 1

n+ 1

n2%

= 2 + limn!!

1

n

1 " limn!!

1

n+ lim

n!!1

n# limn!!

1

n

= 2.

The following property reflects the intuition that if x and z are close to 3, thenanything between x and z should also be close to 3.

Proposition 1.1.4 (Sandwich Rule) If xn " yn " zn and limn!!xn = lim

n!! zn = l,

then limn!!yn = l.

Example 1.1.3 We have

" 1"n" ("1)n"

n" 1"

n, " 1

n" sinn

n" 1

n.

Moreover, we have limn!!

1"n= lim

n!!1

n= 0. By the arithmetic rule, we know lim

n!!&" 1"n' =

limn!!$" 1n% = 0. Therefore we get

limn!!

("1)n"n= lim

n!!sinn

n= 0.


The argument for the first limit can be generalized as follows: If limn!! (xn( = 0, then

limn!!xn = 0. See Exercise 1.1.6.

Example 1.1.4 Since

0 <"n + 2 ""n = ("n + 2 ""n)("n + 2 +"n)"

n + 2 +"n = 2"n + 2 +"n <

2"n,

by the sandwich rule and

limn!!

2"n= 2 # lim

n!!1"n= 0,

we get

limn!!(

"n + 2 ""n) = 0.

Example 1.1.5 We show that

limn!!

n"a = 1, for a > 0.

First assume a # 1. Put xn = n"a " 1. Then xn # 0, and

a = (1 + xn)n = 1 + nxn + n(n " 1)2

x2n +$+ xnn # nxn.This gives us

0 " xn " a

n.

By the sandwich rule and limn!!

a

n= a lim

n!!1

n= 0, we get lim

n!!xn = 0. Hence, by the

arithmetic rule,

limn!!

n"a = lim

n!!(1 + xn) = 1 + limn!!xn = 1.

Next, in the case 0 < a < 1, we put b = 1

a. Then b > 1. By the above discussion, and

the arithmetic rule, we have

limn!!

n"a = lim

n!!1n"b= 1

limn!!

n"b= 1.


Example 1.1.6 The idea of Example 1.1.5 can also be used to show that

limn!!

n"n = 1.

Let xn = n"n " 1. Then xn # 0, and

n = (1 + xn)n = 1 + nxn + n(n " 1)2

x2n +$+ xnn # n(n " 1)2

x2n.

This implies

0 " xn ""2"

n " 1 .By the sandwich rule and Example 1.1.1, we get lim

n!!xn = 0. It follows thatlimn!!

n"n = lim

n!!(1 + xn) = 1.

Example 1.1.7 The following sequences of “n-th root type” can be compared withthe sequences in Examples 1.1.5 and 1.1.6.

1 < n"n + 1 < n

"2n = n

"2 n"n,

1 < n 1n+1 < n

"n,

1 < (n2 " n) nn2!1 < (n2) n

n2!1 < (n2) nn2!2 = ( n

"n)4.

By Examples 1.1.6, 1.1.7, and the arithmetic rule, the sequences on the right hand sideall converge to 1. Then by the sandwich rule, we have

limn!!

n"n + 1 = lim

n!!n1

n+1 = limn!!(n2 " n) n

n2!1 = 1.


3 = n"3n < n

"2n + 3n < n

"2 # 3n = 3 n

"2.

By Example 1.1.5, the limit of the right hand side is 3. By the sandwich rule, we getlimn!!

n"2n + 3n = 3.

Example 1.1.9 We show

limn!!an = 0, for (a( < 1.


First assume 0 < a < 1 and write a = 1

1 + b . Then b > 0 and

0 < an = 1

(1 + b)n =1

1 + nb + n(n " 1)2

b2 +$+ bn" 1

nb.

By limn!!

1

nb= 0, and the sandwich rule, we get lim

n!!an = 0.Next, if "1 < a < 0, then 0 < (a( < 1 and lim

n!! (an( = limn!! (a(n = 0. By the remark in

Example 1.1.3, we get limn!!an = 0.


limn!!

an

n!= 0, for any a.

Here we discuss the special case with a = 5 (the general case being similar).For n > 5, we have

0 < 5n

n!= 5 # 5 # 5 # 5 # 51 # 2 # 3 # 4 # 5 #

5

6# 57$ 5

n" 5 # 5 # 5 # 5 # 51 # 2 # 3 # 4 # 5 #

5

n= 56

5!

1

n.

By limn!!

56

5!

1

n= 56

5!limn!!

1

n= 0, and the sandwich rule, we get lim

n!!5n

n!= 0.

The following property reflects the intuition that larger sequence should have largerlimit.

Proposition 1.1.5 (Order Rule) Suppose limn!!xn and lim

n!!yn exist.

1. If xn # yn for su!ciently large n, then limn!!xn # lim

n!!yn.

2. If limn!!xn > lim

n!!yn, then xn > yn for su!ciently large n.

Note that a property fails for only finitely many n means that it holds for su!cientlylarge n.

By taking yn = k, we get the following properties for a convergent sequence {xn}.1. If xn # k for su!ciently large n, then lim

n!!xn # k.2. If lim

n!!xn > k, then xn > k for su!ciently large n.

Similar statements can be made by taking xn to be constant.

Example 1.1.11 We must pay special attention to the di"erence between inequalityand strict inequality in the two statements of Proposition 1.1.5. For example, although

xn = 1

n> yn = 1

n2, we have lim

n!!xn = limn!!yn.


Example 1.1.12 By limn!!

2n2 + nn2 " n + 1 = 2, we know 1 < 2n2 + n

n2 " n + 1 < 3 for su!ciently

large n, by Statement 2 in Proposition 1.1.5. This implies

1 < n

)2n2 + nn2 " n + 1 < n

"3.

By limn!!

n"3 = 1 and the sandwich rule, we get lim

n!!n

)2n2 + nn2 " n + 1 = 1.

In general, the argument tells us that, if limn!!xn = l > 0, then lim

n!! n"xn = 1.

Example 1.1.13 The sequence xn = 3n(n!)2(2n)! satisfies

limn!!

xnxn"1 = lim

n!!3n2

2n(2n " 1) =3

4= 0.75.

Therefore by the order rule, we havexnxn"1 < 0.8 for su!ciently large n, say for n > N for

a fixed big natural number N (in fact, N = 8 is enough). Then for n > N , we have

0 < xn = xnxn"1

xn"1xn"2$

xN+1xN

xN < 0.8n"NxN = C # 0.8n, C = 0.8"NxN .

By Example 1.1.9, we have limn!!0.8n = 0. Since C is a constant, by the sandwich rule,

we get limn!!xn = 0.

In general, the argument shows that, if limn!!

xnxn"1 = l and (l( < 1, then lim

n!!xn = 0.

Exercises

1.1.1 Find the limits of the sequences (assuming we already know limn!!

1"n= lim

n!!1

n= 0):

1.1"

2n " 1 ;

2.

"2n + 1n " 1 ;

3.1

n23

;

4.1

3"n2 " n + 2 ;

5.1

3"n2 + an + b ;

6.sinn!"

n;

7.("1)n(n + 1)n2 + ("1)n+1 ;

8.55(2n + 1)2 " 1010

10n2 " 5 ;


9.1010n

n2 " 10 ;10.

n2

n!;

11.n2 + 3n + 5

n!;

12.n2 + n # 3n + 5!

n!;

13.n2 + n! + (n " 1)!3n " n! + (n " 1)! ;

14."n + sinn ""n " cosn;

15."n + a ""n + b;

16."n + 1 +"n + 2 " 2"n + 3;

17. 3"n + 1 " 3

"n + 2;

18.*

n

n + 2 ;19. 3

*n

n + 2 ;20.

*n + an + b ;

21.

)n2 + n + 1n2 " n + 2 ;

22.

)a0 n

2 + a1 n + a2b0 n2 + b1 n + b2 , a0, b0 > 0;

23.

"a0 n2 + a1 n + a2

b0 n + b1 , a0 > 0, b0 ! 0;24.

"n("n + 1 ""n);

25."n + a("n + b ""n + c);

26."n2 + 2n + 3 ""n2 " n + 1;

27."n2 + a1 n + a2 ""n2 + b1 n + b2;

28. n("n2 + n + 3 ""n2 + n " 1);29.

3"n2( 3"n + 1 " 3

"n);

30. $n + an + b %

!

;

31. &n2 + a1 n + a2n2 + b1 n + b2 '

!

;

32. &a0 n2 + a1 n + a2b0 n2 + b1 n + b2 '

p

, a0, b0 > 0;33. (2n + 1) 1

n2!1 ;

34. (n + sinn) 1n+cosn ;

35. n"7 # 2n;

36. n"7 + 2n;

37. n"7n + 2n;

38.n"7n+1 + 2n;

39.n"7n + 22n+1;

40.n"7n # 22n+1;

41. n"n # 2n;

42. n"n + 2n;

43. n+(n + 1)2n;

44.n

)n # 23n + 4n+1 + 32n"1

n.

1.1.2 First show thatlimn!!nan = 0, for (a( < 1.

Then find the limits of the following sequences:

1.n + 12n

;

2.n2

2n;

3. n99 # 0.99n;

4.n + 2n3n

;

5.n # 3n(1 + 2n)2 ;

6.5n " n # 6n+132n"1 " 23n+1 ;

7. n3an, (a( < 1;8. npan, (a( < 1;9.

np

an, (a( > 1.

1.1.3 Use limn!!

1

np= 0 for p > 0 to find the following limits:

1. 3"n + 1 " 3

"n;

2.3"n2 + 1 " 3

"n2 " 1;

3.

"n + 1

3"n2 " n + 2 ;

4.

+"n + sinn"

n " cosn .


1.1.4 Explain that limn!!xn = L if and only if lim

n!!(xn " L) = 0. Use this observation to derive

limn!!

)n + 2n= 1 from Example 1.1.4.

1.1.5 If limn!!(xn yn) = 0, can you conclude that either lim

n!!xn = 0 or limn!!yn = 0?

1.1.6 Use the sandwich rule to prove that limn!! (xn( = 0 implies lim

n!!xn = 0. Is the converse true?1.1.7 Prove that if xn converges to a positive number, then lim

n!! n"xn = 1.

1.1.8 Prove that if , xn

xn"1 , " c for a constant c < 1, then limn!!xn = 0.

1.1.9 Prove that if limn!!

xn

xn"1 = l and (l( < 1, then limn!!xn = 0. Use this and Example 1.1.9 to

derive the limit in Example 1.1.10 for any a. Moreover, what happens when (l( # 1?1.1.10 Prove that if lim

n!!n+(xn( = l and (l( < 1, then lim

n!!xn = 0. What happens if (l( # 1?1.1.2 Bounded Sequences

A sequence {xn} is bounded above if xn " B for a constant B (called an upper bound)and all n. Similarly, it is bounded below if xn # B for a constant B (called a lower bound)and all n. If a sequence is bounded above and below, then we say it is bounded.

The order rule implies the following.

Proposition 1.1.6 Any convergent sequence is bounded.

For example, the sequences {n} and !n2 + ("1)nn + 1 # diverge because they are not

bounded above. On the other hand, a bounded sequence does not have to converge, asshown by the sequence 1,"1,1,"1, . . . . Therefore the converse of Proposition 1.1.6 doesnot hold in general.

The converse of Proposition 1.1.6 does hold under some additional assumption.A sequence {xn} is increasing if

x1 " x2 " x3 " $ " xn " xn+1 " $ .

It is strictly increasing if

x1 < x2 < x3 < $ < xn < xn+1 < $ .

The concepts of decreasing and strictly decreasing can be similarly defined. Moreover, asequence is monotonic if it is either increasing or decreasing.

The sequences - 1n., - 1

2n., { n"2} are (strictly) decreasing. The sequences -" 1

n.,

{n} are increasing.

Proposition 1.1.7 Any bounded monotonic sequence converges.

An increasing sequence {xn} is bounded if it is bounded above, because the first termof the sequence is already the lower bound. Similar remark can be made for a decreasingsequence.


Example 1.1.14 The sequence

"2,+2 +"2,

/2 ++2 +"2, . . .

is recursively given byx1 ="2, xn+1 ="2 + xn.

We claim that the sequence {xn} is increasing and bounded above by 2.

We have x1 ="2 < 2 and x2 =+2 +"2 > x1. Assume xn < 2 and xn+1 > xn. Thenxn+1 ="2 + xn <"2 + 2 = 2, xn+2 ="2 + xn+1 >"2 + xn = xn+1.

The claim is therefore proved by induction. By Proposition 1.1.7, the sequence converges.Let l be the limit. Then by taking the limits on both sides of the equality x2n+1 = 2 + xnand applying the arithmetic rule, we get l2 = 2 + l. The solution is l = 2 or "1. Sincexn > 0, by the order rule we must have l # 0. Therefore we conclude that lim

n!!xn = 2.

Figure 1.2 schematically demonstrates the limiting process of a recursively definedsequence.

x

y

x1

f(x1)x2

f(x2)

x3

f(x3)

x4 l

y = f(x)y = x

Figure 1.2: Limit of a recursively defined sequence xn+1 = f(xn).

Example 1.1.15 We give another argument that limn!!an = 0 for 0 < a < 1. See

Example 1.1.9.Since 0 < a < 1, the sequence {an} is decreasing and satisfies 0 < an < 1. Hence, by

Proposition 1.1.7, the sequence converges to a limit l. By the remark in Example 1.1.1,we also have lim

n!!an"1 = l. Then by the arithmetic rule,

l = limn!!an = lim

n!!a # an"1 = a limn!!an"1 = al.

Since a < 1, we get l = 0.


Example 1.1.16 We give another argument that the limit in Example 1.1.13 con-

verges to 0. By limn!!

xnxn"1 = 0.75 < 1 and the order rule, we have

xnxn"1 < 1 for su!ciently

large n. Since all terms are positive, we have xn < xn"1 for su!ciently large n. Thereforeafter finitely many terms, the sequence is decreasing. Moreover, 0 is a lower bound ofthe sequence, so that the sequence converges.

Let limn!!xn = l. Then we also have lim

n!!xn"1 = l. If l ! 0, thenlimn!!

xnxn"1 =

limn!!xn

limn!!xn"1 =

l

l= 1.

But the limit is actually 0.75. The contradiction shows that l = 0.

Example 1.1.17 We show that the sequence -$1 + 1

n%n. converges as n ! %. By

comparing the binomial expansions of two consecutive terms

$1 + 1

n%n = 1 + n 1

n+ n(n " 1)

2!

1

n2+$+ n(n " 1)$3 # 2 # 1

n!

1

nn

= 1 + 1

1!+ 1

2!$1 " 1

n% +$

+ 1

n!$1 " 1

n%$1 " 2

n%$$1 " n " 1

n% ,

$1 + 1

n + 1%n+1 = 1 + 1

1!+ 1

2!$1 " 1

n + 1% +$+ 1

n!$1 " 1

n + 1%$1 "2

n + 1%$$1 "n " 1n + 1%

+ 1

(n + 1)! $1 "1

n + 1%$1 "2

n + 1%$$1 "n

n + 1% ,we see that the sequence is increasing:

$1 + 1

n%n " $1 + 1

n + 1%n+1

.

It remains to show that the sequence is bounded above. By the expansion above, wehave

$1 + 1

n%n < 1 + 1

1!+ 1

2!+$+ 1

n!

< 1 + 1 + 1

1 # 2 +1

2 # 3 +$+1

(n " 1)n= 1 + 1 + $1 " 1

2% + $1

2" 1

3% +$+ $ 1

n " 1 "1

n%

= 1 + 1 + 1 " 1

n< 3.


Hence, by Proposition 1.1.7, the sequence converges. The limit is denoted by e

limn!!$1 + 1

n%n = e.

It is clear that 1 < e " 3 by the order rule. As a matter of fact, it can be shown thate = 2.7182818... is an irrational number.

Example 1.1.18 Consider the sequence {xn}:xn = 1

12+ 1

22+ 1

32+$+ 1

n2.

This sequence is clearly increasing. Moreover, similar to the computation in Exam-ple 1.1.17, we have

xn " 1 + 1

1 # 2 +1

2 # 3 +$+1

(n " 1)n = 2 "1

n< 2.

Hence the sequence is bounded, and converges by Proposition 1.1.7.The limit of the sequence is the sum of the infinite series

!!n=1

1

n2= 1 + 1

22+ 1

32+$+ 1

n2+$ ,

and is actually equal to!2

6(this last fact will not be proved here).

Without the condition of monotonicity, a bounded sequence may not converge. Thecounterexamples, such as 1,"1,1,"1, . . . , suggest that the divergence may be due to thefact that parts of the sequence may still converge, but di"erent parts may converge todi"erent limits. Therefore there is no single value for the sequence to converge to.

To describe the phenomenon that part of a sequence may converge, we introduce thefollowing concept. A subsequence of a sequence {xn} is obtained by choosing infinitelymany terms and still arranging them in the order as in {xn}. The subsequence is then

xn1 , xn2 , . . . , xnk , . . . ,

or {xnk} in short, with the indices satisfying

n1 < n2 < $ < nk < $ .

If the sequence {xn} starts from n = 1, then n1 # 1, which further implies nk # k for allk.


The following are some subsequences of -xn = 1

n., corresponding to the choices nk =

2k, nk = 2k " 1, nk = 2k, and nk = k!, respectively:x2k = 1

2k! 12,1

4,1

6,$ ,

1

2k,$ ;

x2k"1 = 1

2k " 1 !1

1,1

3,1

5,$ ,

1

2k " 1 ,$ ;

x2k = 1

2k! 12,1

4,1

8,$ ,

1

2k,$ ;

xk! = 1

k!! 11,1

2,1

6,$ ,

1

k!,$ .

If something is close to l, then any part of it is also close to l. This leads to thefollowing result.

Proposition 1.1.8 If a sequence converges to l, then every subsequence convergesto l.

Example 1.1.19 By limn!!

2n2 + nn2 " n + 1 = 2, we also know

limn!!

2(n2 " 1)2 + (n2 " 1)(n2 " 1)2 " (n2 " 1) + 1 = lim

n!!2n4 " n2 + 1n4 " 3n2 + 3 = 2,

and

limn!!

2(n!)2 + n!(n!)2 " n! + 1 = 2.

Example 1.1.20 Consider the sequence xn = ("1)n. The subsequence x2k = 1 con-verges to 1, and the subsequence x2k"1 = "1 converges to "1. Since the two subsequenceshave di"erent limits, Proposition 1.1.8 tells us that xn diverges.

Moreover, the expectation that a bounded sequence is made up of a number ofconvergent subsequences (with perhaps di"erent limits) is reflected below.

Theorem 1.1.9 (Bolzano-Weierstrass) Every bounded sequence has a conver-gent subsequence.

The Bolzano-Weierstrass Theorem is a deep result that touches the essential di"er-ence between real numbers and rational numbers.


Example 1.1.21 Let us list all finite decimal expressions in (0,1) as a sequence {xn}:xn ! 0.1,0.2, . . . ,0.9, 0.01,0.02, . . . ,0.99, 0.001,0.002, . . . ,0.999, . . .

The number 0.318309$ is the limit of the sequence

0.3,0.31,0.318,0.3183,0.31830,0.318309, . . . ,

which is clearly a subsequence of {xn}. It is easy to see that any number in [0,1] is thelimit of a convergent subsequence of {xn}.

Exercises

1.1.11 Determine the convergence of the following sequences:

1.("1)nn2

n3 " 1 ;

2.("1)nn2

n2 " 1 ;

3.("1)nn2

n " 1 ;

4.cos2 n"

n;

5. n("1)n ;6.

2n"n!;

7.n!

2n;

8.2n + cosn2

n + ("1)n"n + sinn ;9.

1

2,2

1,2

3,3

2,3

4,4

3, . . . ,

n

n + 1 ,n + 1n

, . . . ;

10.1 # 3 # 5$(2n " 1)2 # 4 # 6$(2n) ;

11.1 # 3 # 5$(2n " 1)

n!;

12.

0111121 +011121

2+01121

3+$+

)1

n;

13.1

13+ 1

23+ 1

33+$+ 1

n3.

1.1.12 Determine the convergence of the following recursively defined sequences.

1. xn+1 = x2n + 12

, x1 > 0;2. xn+1 = 2 " 1

xn, x1 > 0;

3. xn+1 ="a + xn, a > 0, x1 > 0;4. xn+1 = 1

2$xn + a

xn%, a > 0.

1.1.13 Explain the continued fraction expansion

"2 = 1 + 1

2 + 1

2 + 1

2 +$.

What if 2 on the right hand side is changed to some other positive number?

1.1.14 For any a, b > 0, define a sequence by

x1 = a, x2 = b, xn = xn"1 + xn"22

.

Prove that the sequence converges.


1.1.15 The arithmetic and the geometric means of two positive numbers a, b area + b2

and"ab.

By repeating the process, we get two sequences defined by

x1 = a + b2

, y1 ="ab, xn+1 = xn + yn2

, yn+1 ="xnyn.

Prove that xn # xn+1 # yn+1 # yn, and the two sequences converge to the same limit.

1.1.16 The Fibonacci sequence

1,1,2,3,5,8,13,21,34, . . .

is defined by x0 = x1 = 1 and xn+1 = xn + xn"1. Consider the sequence yn = xn+1xn

.

1. Find the relation between yn+1 and yn.

2. Assume the sequence {yn} converges, find the limit l.

3. Use the relation between yn+2 and yn to prove that l is an upper bound of y2k and a lowerbound of y2k+1.

4. Prove that the subsequence {y2k} is increasing and the subsequence {y2k+1} is decreasing.5. Prove that the sequence {yn} converges to l.

1.1.17 Let xn = $1 + 1

n%n+1.

1. Use induction to prove (1 + x)n # 1 + nx for x > "1 and any natural number n.

2. Use the first part to provexn"1xn< 1. This shows that {xn} is decreasing.

3. Prove that limn!!xn = e.

4. Prove that -$1 " 1

n%n. is increasing and converges to e"1.

1.1.18 Prove that for n > k, we have

$1 + 1

n%n # 1 + 1

1!+ 1

2!$1 " 1

n% +$+ 1

k!$1 " 1

n%$1 " 2

n%$$1 " k

n% .

Then use Proposition 1.1.5 to show that

e # 1 + 1

1!+ 1

2!+$+ 1

k!# $1 + 1

k%k .

Finally, prove

limn!!$1 + 1

1!+ 1

2!+$+ 1

n!% = e.

1.1.3 Rigorous Definition

The meaning of the phrases “approaching l” and “going to infinity” in the intuitivedefinition of limit is rather ambiguous. Before we introduce the rigorous definition, let


us study the limit limn!!

1

n= 0 in more detail. When we say that xn = 1

napproaches l = 0

as n goes to infinity, we mean an infinite collection of facts

n > 1 "# (xn " l( < 1,n > 10 "# (xn " l( < 0.1,n > 100 "# (xn " l( < 0.01,&

n > 1000000 "# (xn " l( < 0.000001,&Here, the symbol “"# ” stands for “implies”.

In everyday life, whether certain quantity is considered large or small depends on thecontext. For example, the population of millions of people (n > 100000000) is consideredas large for a city but rather small as a country. On the other hand, the diameter withinone millimeter (d < 1) is considered thin for an electric wire. But the hair is consideredthin only if the diameter is less than 0.05 millimeter (d < 0.05).

Thus large or small is meaningful only when compared with some reference quantity.

For the limit limn!!

1

n= 0, we say that, if n is in the hundreds, then (xn " l( is within the

hundredth, and if n is in the millions, then (xn " l( is within the millionth, etc. Of coursefor a di"erent sequence, the exact relation between the largeness of n and the smallness

of (xn " l( may be di"erent. For example, for xn = 2n

n!and l = 0, we have

n > 10 "# (xn " l( < 0.0003,n > 20 "# (xn " l( < 0.0000000000005,&

But the key observation here is that a limit is an infinite collection of statements of theform “if n is larger than certain large number N , then (xn " l( is smaller than certainsmall number "”. In practice, we cannot verify all such statements one by one. Even ifwe have verified the truth of the first one million statements, there is no guarantee thatthe one million and the first statement is true. To mathematically establish the truth ofall such statements, we have to formulate one statement that includes the considerationfor all N and ".

Which one statement do we need to establish? We note that n > N is the causeof the e"ect (xn " l( < ", and the precise logical relation between the cause and thee"ect varies from limit to limit. For some limit, n in thousands already guarantees that(xn " l( < 0.000001. For some other limit, n has to be in the billions in order to guaranteethat (xn " l( < 0.000001. But the key observation here is that no matter how small " is,the “target” (xn " l( < " can always be guaranteed for su!ciently large n. So although(xn " l( < 0.000001 may not be satisfied for n in the million range, it will probably besatisfied for n in the billions range. In fact, for any given range " > 0, we can always findan integer N such that xn is close to the limit l within the range " whenever n > N . Inother words,

n > N "# (xn " l( < ".For xn = 1

n, we can take N = 1

".


Definition 1.1.10 (Rigorous) A sequence {xn} converges to a finite number l iffor any " > 0, there is N , such that n > N implies (xn " l( < ".

In case N is a natural number (which can always be arranged if needed), the conditionmeans that all the terms xN+1, xN+2, xN+3, . . . lie in the shaded area in Figure 1.3.

l

l + "l " "

Nn

x1

x2

x3

x4

xN+1

xN+2xN+3

xn

Figure 1.3: n > N implies (xn " l( < ".

Example 1.1.22 We rigorously prove

limn!!

1

np= 0, for p > 0.

For any " > 0, choose N = 1

"1p

. Then

n > N "# , 1np" 0, = 1

np< 1

Np= ".

Since we will carry out rigorous arguments, we need to be more specific about theirlogical foundations . We need to assume the basic knowledge of real numbers. Thisinclude

1. The basic arithmetic operations: x + y, x " y, xy, xy.

2. Exponentiation: for x > 0 and y any real number, we know how to “raise x to they-th power” to obtain xy.

3. The order relations: for two real numbers x, y, we know the meaning of x < y andx " y, and the properties associated to such inequalities. For example, we assume that

we already know x > y > 0 implies1

x< 1

yand xp > yp for p > 0. These properties have

already been used in the argument in the example above.On the other hand, at this point we do not assume that we know anything about

the logarithm. The logarithm and its properties will be rigorously established in a latersection.

Example 1.1.23 To rigorously prove the claimed conclusion in Example 1.1.4, weestimate the di"erence between the n-th term of the sequence and the expected limit:

("n + 2 ""n " 0( = (n + 2) " n"n + 2 +"n <

2"n.


Therefore for any " > 0, it su!ces to have2"n< ", or n > 4

"2. In other words, we should

choose N = 4

"2.

The discussion above is an analysis of the problem, which you should probably writeon your scratch paper. The following is the formal rigorous argument that you are

supposed to present: For any " > 0, choose N = 4

"2. Then

n > N "# ("n + 2 ""n " 0( = 2"n + 2 +"n <

2"n< 2"

N= ".

Example 1.1.24 Consider the sequencen2 " 1n2 + 1. For any " > 0, we have

n > N =)

2

"" 1 "# 3n2 " 1

n2 + 1 " 13 =2

n2 + 1 <2

N2 + 1 =2

$2"" 1% + 1 = ".

Therefore the sequence converges to 1.

How did we choose N =)

2

"" 1? First note that we want to achieve

3n2 " 1n2 + 1 " 13 < ".

This is equivalent to the inequality

n >)

2

"" 1.

Therefore, choosing N =)

2

"" 1 does the trick.

In Examples 1.1.24, the formula for N is obtained by solving 3n2 " 1n2 + 1 " 13 < " in an

exact way. However, this may not be so easy in general. For example, for the limit inExample 1.1.2, we need to solve

3 2n2 + nn2 " n + 1 " 23 =

3n " 2n2 " n + 1 < ".

While the exact solution can be found, the formula for N is rather complicated. Formore complicated example, it may not even be possible to find the formula for the exactsolution.


We note that finding the exact solution of (xn" l( < " is the same as finding N = N("),such that

n > N $# (xn " l( < ".However, in order to rigorously prove that the limit is what we claim to be, we only needto establish one direction of the implication, namely“ "# ”. In other words, there isno need to find the “exact” N . All we need to do is to find one (any) N such that theimplication “"# ” is true.

Furthermore, in the statement “n > N "# (xn " l( < "”, the inequality (xn " l( < "holds under the condition n > N . In other words, (xn " l( < " is a conditional inequality.We have seen many “un-conditional” inequalities before. For example, 1 " 1 + x2 and(x( " "x2 + 1 are un-conditional inequalities. Hence, proving lim

n!!xn = l is to find N so

that the inequality (xn " l( < " holds under the the condition n > N .These observations lead to a so-called “loose-and-track” approach for proving lim

n!!xn =l. We demonstrate it in the following examples.

Example 1.1.25 To prove limn!!

n2 " 1n2 + 1 = 1, we need to find a condition under which

the inequality

3n2 " 1n2 + 1 " 13 < "

holds. This can be done by “loosing” up the term on the left hand side in the followingsequence of inequalities:

3n2 " 1n2 + 1 " 13 =

2

n2 + 11!< 2

n

2!< 2

N

3!" ",

where 1!, 2!, and 3! indicates the condition under which the corresponding inequalityholds. More explicitly, we have

1! ! for all natural integers n;

2! ! n > N ;

3! ! N # 24".Thus, the whole sequence of inequalities hold when n > N , with N a number greaterthan or equal to 24". Hence, by combining 1!, 2!, and 3!, we have

n > 24" "# 3n2 " 1n2 + 1 " 13 < ".

Therefore, we get limn!!

n2 " 1n2 + 1 = 1.

Example 1.1.26 Example 1.1.5 tells us that ( n"a " 1( " a

nfor a # 1. Hence

( n"a " 1( 1!" a

n

2!< a

N

3!= ",


where

1! ! for all natural integers n;

2! ! n > N ;

3! ! N = a4".Hence, for a # 1, we have

n > a4" "# ( n"a " 1( < ".This rigorously proves that lim

n!!n"a = 1 in the case a # 1.

Similarly, the estimation in Example 1.1.6 can be turned into a rigorous proof oflimn!!

n"n = 1. In fact, we have

( n"n " 1( 1!""2"

n " 12!<"2"

N " 13!< 2"

N " 14!= ",

where

1! ! for all natural integers n > 1;2! ! n > N ;

3! ! for any N > 1;4! ! N = 44"2 + 1.

Hence, by combining all the conditions, we have

n > N = 4

"2+ 1 "# ( n"n " 1( < ".

Example 1.1.27 We try to rigorously prove limn!!n2an = 0 when (a( < 1.

Using the idea of Example 1.1.9, we write (a( = 1

1 + b . Then b > 0, since (a( < 1. For

n # 4, we have

(n2an " 0( = n2(a(n = n2

(1 + b)n =n2

1 + nb + n(n " 1)2!

b2 + n(n " 1)(n " 2)3!

b3 +$+ bn

1!< n2

n(n " 1)(n " 2)3!

b3

2!< 3!n2

nn

2

n

2b3= 3!22

nb33!< 3!22

Nb34!= ",

where

1! ! n > 2;2! ! n # 4;3! ! n > N ;

4! ! N = 3!22

b3".


Hence,

n >max! 24

((a("1 " 1)3" ,4# "# (n2an " 0( < ",so that lim

n!!n2an = 0 when (a( < 1.More generally, it is clear from the proof that we have

limn!!npan = 0, for any p and (a( < 1.

After familiarizing ourselves with the “loose-and-track” approach, we may skip theintermediate steps to give rigorous proofs, as demonstrated in the following examples.

Example 1.1.28 We rigorously prove limn!!

an

n!= 0 in Example 1.1.10 for general a.

Choose a natural number M , such that (a( <M . Then for n >M , we have

,ann!, < Mn

n!= M #M$M

1 # 2$M # M

M + 1 #M

M + 2$M

n" MM

M !# Mn= MM+1

M !# 1n.

Therefore for any " > 0, we have

n >max!MM+1M !"

,M# "# ,ann!" 0, < MM+1

M !# 1n< MM+1

M !# 1

MM+1M !"

= ",

so that limn!!

an

n!= 0 for any fixed a.

The rigorous definition of limit allows us to prove (rigorously) some properties oflimits.

Example 1.1.29 Assume limn!!xn = l > 0. We prove that lim

n!!"xn ="l.

The question is the following. We already know

For any " > 0, there is N , such that n > N "# (xn " l( < ".We want to show

For any " > 0, there is N , such that n > N "# ("xn ""l( < ".In short, we need to prove that the first implication implies the second implication.

Note that

("xn ""l( = (("xn ""l)("xn +"l)("

xn +"l = (xn " l("xn +"l "

(xn " l("l

.


Hence, for any given " > 0, the second implication will hold if(xn " l("

l< ", or (xn" l( <"l".

But the inequality (xn " l( <"l" follows from the first implication, provided we apply thefirst implication to

"l" in place of ".

The analysis above gives the following formal proof. Let " > 0. By applying thedefinition of lim

n!!xn = l to the positive number"l" > 0, there is N , such that n > N

implies (xn " l( <"l". Thenn > N "# (xn " l( <"l"

"# ("xn ""l( = (("xn ""l)("xn +"l)("

xn +"l = (xn " l("xn +"l "

(xn " l("l< ".

A key point here is to take advantage of the fact that the definition of limit canbe applied to any positive number,

"l" for example, not necessarily the given positive

number ".

Example 1.1.30 We prove the arithmetic rule limn!!(xn + yn) = lim

n!!xn + limn!!yn in

Proposition 1.1.3.Let lim

n!!xn = l and limn!!yn = k. Then for any "1 > 0, "2 > 0, there are N1, N2, such

that

n > N1 "# (xn " l( < "1,n > N2 "# (yn " k( < "2.

The reason for specifying "1, "2 instead of " is that we expect to choose them as somemodification of ", as demonstrated in Example 1.1.29.

Let N =max{N1,N2}. Thenn > N "# n > N1, n > N2"# (xn " l( < "1, (yn " k( < "2"# ((xn + yn) " (l + k)( " (xn " l( + (yn " k( < "1 + "2.

If "1 + "2 " ", then this rigorously proves limn!!(xn + yn) = l + k. Of course this means that

we may choose "1 = "2 = "

2at the first step of the argument.

The analysis above gives the following formal proof. For any " > 0, apply the definition

of limn!!xn = l and lim

n!!yn = k to"

2> 0. We know that there are N1 and N2, such that

n > N1 "# (xn " l( < "

2,

n > N2 "# (yn " k( < "

2.

Then

n > N =max{N1,N2} "# (xn " l( < "

2, (yn " k( < "

2

"# ((xn + yn) " (l + k)( " (xn " l( + (yn " k( < "

2+ "

2= ".


Example 1.1.31 We prove the arithmetic rule limn!!(xn yn) = lim

n!!xn # limn!!yn in Propo-

sition 1.1.3.Let lim

n!!xn = l and limn!!yn = k. Then for any "1 > 0, "2 > 0, there are N1, N2, such

that

n > N1 "# (xn " l( < "1,n > N2 "# (yn " k( < "2.

Then for n > N =max{N1,N2}, we have (see Figure 1.4)

(xn yn " l k( = ((xn " l)yn + l(yn " k)(" (xn " l( # (yn( + (l( # (yn " k(< "1((k( + "2) + (l("2,where we use (yn " l( < "2 implies (yn( < (k( + "2. The proof of lim

n!!(xn yn) = l k will be

complete if, for any " > 0, we can choose "1 > 0 and "2 > 0, such that

"1((k( + "2) + (l("2 " ".This can be achieved by choosing "1, "2 satisfying

"2 " 1, "1((k( + 1) " "

2, ((l( + 1)"2 " "

2.

(In the last inequality, we added 1 to (l( to avoid a possible zero denominator in thefollowing expression of "2 in the case of l = 0.) Therefore, if we choose

"1 = "

2((k( + 1) , "2 =min!1, "

2((l( + 1)#at the very beginning of the proof, then we get a rigorous proof of the arithmetic rule.The formal writing of the proof is left to the reader.

area= "l""yn ! k"

area="x

n!l""y n"

xn

l

ynk

Figure 1.4: The shaded region represents the error term.


Example 1.1.32 We prove the sandwich rule of Proposition 1.1.4.

Suppose xn " yn " zn and limn!!xn = lim

n!! zn = l. For any " > 0, there are N1 and N2,

such that

n > N1 "# (xn " l( < ",n > N2 "# (zn " l( < ".

Then

n > N =max{N1,N2} "# (xn " l( < ", (zn " l( < ""# l " " < xn, zn < l + ""# l " " < xn " yn " zn < l + ""# (yn " l( < ".

Example 1.1.33 We prove Proposition 1.1.6, which says that a convergent sequencemust be bounded.

Let limn!!xn = l. Then for " = 1 > 0, there is N , such that n > N implies (xn " l( < 1.

By choosing a larger N if necessary, we may further assume that N is a natural number.Then l + 1 is an upper bound of all xn, n > N . This implies that max{x1, . . . , xN , l + 1}is an upper bound of the whole sequence. A lower bound can be found in a similar way.

Note that we may always choose N to be a natural number in the definition of limit.Such choice is needed here because we also use N as the index of a term in the sequence.

Example 1.1.34 We prove the order rule in Proposition 1.1.5.

Assume that xn # yn for su!ciently large n, limn!!xn = l, and lim

n!!yn = k. Let " > 0 be

given. Then, there are N1,N2, and N3 such that

n > N1 "# xn # yn;n > N2 "# (xn " l( < ";n > N3 "# (yn " k( < ".

Hence, if n >max{N1,N2,N3}, then we have

l + " > xn # yn > k " ".Since " > 0 is arbitrary, the inequality l + " > k " " implies l # k; otherwise l < k leads to acontradiction

l + k " l2> k " k " l

2,

if we choose " = k " l2

.


Conversely, we assume limn!!xn = l, lim

n!!yn = k, and l > k. For any " > 0, there is N ,

such that n > N implies (xn " l( < " and (yn " k( < ". Thenn > N "# xn > l " ", yn < k + " "# xn " yn > (l " ") " (k + ") = l " k " 2".

By choosing " = l " k2> 0 at the beginning of the argument, we conclude that xn > yn for

n > N .

The choicel " k2

can also be seen from intuition. For large n, xn is close to l and yn

is close to k. If the closeness is within half of the distance between l and k, i.e., within

" = l " k2

, then we see that, for large n, xn is above the midpointl + k2

between l and k,

and yn is below the midpoint.

Example 1.1.35 We prove Proposition 1.1.8 about the limit of subsequences.

Suppose limn!!xn = l. Then for any " > 0, there is N , such that n > N implies

(xn " l( < ". Also recall that, if we assume the sequence starts from n = 1, then the indexof the subsequence satisfies nk # k (prove this by induction!). Therefore

k > N "# nk # k > N "# (xnk " l( < ".

The definition of limit makes explicit use of the limit value l. Therefore if we wantto show the convergence of a sequence by the definition, we need to first know the valuel and then find suitable N for each ". In many cases, however, it is hard (or evenimpossible) to find the limit of a convergent sequence.

Proposition 1.1.7 gives an example where we know the convergence of a sequencewithout knowing the actual limit value. The following criterion provides a method forthe general case.

Definition 1.1.11 A sequence {xn} is called a Cauchy sequence if for any " > 0,there is N , such that m,n > N implies (xn " xm( < ".

Theorem 1.1.12 (Cauchy Criterion) A sequence converges if and only if it is aCauchy sequence.

Theorem 1.1.12 is called the Cauchy criterion for limits of sequences, or just theCauchy criterion for short.

In fact, suppose {xn} converges to l. For any " > 0, there is an N , such that

n # N "# (xn " l( < "

2.


Then

m,n > N "# (xm " l( < "

2, (xn " l( < "

2

"# (xn " xm( = ((xn " l) " (xm " l)( " (xn " l( + (xm " l( < "

2+ "

2= ".

The proof of the converse of the Cauchy criterion is much more di!cult. It could beaccomplished in the following three steps.

1. Any Cauchy sequence is bounded.

2. By the Bolzano-Weierstrass Theorem, any bounded sequence has a convergentsubsequence.

3. If a Cauchy sequence has a subsequence converging to l, then the whole sequenceconverges to l.

We leave the proof as an exercise to the readers.

Example 1.1.36 In Example 1.1.18, we proved that the sequence {xn},xn = 1

12+ 1

22+ 1

32+$+ 1

n2.

is increasing and bounded, so that it converges.Alternatively, we can prove that {xn} is a Cauchy sequence. In fact, for any " > 0,n #m > 1

"+ 1 "# (xn " xm( = 3 1

(m + 1)2 +$+1

n23

" 1

(m " 1)m +1

m(m + 1) +$+1

(n " 1)n= $ 1

m " 1 "1

m% + $ 1

m" 1

m + 1% +$+ $1

n " 1 "1

n%

= 1

m " 1 "1

n" 1

m " 1 < ".Hence {xn} coverges by the Cauchy criterion.

Example 1.1.37 We prove that the sequence, called the harmonic series,

xn = 1 + 1

2+ 1

3+$ + 1

n,

diverges by showing that the Cauchy criterion fails. In fact, we show that there is " > 0,such that for any N , we can find m,n > N satisfying (xm " xn( # ".

For any n, we have

x2n " xn = 1

n + 1 +1

n + 2 +$+1

2n# 1

2n+ 1

2n+$+ 1

2n= 1

2.


Choose " = 1

2. For any N , we may find a natural number n > N . Then we also have

m = 2n > N and

(xm " xn( = x2n " xn # 1

2= ".

Therefore, the sequence fails the Cauchy criterion and diverges.

Exercises

1.1.19 Rigorously prove the limits:

1. limn!!

sin 2n + cosnn

= 0;2. lim

n!!n + sin 2nn " cosn = 1;

3. limn!!

2n + 32n ""n " 1 = 1;

4. limn!!

n!

nn= 0;

5. limn!!(

"n + a ""n + b) = 0;

6. limn!!( 3

"n " 3"n " 1) = 0;

7. limn!!

n!

an= 0, (a( > 1;

8. limn!!

np

n!= 0.

1.1.20 Prove that if limn!!xn = l and lim

n!!xn = k, then l = k.1.1.21 Prove that if lim

n!!xn = l, then limn!! (xn( = (l(.

1.1.22 Prove that limn!! (xn " l( = 0 if and only if lim

n!!xn = l.1.1.23 Prove that if lim

n!!xn = l, then limn!!(cxn) = c l.

1.1.24 Prove that if limn!!xn = l and lim

n!!xn = k ! 0, then limn!!

xn

yn= l

k.

1.1.25 Prove that a sequence {xn} converges if and only if the subsequences {x2n} and {x2n+1}converge to the same limit.

1.1.26 Prove that a monotonic sequence converges if and only if there is a convergent subse-quence.

1.1.27 Prove that if a bounded sequence {xn} diverges, then there are two subsequences thatconverge to di"erent finite limits.

1.1.28 Prove Theorem 1.1.12, the Cauchy criterion for limits of sequences.

1.1.29 If {xn} is a Cauchy sequence, can you conclude that {(xn(} is also a Cauchy sequence?Why? What about the converse?

1.1.30 Suppose the sequence {xn} satisfies(xn+1 " xn( " 14n2, n = 1,2,3, . . . .

Prove that {xn} is convergent.1.1.31 If

xn " xn"1 = sin(!4n), n = 1,2,3, . . . ,does the sequence {xn} converge? Explain.

1.1.32 Suppose xn > 0 for all n. Prove that if limn!!

xn+1xn= l, then lim

n!! n"xn = l.


1.1.4 Infinite Limits

A sequence may diverge for various reasons. For example, the sequence {n} divergesbecause it become arbitrarily large, so that the sequence is not bounded. For anotherexample, the sequence {("1)n} is bounded but diverges because it has two subsequenceswith di"erent finite limits. The first case may be summarized by the following definition.

Definition 1.1.13 A sequence diverges to infinity, denoted limn!!xn = %, if for any

B, there is N , such that n > N implies (xn( > B.

In the definition, the infinity means that the absolute value (or the magnitude) ofthe sequence can get arbitrarily large. If we further take into account of the signs, thenwe get the definition of lim

n!!xn = +% by requiring xn > B, and get the definition of

limn!!xn = "% by requiring xn < B. For example, we have lim

n!!n = +%.

The meaning of limn!!xn = +% is illustrated in Figure 1.5.

B

Nn

x1

x2

x3

x4

xN+1xN+2

xN+3xn

Figure 1.5: n > N implies xn > B.


limn!!np = +%, for p > 0.

Here is the rigorous proof. For any B > 0, we have

n > B 1p "# np > B.

Note that in the definition of limn!!xn = +%, we may additionally assume B > 0 without

loss of generality. Similarly, in the definition of limn!!xn = "%, we may also assume B < 0

(or B < "100 if needed).

Example 1.1.39 The idea for Example 1.1.9 may also be used to show that limn!!an =

% for (a( > 1. Specifically, let (a( = 1 + b. Then (a( > 1 implies b > 0, and we have

(an( = (1 + b)n = 1 + nb + n(n " 1)2

b2 +$+ bn > nb.


For any B, choose N = B

b. Then we have

n > N "# (an( > nb > Nb = B.

This proves that limn!!an =%. If we take the sign into account, this also proves lim

n!!an =+% for a > 1.

Example 1.1.40 Assume xn ! 0. We prove that limn!!xn = 0 implies lim

n!!1

xn= %.

Actually the converse is also true and the proof is left to the reader.

If limn!!xn = 0, then for any B > 0, we apply the definition of the limit to

1

B> 0. Thus

there is N , such that

n > N "# (xn( < 1

B"# , 1

xn, > B.

Therefore we conclude limn!!

1

xn=%.

Applying what we just proved to the limit in Example 1.1.9, we get another proof oflimn!!an =% for (a( > 1.

A sequence {xn} is an infinitesimal if limn!!xn = 0. Example 1.1.40 shows that a

nonzero sequence is an infinitesimal if and only if its reciprocal diverges to infinity.Many properties of the finite limit can be extended. The following are some examples

of the extended arithmetic rules. Example 1.1.40 is included in the first two rules.

•l

0=% for l ! 0: If lim

n!!xn = l ! 0 and limn!!yn = 0, then lim

n!!xnyn=%.

•l

% = 0: If limn!!xn = l and lim

n!!yn =%, then limn!!

xnyn= 0.

• (+%) + (+%) = +%: If limn!!xn = +% and lim

n!!yn = +%, then limn!!(xn + yn) = +%.

• ("%) + l = "%: If limn!!xn = "% and lim

n!!yn = l, then limn!!(xn + yn) = "%.

• (+%) # l = "% for l < 0: If limn!!xn = +% and lim

n!!yn = l < 0, then limn!!xnyn = "%.

•l

0+ = +% for l > 0: If limn!!xn = l > 0, lim

n!!yn = 0 and yn > 0, then limn!!

xnyn= +%.

The notations such asl

% = 0 are simply used as symbols for the underlying statements

about limits. One should also be cautious not to overly extend the arithmetic properties.For example, the following properties are actually wrong

% +% =%,+%"% = "1, 0 # % = 0, 0 # % =%.


A counterexample for the first equality is xn = n and yn = "n, for which we havelimn!!xn =%, lim

n!!yn =% and limn!!(xn + yn) = 0.

The following are the extensions of some other properties.

• Sandwich Rule: If xn # yn and limn!!yn = +%, then lim

n!!xn = +%.

• Order Rule: If limn!!xn = +% and lim

n!!yn = l is finite, then xn > yn for su!ciently

large n.

The following complements Proposition 1.1.7.

Proposition 1.1.14 If a monotonic sequence {xn} is unbounded, then limn!!xn =%.

Specifically, an increasing sequence {xn} is bounded below by x1. Hence only twoexclusive cases can happen: {xn} is bounded above, or {xn} is not bounded above.

If {xn} is bounded above, then Proposition 1.1.7 says that {xn} converges.If {xn} is not bounded above, then any number B is not an upper bound. In other

words, there is some term aN > B. Then the sequence is increasing, we have

n > N "# xn # aN > B.

This proves that {xn} diverges to +%.Similar discussion can be made to decreasing sequences. If a decreasing sequence is

bounded, then it converges. If the sequence is not bounded, then it diverges to "%.

Example 1.1.41 By limn!!n = +% and the extended arithmetic rule, we have lim

n!!nk =+% for k # 1. Then by the extended arithmetic rule again, we have

limn!!(n3 " 3n + 1) = lim

n!!n3 $1 " 3

n2+ 1

n3% = (+%) # 1 = +%.

Example 1.1.42 For a > 1, the sequence {an} is increasing. So either the sequence isbounded and converges to a finite limit l, or the sequence is unbounded and diverges to+%. If lim

n!!an = l, thena l = a lim

n!!an = limn!!an+1 = l.

Since the sequence is increasing, we have l # a > 1. Therefore it is impossible to havea l = l. We conclude that lim

n!!an = +%.

Exercises

1.1.33 Prove the following results:

1.2. Limits of Functions 33

1. limn!!

an

n=%, for (a( > 1;

2. limn!!n sina =%, for a ! 0,±!,±2!, . . . ;

3. limn!!

2n2 " 13n +"n = +%;

4. limn!!n("n + 1 ""n " 1) = +%.

1.1.34 Prove that if limn!!xn = +% and p > 0, then lim

n!!xpn = +%. What about the case p < 0?

1.1.35 Construct sequences {xn} and {yn}, such that both diverge to infinity, but {xn + yn}can have any of the following behaviors.

1. limn!!(xn + yn) =%.

2. limn!!(xn + yn) = 2.

3. {xn + yn} is bounded but does not converge.

4. {xn + yn} is not bounded and does not diverge to infinity.

This exercise shows that % +% has no definite meaning.

1.1.36 Suppose limn!!xn = +%. Show that the following arithmetic rules hold.

1. If {yn} is bounded below, then limn!!(xn + yn) = +%.

2. If c > 0, then limn!!(cxn) = +%.

3. If yn >M for some M > 0, for su!ciently large n, then limn!!(xn yn) = +%.

4. If {yn} is bounded and xn ! 0 for all n, then limn!!

ynxn= 0.

1.2 Limits of Functions

1.2.1 The Concept of Limit and Its Properties

We have seen various behaviors that sequences may disply as the index goes to infin-ity. Similar observations can be made for functions near a point. For example, as xis approaching 0, the functions x2 and (x( also approach 0, x"1 gets larger and doesnot approach a fixed finite number, and sin(x"1) swings between "1 and 1, and neverapproaches any one specific finite number.

Among the three functions, only x2 and (x( converge as x goes to 0. We also notethat, as we talk about the behavior as x approaches 0, the function does not have to bedefined at 0. Analogously, as the index of a sequence goes to infinity, the index is neverequal to infinity.

Definition 1.2.1 (Non-rigorous) If f(x) approaches a finite number l when x ap-proaches (but not equal to) a, then we say the function f(x) converges to the limit lat a and write

limx!a

f(x) = l.The function diverges at a if it does not approach a specific finite number when xapproaches a.

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