Smith R., Minton R. Calculus

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A-3 APPENDIX A

Proofs of Selected Theorems 3

Notice that the ﬁrst term in (A.4) is slightly more complicated, as we must also estimate

the size of |g(x)|. Notice that

|g(x)|=|g(x) − L

+ L

|≤|g(x) − L

|+|L

|. (A.6)

Since lim

x→a

g(x) = L

, there is a number δ

> 0, such that

|g(x) − L

| < 1, whenever 0 < |x − a| <δ

From (A.6), we now have that

|g(x)|≤|g(x) − L

|+|L

| < 1 +|L

Returning to the ﬁrst term in (A.4), we have that for 0 < |x − a| <δ

| f (x) − L

||g(x)| < | f (x) − L

|(1 +|L

|). (A.7)

Now, since lim

x→a

f (x) = L

, given any ε>0, there is a number δ

> 0 such that for

0 < |x − a| <δ

| f (x) − L

| <

2(1 +|L

From (A.7), we then have

| f (x) − L

||g(x)| < | f (x) − L

|(1 +|L

2(1 +|L

(1 +|L

whenever0 < |x − a| <δ

and 0 < |x − a| <δ

.Togetherwith (A.4)and(A.5),this tells

us that for δ = min{δ

,δ

},if0< |x − a| <δ, then

| f (x)g(x) − L

|≤|f (x) − L

||g(x)|+|L

||g(x) − L

= ε,

which proves (iii).

(iv) We ﬁrst show that

lim

x→a

g(x)

for L

= 0. Notice that in this case, we need to show that we can make



g(x)

−



as small

as desired. We have



g(x)

−



− g(x)

g(x)



. (A.8)

Of course, since lim

x→a

g(x) = L

, we can make the numerator of the fraction on the right-

hand side as small as needed. We must also consider what happens to the denominator,

though. Recall that given any ε

> 0, there is a δ

> 0 such that

|g(x) − L

| <ε

, whenever 0 < |x − a| <δ

In particular, for ε

, this says that

|g(x) − L

| <

Notice that by the triangle inequality, we can now say that

|=|L

− g(x) + g(x)|≤|L

− g(x)|+|g(x)| <

+|g(x)|.

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4 APPENDIX A

Proofs of Selected Theorems A-4

Subtracting

from both sides now gives us

< |g(x)|,

so that

|g(x)|

From (A.8), we now have that for 0 < |x − a| <δ



g(x)

−



− g(x)

g(x)



2|L

− g(x)|

. (A.9)

Further, given any ε>0, there is a δ

> 0 so that

− g(x)| <

εL

, whenever 0 < |x − a| <δ

From (A.9), we now have that for δ = min{δ

,δ



g(x)

−



2|L

− g(x)|

<ε,

whenever 0 < |x − a| <δ, as desired. We have now established that

lim

x→a

g(x)

From (iii), we now have that

lim

x→a

f (x)

g(x)

= lim

x→a



f (x)

g(x)





lim

x→a

f (x)





lim

x→a

g(x)



= L





which proves the last part of the theorem.

The following result appeared as Theorem 3.3 in section 1.3.

THEOREM A.2

Suppose that lim

x→a

f (x) = L and n is any positive integer. Then,

lim

x→a



f (x) =



lim

x→a

f (x) =

√

where for n even, we assume that L > 0.

PROOF

(i) We ﬁrst give the proof for the case where L > 0. Since lim

x→a

f (x) = L, we know that

given any ε

> 0, there is a δ

> 0, so that

| f (x) − L| <ε

, whenever 0 < |x − a| <δ

To show that lim

x→a



f (x) =

√

L, we need to show that given any ε>0, there is a δ>0

such that





f (x) −

√



<ε,whenever 0 < |x −a| <δ.

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A-5 APPENDIX A

Proofs of Selected Theorems 5

Notice that this is equivalent to having

√

L − ε<



f (x) <

√

L + ε.

Now, raising all sides to the nth power, we have



√

L − ε



< f (x) <



√

L + ε



Subtracting L from all terms gives us



√

L − ε



− L < f (x) − L <



√

L + ε



− L.

Since ε is taken to be small, we now assume that ε<

√

L. Observe that in this case,

0 <

√

L − ε<

√

L. Let ε

= min





√

L + ε



− L, L −



√

L − ε





> 0. Then, since

lim

x→a

f (x) = L, we know that there is a number δ>0, so that

−ε

< f (x) − L <ε

, whenever 0 < |x − a| <δ.

It then follows that

(

√

L − ε)

− L ≤−ε

< f (x) − L <ε

≤ (

√

L + ε)

− L,

whenever 0 < |x − a| <δ. Reversing the above sequence of steps gives us

lim

x→a

√

f (x) =

√

L, as desired.

(ii) If L < 0 and n is odd and lim

x→a

f (x) = L < 0, then

lim

x→a

[− f (x)] =−L > 0.

It then follows from part (i) that

−lim

x→a



f (x) = lim

x→a



− f (x) =

√

−L =−

√

from which the result follows.

(iii) If L = 0 and n isodd and lim

x→a

f (x) = L = 0, then we have that givenany ε

> 0,

there is a δ>0 so that

| f (x)| <ε

, whenever 0 < |x −a| <δ.

It follows that





f (x) −0





f (x)





| f (x)| <ε

if and only if | f (x)| <ε

. Taking ε

= ε

gives us the result whenever

0 < |x − a| <δ.

The following result appeared in section 1.3, as Theorem 3.5.

THEOREM A.3 (Squeeze Theorem)

Suppose that

f (x) ≤ g(x) ≤ h(x), (A.10)

for all x in some interval (c, d), except possibly at the point a ∈ (c, d) and that

lim

x→a

f (x) = lim

x→a

h(x) = L,

for some number L. Then, it follows that

lim

x→a

g(x) = L, also.

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6 APPENDIX A

Proofs of Selected Theorems A-6

PROOF

To show that lim

x→a

g(x) = L, we must prove that given any ε>0, there is a δ>0, such that

|g(x) − L| <ε, whenever 0 < |x − a| <δ.

Since lim

x→a

f (x) = L, we have that given any ε>0, there is a δ

> 0, such that

| f (x) − L| <ε, whenever 0 < |x − a| <δ

Likewise, since lim

x→a

h(x) = L, we have that given any ε>0, there is a δ

> 0, such that

|h(x) − L| <ε, whenever 0 < |x − a| <δ

Now, choose δ = min{δ

,δ

}. Then, if 0 < |x −a| <δ, it follows that 0 < |x − a| <δ

and 0 < |x − a| <δ

, so that

| f (x) − L| <ε and |h(x) − L| <ε.

Equivalently, we can say that

L − ε< f (x) < L +ε and L −ε<h(x) < L +ε. (A.11)

It now follows from (A.10) and (A.11) that if 0 < |x −a| <δ, then

L − ε< f (x) ≤ g(x) ≤ h(x) < L +ε,

which gives us

L − ε<g(x) < L +ε

or |g(x) − L| <εand it follows that lim

x→a

g(x) = L, as desired.

The following result appeared as Theorem 4.3 in section 1.4.

THEOREM A.4

Suppose lim

x→a

g(x) = L and f is continuous at L. Then,

lim

x→a

f (g(x)) = f



lim

x→a

g(x)



= f (L).

PROOF

To prove the result, we must show that given any number ε>0, there is a number δ>0

for which

| f (g(x)) − f (L)| <ε, whenever 0 < |x −a| <δ.

Since f is continuous at L, we know that lim

t→L

f (t) = f (L). Consequently, given any ε>0,

there is a δ

> 0 for which

| f (t) − f (L)| <ε, whenever 0 < |t − L| <δ

Further, since lim

x→a

g(x) = L, we can make g(x) as close to L as desired, simply by making x

sufﬁcientlyclosetoa.Inparticular,theremustbeanumberδ>0 forwhich|g(x) − L| <δ

whenever 0 < |x −a| <δ. It now follows that if 0 < |x −a| <δ, then |g(x) − L| <δ

so that

| f (g(x)) − f (L)| <ε,

as desired.

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A-7 APPENDIX A

Proofs of Selected Theorems 7

The following result appeared as Theorem 5.1 in section 1.5.

THEOREM A.5

For any rational number t > 0,

lim

x→±∞

= 0,

where for the case where x →−∞, we assume that t =

, where q is odd.

PROOF

We ﬁrst prove that lim

x→∞

= 0. To do so, we must show that given any ε>0, there is

an M > 0 for which



− 0



<ε, whenever x > M. Since x →∞, we can take x to be

positive, so that



− 0



<ε,

which is equivalent to

<ε

1/t

< x.

Notice that taking M to be any number greater than

1/t

, we will have



− 0



<εwhen-

ever x > M, as desired.

For the case lim

x→−∞

= 0, we must show that given any ε>0, there is an N < 0 for

which



− 0



<ε, whenever x < N. Since x →−∞, we can take x to be negative, so

that



− 0



<ε,

which is equivalent to

|x|

<ε

1/t

< |x|=−x,

since x < 0. Multiplying both sides of the inequality by −1, we get

−

1/t

> x.

NoticethattakingN tobeanynumberlessthan−

1/t

,wewillhave



− 0



<ε,whenever

x < N, as desired.

In section 2.8, we presented Rolle’s Theorem (Theorem 8.1), but only gave a graphical

idea of the proof. Now, with the addition of the Extreme Value Theorem, we can give a

complete proof of this result.

THEOREM A.6 (Rolle’s Theorem)

Suppose that f is continuous on the interval [a, b], differentiable on the interval

(a, b) and f (a) = f (b). Then there is a number c ∈ (a, b) such that f



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8 APPENDIX A

Proofs of Selected Theorems A-8

PROOF

There are three cases to be considered here.

(i) If f (x) is constant on [a, b], then, f



(x) = 0 throughout (a, b).

(ii) Suppose that f (x) < f (a), for some x ∈ (a, b). Since f is continuous on [a, b],

we have by the Extreme Value Theorem (Theorem 2.1 in section 3.2) that f attains

an absolute minimum on [a, b]. Since f (x) < f (a) = f (b), for some x ∈ (a, b), the

absolute minimum occurs at some point c ∈ (a, b). Then by Fermat’s Theorem (Theo-

rem 2.2 in section 3.2), c must be a critical number of f . Finally, since f is differentiable

on (a, b), we must have f



(iii) Similarly, suppose that f (x) > f (a), for some x ∈ (a, b). Then as in (ii), we

have by the Extreme Value Theorem that f attains an absolute maximum on [a, b]. Since

f (x) > f (a) = f (b), for some x ∈ (a, b), the absolute maximum occurs at some point

c ∈ (a, b). Then, by Fermat’s Theorem, we must have f



In section 7.6, we prove l’Hˆopital’s Rule only for a special case. Here, we present a

general proof for the

case. First, we need the following generalization of the Mean Value

Theorem.

THEOREM A.7 (Generalized Mean Value Theorem)

Suppose that f and g are continuous on the interval [a, b] and differentiable on

the interval (a, b) and that g



(x) = 0, for all x on (a, b). Then, there is a number

z ∈ (a, b), such that

f (b) − f (a)

g(b) − g(a)



(z)



(z)

Notice that the Mean Value Theorem (Theorem 8.4in section 2.8) is simply thespecial

case of Theorem A.7 where g(x) = x.

PROOF

First, observe that since g



(x) = 0, for all x on (a, b), we must have that g(b) − g(a) = 0.

This follows from Rolle’s Theorem (Theorem A.6), since if g(a) = g(b), there would be

some number c ∈ (a, b) for which g



h(x) = [ f (b) − f (a)]g(x) − [g(b) − g(a)] f (x).

Notice that h is continuous on [a, b] and differentiable on (a, b), since both f and g are

continuous on [a, b] and differentiable on (a, b). Further, we have

h(a) = [ f (b) − f (a)]g(a) −[g(b) − g(a)] f (a)

= f (b)g(a) − g(b) f (a)

and h(b) = [ f (b) − f (a)]g(b) −[g(b) − g(a)] f (b)

= g(a) f (b) − f (a)g(b),

so that h(a) = h(b). In view of this, Rolle’s Theorem says that there must be a number

z ∈ (a, b) for which

0 = h



(z) = [ f (b) − f (a)]g



(z) − [g(b) − g(a)] f



(z)

f (b) − f (a)

g(b) − g(a)



(z)



(z)

as desired.

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A-9 APPENDIX A

Proofs of Selected Theorems 9

We can now give a general proof of l’Hˆopital’s Rule for the

case. The proof of the

∞

case can be found in a more advanced text.

THEOREM A.8 (l’H

opital’s Rule)

Suppose that f and g are differentiable on the interval (a, b), except possibly at some

ﬁxed point c ∈ (a, b) and that g



(x) = 0, on (a, b), except possibly at x = c. Suppose

further that lim

x→c

f (x)

g(x)

has the indeterminate form

∞

and that lim

x→c



(x)



(x)

= L

(or ±∞). Then,

lim

x→c

f (x)

g(x)

= lim

x→c



(x)



(x)

PROOF



case



In this case, we have that lim

x→c

f (x) = lim

x→c

g(x) = 0. Deﬁne

F(x) =



f (x)ifx = c

0ifx = c

and G(x) =



g(x)ifx = c

0ifx = c

Notice that lim

x→c

F(x) = lim

x→c

f (x) = 0 = F(c)

and lim

x→c

G(x) = lim

x→c

g(x) = 0 = G(c),

so that both F and G are continuous on all of (a, b). Further, observe that for x = c,



(x) = f



(x) and G



(x) = g



(x) and so, both F and G are differentiable on each of the

intervals (a, c) and (c, b). We ﬁrst consider the interval (c, b). Notice that F and G are

continuous on [c, b] and differentiable on (c, b) and so, by the Generalized Mean Value

Theorem,forany x ∈ (c, b),we havethat there is some number z, withc < z < x, for which



(z)



(z)

F(x) − F(c)

G(x) − G(c)

F(x)

G(x)

f (x)

g(x)

where we have used the fact that F(c) = G(c) = 0. Notice that as x → c

, z → c

, also,

since c < z < x. Taking the limit as x → c

,wenowhave

lim

x→c

f (x)

g(x)

= lim

z→c



(z)



(z)

= lim

z→c



(z)



(z)

= L.

Similarly, by focusing on the interval (a, c), we can show that lim

x→c

−

f (x)

g(x)

= L, which proves

that lim

x→c

f (x)

g(x)

= L (since both one-sided limits agree).

The following theorem corresponds to Theorem 6.1 in section 9.6.

THEOREM A.9

Given any power series,

∞



k=0

(x − c)

, there are exactly three possibilities:

(i) the series converges for all x ∈ (−∞, ∞) and the radius of convergence is

r =∞;

(ii) the series converges only for x = c (and diverges for all other values of x)

and the radius of convergence is r = 0or

(iii) the series converges for x ∈ (c −r, c +r) and diverges for x < c −r and for

x > c +r, for some number r with 0 < r < ∞.

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10 APPENDIX A

Proofs of Selected Theorems A-10

In order to prove Theorem A.9, we ﬁrst introduce and prove two simpler results.

THEOREM A.10

(i) If the power series

∞



k=0

converges for x = a = 0, then it also converges for

all x with |x| < |a|.

(ii) If the power series

∞



k=0

diverges for x = d, then it also diverges for all x

with |x| > |d|.

PROOF

(i) Suppose that

∞



k=0

converges. Then, by Theorem 2.2 in section 9.2, lim

k→∞

= 0.

For this to occur, we must be able to make |b

| as small as desired, just by making k

sufﬁciently large. In particular, there must be a number N > 0, such that |b

| < 1, for all

k > N. So, for k > N, we must have









If |x| < |a|, then



< 1 and so,

∞



k=0



is a convergent geometric series. It then follows

from the Comparison Test (Theorem 3.3 in section 9.3) that

∞



k=0

|converges and hence,

∞



k=0

converges absolutely.

(ii) Suppose that

∞



k=0

diverges. Notice that if x is any number with |x| > |d|, then

∞



k=0

must diverge, since if it converged, we would have by part (i) that

∞



k=0

would

also converge, which contradicts our assumption.

Next, we state and prove a slightly simpler version of Theorem A.9.

THEOREM A.11

Given any power series,

∞



k=0

, there are exactly three possibilities:

(i) the series converges for all x ∈ (−∞, ∞) and the radius of convergence is

r =∞;

(ii) the series converges only for x = 0 (and diverges for all other values of x)

and the radius of convergence is r = 0or

(iii) the series converges for x ∈ (−r, r) and diverges for x < −r and x > r, for

some number r with 0 < r < ∞.

PROOF

If neither (i) nor (ii) is true, then there must be nonzero numbers a and d such that the series

converges for x = a and diverges for x = d. From Theorem A.10, observe that

∞



k=0

diverges for all values of x with |x| > |d|. Deﬁne the set S to be the set of all values of x for

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A-11 APPENDIX A

Proofs of Selected Theorems 11

which the series converges. Since the series converges for x = a, S is nonempty. Further,

|d|is an upper bound on S, since the series diverges for all values of x with |x| > |d|. By the

Completeness Axiom (see section 9.1), S must have a least upper bound r. So, if |x| > r,

then

∞



k=0

diverges. Further, if |x| < r, then |x| is not an upper bound for S and there

must be a number t in S with |x| < t.Since t ∈ S,

∞



k=0

converges and by Theorem A.10,

∞



k=0

converges since |x| < t ≤|t|. This proves the result.

We can now prove the original result (Theorem A.9).

PROOF OF THEOREM A.9

Let t = x − c and the power series

∞



k=0

(x − c)

becomes simply

∞



k=0

. By Theo-

rem A.11, we know that either the series converges for all t (i.e., for all x) or only for t = 0

(i.e., only for x = c) or there is a number r > 0 such that the series converges for |t| < r

(i.e., for |x − c| < r)and divergesfor |t| > r (i.e., for |x −c| > r).This provesthe original

result.

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