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n l i 1	n		n				n
n	3		3i	3			3i
46. lim		1			2	1

47.Prove the formula for the sum of a ﬁnite geometric series with ﬁrst term a and common ratio r 1:

n	a r	n	1
ar i 1 a ar ar 2 ar n 1
	r 1
i 1

		APPENDIX F PROOFS OF THEOREMS \|\|\|\| A39
	n			3
48.	Evaluate			3	.

		2		i 1
	i 1	2
	n
49.	Evaluate 2i 2i .
	i 1
	m			n
50.	Evaluate				i j .
	i 1		j 1

F PROOFS OF THEOREMS

In this appendix we present proofs of several theorems that are stated in the main body of the text. The sections in which they occur are indicated in the margin.

SECTION 2.3	LIMIT LAWS			Suppose that c is a constant and the limits
						lim f x L	and	lim t x M
						x la		x la
	exist. Then
	1.	lim f x t x L M					2.	lim f x t x L M
		x la					x la
	3.	lim cf x cL					4.	lim f x t x LM
		x la					x la
	5.	lim	f x		L	if M 0
	5.	lim	t x			if M 0
		x la	t x		M

PROOF OF LAW 4 Let 0 be given. We want to ﬁnd 0 such that

if 0 x a

then f x t x LM

In order to get terms that contain f x L and t x M , we add and subtract Lt x as follows:

f x t x LM f x t x Lt x Lt x LM

f x L t x L t x M

f x L t x L t x M (Triangle Inequality)

f x L t x L t x M

We want to make each of these terms less than 2.

Since limx la t x M, there is a number 1 0 such that

0 x a

then t x M

		L )
2(1

Also, there is a number

0 such that if 0 x a

2, then

t x M 1

and therefore

t x t x M M t x M M 1 M

A40 |||| APPENDIX F PROOFS OF THEOREMS

Since limx la f x

L, there is a number

0 such that

if	0 x a 3	then	f x L

				2(1	M )
				2(1	M )
Let min 1, 2, 3 . If 0 x a , then we have 0 x a 1,
0 x a	2, and 0 x a	3, so we can combine the inequalities to obtain
f x t x LM f x L t x L t x M
			(1 M ) L

		M )		2(1	L )
	2(1	M )		2(1	L )

2 2

This shows that limx la f x t x LM.							M
PROOF OF LAW 3 If we take t x c in Law 4, we get
lim cf x lim t x f x lim t x lim f x
x la	x la			x la	x la
	lim c lim f x
	x la	x la
	c lim f x		(by Law 7)				M
	x la
PROOF OF LAW 2 Using Law 1 and Law 3 with c 1, we have
lim f x t x lim f x 1 t x lim f x lim 1 t x
x la	x la				x la	x la
	lim f x 1 lim t x lim f x lim t x						M
	x la		x la		x la	x la
PROOF OF LAW 5 First let us show that
	lim	1		1
	lim			M
	x la t x			M

To do this we must show that, given 0, there exists

0 x a

then

Observe that

M t x

M x

0 such that

We know that we can make the numerator small. But we also need to know that the

denominator is not small when x is near a. Since limx la t x M, there is a number
1 0 such that, whenever 0 x a 1, we have
t x M		2
		M

and therefore M M t x t x M t x t x

	M	t x
	2

APPENDIX F PROOFS OF THEOREMS |||| A41

This shows that

0 x a

then t x		M
then t x		2

and so, for these values of x,
	1	1
		M t x
	Mt x
Also, there exists 2 0 such that
if	0 x a 2

then

1	2	2
M
M	M	M	2

t x M M22

Let min 1,

2 . Then, for 0 x a , we have

M t x

2 M2

t x

Mt x

It follows that limx la 1 t x 1 M. Finally, using Law 4, we obtain

lim

f x

lim f x

lim f x lim

t x

x la t x

x la

x la t x

M M

2 THEOREM

If f x t x for all x in an open interval that contains a (except

possibly at a) and

lim f x L

and

lim t x M

x la

then L M.

PROOF We use the method of proof by contradiction. Suppose, if possible, that L M. Law 2 of limits says that

lim t x f x M L

x la

Therefore, for any 0, there exists 0 such that

if 0 x a

then t x f x M L

In particular, taking L M (noting that L M 0 by hypothesis), we have a number 0 such that

if	0 x a	then	t x f x M L L M
Since a a for any number a, we have
if	0 x a	then	t x f x M L L M

which simpliﬁes to

0 x a

then t x f x

But this contradicts f x t x . Thus the inequality L M must be false. Therefore
L M.	M

A42 |||| APPENDIX F PROOFS OF THEOREMS

3 THE SQUEEZE THEOREM	If f x t x h x for all x in an open interval that
contains a (except possibly at a) and
	lim f x lim h x L
	x la	x la
then		lim t x L
		x la

PROOF Let

that is,

0 be given. Since limx la f x L, there is a number 1 0 such that

if 0 x a 1	then	f x L
0 x a 1	then	L f x L

Since limx la h x

L, there is a number

0 x a

2 0 such that

then h x L

that is,

0 x a

then L h x L

	Let	min 1, 2 . If 0 x a , then 0 x a 1 and 0 x a		2,
	so
			L f x t x h x L
	In particular,		L t x L
	and so t x L . Therefore limx la t x L.			M

SECTION 2.5	THEOREM		If f is a one-to-one continuous function deﬁned on an interval a, b ,
	then its inverse function f 1 is also continuous.

	PROOF	First we show that if f is both one-to-one and continuous on a, b , then it must
	be either increasing or decreasing on a, b . If it were neither increasing nor decreasing,
	then there would exist numbers x1, x2, and x3 in a, b with x1 x2 x3 such that f x2
	does not lie between f x1 and f x3 . There are two possibilities: either (1) f x3 lies
	between f x1 and f x2 or (2) f x1 lies between f x2 and f x3 . (Draw a picture.) In
	case (1) we apply the Intermediate Value Theorem to the continuous function f to get a
	number c between x1 and x2 such that f c f x3 . In case (2) the Intermediate Value
	Theorem gives a number c between x2 and x3 such that f c f x1 . In either case we
	have contradicted the fact that f is one-to-one.
	Let us assume, for the sake of deﬁniteness, that f is increasing on a, b . We take any
	number y0 in the domain of f 1 and we let f 1 y0 x0; that is, x0 is the number in
	a, b such that f x0 y0. To show that f 1 is continuous at y0 we take any 0 such
	that the interval x0 , x0 is contained in the interval a, b . Since f is increasing,
	it maps the numbers in the interval x0 , x0 onto the numbers in the interval
	f x0 , f x0 and f 1 reverses the correspondence. If we let denote the
	smaller of the numbers 1 y0 f x0 and 2 f x0 y0, then the interval
	y0	, y0 is contained in the interval f x0 , f x0 and so is mapped
	into the interval x0 , x0 by f 1. (See the arrow diagram in Figure 1.) We have

APPENDIX F PROOFS OF THEOREMS |||| A43

	therefore found a number 0 such that
	if	y y0		then
		f(x¸-∑)	y¸
		{	y¸
		{
			∂¡
		f	f–!
	{	{		x¸
FIGURE 1	a	x¸-∑		x¸

f 1 y f 1 y0

f(x¸+∑)

}

∂™

	f
}	}	x
x¸+∑	b	x
x¸+∑	b

This shows that limy l y0	f 1 y f 1 y0 and so f 1 is continuous at any number y0 in
its domain.	M

8 THEOREM If f is continuous at

lim

x la

b and limx la t x b, then f t x f b

SECTION 3.3

PROOF Let 0 be given. We want to ﬁnd a number 0 such that

if	0 x a	then	f t x f b
Since f is continuous at b, we have
	lim f y f b
	y lb
and so there exists	1 0 such that
if	0 y b 1	then	f y f b
Since limx la t x b, there exists 0 such that
if	0 x a	then	t x b 1

Combining these two statements, we see that whenever 0 x a

t x b 1, which implies that f t x f b . Therefore we that limx la f t x f b .

The proof of the following result was promised when we proved that lim

we have have proved

sin		1.
		1.

THEOREM If 0 2, then tan .

PROOF Figure 2 shows a sector of a circle with center O, central angle , and radius 1. Then

AD OA tan tan

We approximate the arc AB by an inscribed polygon consisting of n equal line segments

A44 |||| APPENDIX F PROOFS OF THEOREMS

Dand we look at a typical segment PQ. We extend the lines OP and OQ to meet AD in the points R and S. Then we draw RT PQ as in Figure 2. Observe that

Q	°	°
	°

°°R P

FIGURE 2

SECTION 4.3

y y=ƒ

ƒf(a)+fª(a)(x-a)

FIGURE 3

RTO PQO 90

and so RTS 90 . Therefore we have

PQ RT RS

If we add n such inequalities, we get

Ln AD tan

where Ln is the length of the inscribed polygon. Thus, by Theorem 2.3.2, we have

lim Ln tan

n l

But the arc length is deﬁned in Equation 8.1.1 as the limit of the lengths of inscribed polygons, so

lim Ln tan	M
n l

CONCAVITY TEST

(a)If f x 0 for all x in I, then the graph of f is concave upward on I.

(b)If f x 0 for all x in I, then the graph of f is concave downward on I.

PROOF OF (a) Let a be any number in I. We must show that the curve y f x lies above the tangent line at the point a, f a . The equation of this tangent is

y f a f a x a

So we must show that

f x f a f a x a

whenever x I x a . (See Figure 3.)

First let us take the case where x a. Applying the Mean Value Theorem to f on the interval a, x , we get a number c, with a c x, such that

1	f x f a f c x a

Since f 0 on I, we know from the Increasing/Decreasing Test that f is increasing on I. Thus, since a c, we have

f a f c

and so, multiplying this inequality by the positive number x a, we get

2	f a x a f c x a

SECTION 4.4

N See the biographical sketch of Cauchy on page 113.

APPENDIX F PROOFS OF THEOREMS |||| A45

Now we add f a to both sides of this inequality:

	f a f a x a f a f c x a
But from Equation 1 we have f x f a f c x a . So this inequality becomes
3	f x f a f a x a
which is what we wanted to prove.
For the case where x a we have f c f a , but multiplication by the negative
number x a reverses the inequality, so we get (2) and (3) as before.		M

In order to give the promised proof of l’Hospital’s Rule, we ﬁrst need a generalization of the Mean Value Theorem. The following theorem is named after another French mathematician, Augustin-Louis Cauchy (1789–1857).

1 CAUCHY’S MEAN VALUE THEOREM Suppose that the functions f and t are continuous on a, b and differentiable on a, b , and t x 0 for all x in a, b . Then there is a number c in a, b such that

f c f b f a t c t b t a

Notice that if we take the special case in which t x x, then t c 1 and Theorem 1 is just the ordinary Mean Value Theorem. Furthermore, Theorem 1 can be proved in a similar manner. You can verify that all we have to do is change the function h given by Equation 4.2.4 to the function

h x f x f a	f b f a	t x t a
	t b t a

and apply Rolle’s Theorem as before.

L’HOSPITAL’S RULE Suppose f and t are differentiable and t x 0 on an open interval I that contains a (except possibly at a). Suppose that

	lim f x 0	and	lim t x 0
	x la		x la
or that	lim f x	and	lim t x
	x la		x la

(In other words, we have an indeterminate form of type 00 or .) Then

lim	f x	lim	f x

x la t x		x la t x

if the limit on the right side exists (or is or ).

A46 |||| APPENDIX F PROOFS OF THEOREMS

PROOF OF L’HOSPITAL’S RULE We are assuming that limx la f x 0 and limx la t x 0.

Let

L lim

f x

x la t x

We must show that limx la

f x t x L. Deﬁne

F x

f x if x a

G x

t x

if x a

Then F is continuous on I since f is continuous on x I x a and

lim F x lim f x 0 F a

x la

Likewise, G is continuous on I. Let x I and x a. Then F and G are continuous on

a, x and differentiable on a, x and G 0 there (since F f and G t ). There-

fore, by Cauchy’s Mean Value Theorem, there is a number y such that a y x and

F y

F x F a

F x

G y

G x G a

G x

Here we have used the fact that, by deﬁnition, F a 0 and G a 0. Now, if we let

x l a , then y l a (since a y x), so

lim

f x

lim

F x

lim

F y

lim

f y

x la t x

x la G x

yla G y

yla t y

A similar argument shows that the left-hand limit is also L. Therefore

lim

f x

x la

t x

This proves l’Hospital’s Rule for the case where a is ﬁnite.

If a is inﬁnite, we let t 1 x. Then t l 0 as x l , so we have

lim

f x

lim

f 1 t

x l t x

t l0 t 1 t

lim

f 1 t 1 t2

(by l’Hospital’s Rule for ﬁnite a)

t l0 t 1 t 1 t2

lim

f 1 t

lim

f x

t l0 t 1 t

x l t x

SECTION 11.8

In order to prove Theorem 11.8.3, we ﬁrst need the following results.

THEOREM

1.If a power series cn xn converges when x b (where b 0), then it converges whenever x b .

2.If a power series cn xn diverges when x d (where d 0), then it diverges whenever x d .

PROOF OF 1

APPENDIX F PROOFS OF THEOREMS |||| A47

Suppose that cnbn converges. Then, by Theorem 11.2.6, we have

limn l cnbn 0. According to Deﬁnition 11.1.2 with 1, there is a positive integer N such that cnbn 1 whenever n N. Thus, for n N, we have

cn xn cnbbnnxn cnbn bx n bx n

If x b , then x b 1, so x b n is a convergent geometric series. Therefore,
by the Comparison Test, the series n N cn xn is convergent. Thus the series	cn xn is
absolutely convergent and therefore convergent.	M

PROOF OF 2 Suppose that cndn diverges. If x is any number such that x d , then
cn xn cannot converge because, by part 1, the convergence of cn xn would imply the
convergence of cndn. Therefore	cn xn diverges whenever x d .	M

THEOREM For a power series	cn xn there are only three possibilities:

1.The series converges only when x 0.

2.The series converges for all x.

3.There is a positive number R such that the series converges if x R and diverges if x R.

PROOF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers b and d such that cn xn converges for x b and diverges for x d. Therefore the set

S x cn xn converges is not empty. By the preceding theorem, the series diverges ifx d , so x d for all x S. This says that d is an upper bound for the set S.

Thus, by the Completeness Axiom (see Section 11.1), S has a least upper bound R. If
x R, then x S, so cnxn diverges. If x R, then x is not an upper bound for
S and so there exists b S such that b x . Since b S,	cnbn converges, so by the
preceding theorem cn xn converges.	M

3THEOREM For a power series cn x a n there are only three possibilities:

1.The series converges only when x a.

2.The series converges for all x.

3.There is a positive number R such that the series converges if x a R and diverges if x a R.

PROOF If we make the change of variable u x a, then the power series becomes
cnun and we can apply the preceding theorem to this series. In case 3 we have con-
vergence for u R and divergence for	u R. Thus we have convergence for
x a R and divergence for x a	R.	M

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