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A
Proofs of Selected Theorems
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Proofs of Selected Theorems
THEOREM 1.2
Properties of Limits (Properties 2, 3, 4, and 5)
(page 59)
Let b and c be real numbers, let n be a positive integer, and let f and g be
functions with the limits
lim f 共x兲 L
lim g 共x兲 K.
and
x→c
x→c
lim 关 f 共x兲 ± g共x兲兴 L ± K
2. Sum or difference:
x→c
lim 关 f 共x兲g共x兲兴 LK
3. Product:
x→c
f 共x兲
L
, K0
g共x兲 K
n
lim 关 f 共x兲兴 Ln
4. Quotient:
lim
x→c
5. Power:
x→c
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof To prove Property 2, choose > 0. Because 兾2 > 0, you know that there
exists 1 > 0 such that 0 < x c < 1 implies f 共x兲 L < 兾2. You also know
that there exists 2 > 0 such that 0 < x c < 2 implies g共x兲 K < 兾2. Let be
the smaller of 1 and 2; then 0 < x c < implies that
ⱍ
ⱍ
ⱍ
ⱍ f 共x兲 Lⱍ < 2
and
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍg共x兲 Kⱍ < 2.
So, you can apply the triangle inequality to conclude that
ⱍ关 f 共x兲 g共x兲兴 共L K兲ⱍ ⱍ f 共x兲 Lⱍ ⱍg共x兲 Kⱍ < 2 2 which implies that
lim 关 f 共x兲 g共x兲兴 L K lim f 共x兲 lim g共x兲.
x→c
x→c
x→c
The proof that
lim 关 f 共x兲 g共x兲兴 L K
x→c
is similar.
To prove Property 3, given that
lim f 共x兲 L and
x→c
lim g共x兲 K
x→c
you can write
f 共x兲g共x兲 关 f 共x兲 L兴 关g共x兲 兴 关Lg共x兲 f 共x兲兴 LK.
Because the limit of f 共x兲 is L, and the limit of g共x兲 is K, you have
lim 关 f 共x兲 L兴 0
x→c
and
lim 关g共x兲 K兴 0.
x→c
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Proofs of Selected Theorems
ⱍ
ⱍ
Let 0 < < 1. Then there exists > 0 such that if 0 < x c < , then
ⱍ f 共x兲 L 0ⱍ < and
ⱍg共x兲 K 0ⱍ < which implies that
ⱍ关 f 共x兲 L兴 关g共x兲 K兴 0ⱍ ⱍ f 共x兲 Lⱍ ⱍg共x兲 Kⱍ < < .
So,
lim [ f 共x兲 L兴 关g共x兲 K兴 0.
x→c
Furthermore, by Property 1, you have
lim Lg共x兲 LK and
x→c
lim Kf 共x兲 KL.
x→c
Finally, by Property 2, you obtain
lim f 共x兲g共x兲 lim 关 f 共x兲 L兴 关g共x兲 K兴 lim Lg共x兲 lim Kf 共x兲 lim LK
x→c
x→c
x→c
x→c
x→c
0 LK KL LK
LK.
To prove Property 4, note that it is sufficient to prove that
lim
x→c
1
1
.
g共x兲 K
Then you can use Property 3 to write
lim
x→c
f 共x兲
1
1
L
lim f 共x兲
lim f 共x兲 lim
.
x→c g共x兲
g共x兲 x→c
g共x兲 x→c
K
Let > 0. Because lim g共x兲 K, there exists 1 > 0 such that if
x→c
ⱍ
ⱍ
K
ⱍ ⱍ2ⱍ
ⱍ
0 < x c < 1, then g共x兲 K <
which implies that
ⱍKⱍ ⱍg共x兲 关ⱍKⱍ g共x兲兴ⱍ ⱍg共x兲ⱍ ⱍⱍKⱍ g共x兲ⱍ < ⱍg共x兲ⱍ ⱍ 2 ⱍ.
K
ⱍ
ⱍ
That is, for 0 < x c < 1,
ⱍKⱍ < ⱍg共x兲ⱍ
2
or
1
2 .
<
g共x兲
K
ⱍ
ⱍ ⱍ ⱍ
ⱍ
ⱍ
Similarly, there exists a 2 > 0 such that if 0 < x c < 2, then
ⱍ ⱍ ⱍ
K2
g共x兲 K <
.
2
ⱍ
ⱍ
ⱍ
Let be the smaller of 1 and 2. For 0 < x c < , you have
ⱍ
ⱍ ⱍ
ⱍ
1
1
K g共x兲
1
g共x兲 K
g共x兲K
K
So, lim
x→c
1
1
.
g共x兲 K
ⱍ
1
K g共x兲ⱍ
ⱍ ⱍg共x兲ⱍ ⱍ
<
1
K
ⱍ ⱍ
2 K2
.
K 2
ⱍ ⱍ ⱍ ⱍ
Finally, the proof of Property 5 can be obtained by a straightforward application of
mathematical induction coupled with Property 3.
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Proofs of Selected Theorems
A3
THEOREM 1.4
The Limit of a Function Involving a Radical
(page 60)
Let n be a positive integer. The limit below is valid for all c when n is odd,
and is valid for c > 0 when n is even.
n x 冪
n c.
lim 冪
x→c
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Consider the case for which c > 0 and n is any positive integer. For a given
> 0, you need to find > 0 such that
ⱍ
ⱍ
ⱍ
n x 冪
n c < 冪
ⱍ
whenever 0 < x c < which is the same as saying
n
n
< 冪
x冪
c < whenever < x c < .
n
n
n
c, which implies that 0 < 冪
c < 冪
c. Now, let be the smaller of
Assume < 冪
the two numbers.
n
c 共冪
c 兲
n
n
c 兲
共冪
n
and
c
Then you have
关c 共
n c
冪
< x c
n
兲 兴 < x c
< n c < 共冪
兲 c
n c 共冪
兲 c < xc
n
n
共冪c 兲 < x
n
n c < 共冪
兲 c
n
n
n c < 共冪
兲
n
n c < 冪
n x
冪
n c < 冪
n x 冪
n c < .
< 冪
THEOREM 1.5 The Limit of a Composite Function (page 61)
If f and g are functions such that lim g共x兲 L and lim f 共x兲 f 共L兲, then
x→c
冢
x→L
冣
lim f 共g 共x兲兲 f lim g共x兲 f 共L兲.
x→c
x→c
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
For a given > 0, you must find > 0 such that
ⱍ f 共g共x兲兲 f 共L兲ⱍ < whenever
ⱍ
ⱍ
0 < x c < .
Because the limit of f 共x兲 as x → L is f 共L兲, you know there exists 1 > 0 such that
ⱍ f 共u兲 f 共L兲ⱍ < whenever
ⱍu Lⱍ < 1.
Moreover, because the limit of g共x兲 as x → c is L, you know there exists > 0 such that
ⱍg共x兲 Lⱍ < 1
ⱍ
ⱍ
whenever 0 < x c < .
Finally, letting u g共x兲, you have
ⱍ f 共g共x兲兲 f 共L兲ⱍ < whenever
ⱍ
ⱍ
0 < x c < .
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THEOREM 1.7
Functions That Agree at All But One Point
(page 62)
Let c be a real number, and let f 共x兲 g共x兲 for all x c in an open interval
containing c. If the limit of g共x兲 as x approaches c exists, then the limit of f 共x兲
also exists and
lim f 共x兲 lim g共x兲.
x→c
x→c
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Let L be the limit of g共x兲 as x → c. Then, for each > 0 there exists a > 0
such that f 共x兲 g共x兲 in the open intervals 共c , c兲 and 共c, c 兲, and
ⱍg共x兲 Lⱍ < whenever 0 < x c < .
ⱍ
ⱍ
ⱍ f 共x兲 Lⱍ < whenever
ⱍ
ⱍ
Because f 共x兲 g共x兲 for all x in the open interval other than x c, it follows that
0 < x c < .
So, the limit of f 共x兲 as x → c is also L.
THEOREM 1.8 The Squeeze Theorem (page 65)
If h共x兲 f 共x兲 g共x兲 for all x in an open interval containing c, except possibly
at c itself, and if
lim h共x兲 L lim g共x兲
x→c
x→c
then lim f 共x兲 exists and is equal to L.
x→c
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
For > 0 there exist 1 > 0 and 2 > 0 such that
ⱍh共x兲 Lⱍ < whenever 0 < x c < 1
ⱍ
ⱍ
ⱍg共x兲 Lⱍ < whenever 0 < x c < 2.
ⱍ
ⱍ
and
Because h共x兲 f 共x兲 g共x兲 for all x in an open interval containing c, except possibly
at c itself, there exists 3 > 0 such that h共x兲 f 共x兲 g共x兲 for 0 < x c < 3. Let
be the smallest of 1, 2, and 3. Then, if 0 < x c < , it follows that
h共x兲 L < and g共x兲 L < , which implies that
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
ⱍ
< h共x兲 L < and < g共x兲 L < L < h共x兲 and g共x兲 < L .
Now, because h共x兲 f 共x兲 g共x兲, it follows that L < f 共x兲 < L , which implies
that f 共x兲 L < . Therefore,
ⱍ
ⱍ
lim f 共x兲 L.
x→c
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THEOREM 1.11 Properties of Continuity (page 75)
If b is a real number and f and g are continuous at x c, then the functions
listed below are also continuous at c.
1. Scalar multiple: bf
2. Sum or difference: f ± g
3. Product: fg
f
4. Quotient: , g共c兲 0
g
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Because f and g are continuous at x c, you can write
Proof
lim f 共x兲 f 共c兲 and
x→c
lim g共x兲 g共c兲.
x→c
For Property 1, when b is a real number, it follows from Theorem 1.2 that
lim 关共bf 兲共x兲兴 lim 关bf 共x兲兴 b lim 关 f 共x兲兴 b f 共c兲 共bf 兲共c兲.
x→c
x→c
x→c
Thus, bf is continuous at x c.
For Property 2, it follows from Theorem 1.2 that
lim 共 f ± g兲共x兲 lim 关 f 共x兲 ± g共x兲兴
x→c
x→c
lim 关 f 共x兲兴 ± lim 关g共x兲兴
x→c
x→c
f 共c兲 ± g共c兲
共 f ± g兲共c兲.
Thus, f ± g is continuous at x c.
For Property 3, it follows from Theorem 1.2 that
lim 共 fg兲共x兲 lim 关 f 共x兲g共x兲兴
x→c
x→c
lim 关 f 共x兲兴 lim 关g共x兲兴
x→c
x→c
f 共c兲g共c兲
共 fg兲共c兲.
Thus, fg is continuous at x c.
For Property 4, when g共c兲 0, it follows from Theorem 1.2 that
f
f 共x兲
lim 共x兲 lim
x→c g
x→c g共x兲
lim f 共x兲
x→c
lim g共x兲
x→c
f 共c兲
g共c兲
f
共c兲.
g
Thus,
f
is continuous at x c.
g
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THEOREM 1.14 Vertical Asymptotes (page 85)
Let f and g be continuous on an open interval containing c. If f 共c兲 0,
g共c兲 0, and there exists an open interval containing c such that g共x兲 0 for
all x c in the interval, then the graph of the function
h 共x兲 f 共x兲
g共x兲
has a vertical asymptote at x c.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Consider the case for which f 共c兲 > 0, and there exists b > c such that
c < x < b implies g共x兲 > 0. Then for M > 0, choose 1 such that
0 < x c < 1
implies that
f 共c兲
3f 共c兲
< f 共x兲 <
2
2
and 2 such that
0 < x c < 2
implies that 0 < g共x兲 <
f 共c兲
.
2M
Now let be the smaller of 1 and 2. Then it follows that
0 < xc < implies that
f 共x兲
f 共c兲 2M
>
M.
g共x兲
2
f 共c兲
冤 冥
So, it follows that
lim
x→c
f 共x兲
g共x兲 and the line x c is a vertical asymptote of the graph of h.
Alternative Form of the Derivative (page 101)
The derivative of f at c is
f共c兲 lim
x→c
f 共x兲 f 共c兲
xc
provided this limit exists.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
The derivative of f at c is given by
f共c兲 lim
x→0
f 共c x兲 f 共c兲
.
x
Let x c x. Then x → c as x → 0. So, replacing c x by x, you have
f共c兲 lim
x→0
f 共c x兲 f 共c兲
f 共x兲 f 共c兲
lim
.
x→c
x
xc
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A7
THEOREM 2.10 The Chain Rule (page 130)
If y f 共u兲 is a differentiable function of u, and u g共x兲 is a differentiable
function of x, then y f 共g共x兲兲 is a differentiable function of x and
dy
dy
dx du
du
dx
or, equivalently,
d
关 f 共g共x兲兲兴 f共g共x兲兲g共x兲.
dx
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof In Section 2.4, you let h共x兲 f 共g共x兲兲 and used the alternative form of the
derivative to show that h共c兲 f共g共c兲兲g共c兲, provided g共x兲 g共c兲 for values of x other
than c. Now consider a more general proof. Begin by considering the derivative of f.
f共x兲 lim
x→0
f 共x x兲 f 共x兲
y
lim
x→0 x
x
For a fixed value of x, define a function such that
冦
x 0
0,
共x兲 y
f共x兲,
x
x 0.
Because the limit of 共x兲 as x → 0 does not depend on the value of 共0兲, you have
lim 共x兲 lim
x→0
x→0
f共x兲冥 0
冤 y
x
and you can conclude that is continuous at 0. Moreover, because y 0 when
x 0, the equation
y x共x兲 xf共x兲
is valid whether x is zero or not. Now, by letting u g共x x兲 g共x兲, you can use
the continuity of g to conclude that
lim u lim 关g共x x兲 g共x兲兴 0
x→0
x→0
which implies that
lim 共u兲 0.
x→0
Finally,
y u共u兲 uf共u兲 →
y u
u
共u兲 f共u兲,
x
x
x
and taking the limit as x → 0, you have
冤
冥
dy du
du
lim 共u兲 f共u兲
dx
dx x→0
dx
du
du
共0兲 f共u兲
dx
dx
du
f共u兲
dx
du dy
.
dx du
x 0
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Proofs of Selected Theorems
Concavity Interpretation (page 187)
1. Let f be differentiable on an open interval I. If the graph of f is concave
upward on I, then the graph of f lies above all of its tangent lines on I.
2. Let f be differentiable on an open interval I. If the graph of f is concave
downward on I, then the graph of f lies below all of its tangent lines on I.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Assume that f is concave upward on I 共a, b兲. Then, f is increasing on 共a, b兲.
Let c be a point in the interval I 共a, b兲. The equation of the tangent line to the graph
of f at c is given by
g共x兲 f 共c兲 f共c兲共x c兲.
If x is in the open interval 共c, b兲, then the directed distance from point 共x, f 共x兲兲 (on the
graph of f) to the point 共x, g共x兲兲 (on the tangent line) is given by
d f 共x兲 关 f 共c兲 f共c兲共x c兲兴
f 共x兲 f 共c兲 f共c兲共x c兲.
Moreover, by the Mean Value Theorem there exists a number z in 共c, x兲 such that
f共z兲 f 共x兲 f 共c兲
.
xc
So, you have
d f 共x兲 f 共c兲 f共c兲共x c兲
f共z兲共x c兲 f共c兲共x c兲
关 f共z兲 f共c兲兴共x c兲.
The second factor 共x c兲 is positive because c < x. Moreover, because f is increasing,
it follows that the first factor 关 f共z兲 f共c兲兴 is also positive. Therefore, d > 0 and you
can conclude that the graph of f lies above the tangent line at x. If x is in the open
interval 共a, c兲, a similar argument can be given. This proves the first statement. The
proof of the second statement is similar.
THEOREM 3.7 Test for Concavity (page 188)
Let f be a function whose second derivative exists on an open interval I.
1. If f 共x兲 > 0 for all x in I, then the graph of f is concave upward in I.
2. If f 共x兲 < 0 for all x in I, then the graph of f is concave downward in I.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof For Property 1, assume f 共x兲 > 0 for all x in 共a, b兲. Then, by Theorem 3.5, f
is increasing on 关a, b兴. Thus, by the definition of concavity, the graph of f is concave
upward on 共a, b兲.
For Property 2, assume f 共x兲 < 0 for all x in 共a, b兲. Then, by Theorem 3.5, f is decreasing
on 关a, b兴. Thus, by the definition of concavity, the graph of f is concave downward on
共a, b兲.
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THEOREM 3.10 Limits at Infinity (page 196)
If r is a positive rational number and c is any real number, then
lim
x→ c
0.
xr
Furthermore, if x r is defined when x < 0, then lim
x→
c
0.
xr
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
Begin by proving that
lim
x→ 1
0.
x
For > 0, let M 1兾. Then, for x > M, you have
x > M
1
< x
1
ⱍ ⱍ
1
0 < .
x
So, by the definition of a limit at infinity, you can conclude that the limit of 1兾x as
x → is 0. Now, using this result, and letting r m兾n, you can write the following.
lim
x→ c
c
lim
x r x→ x m兾n
冤
c lim
x→ 冢 冣冥
1
n
冪x
m
冢 冪冣
1
c冢冪 lim 冣
x
c lim
x→ n
1
x
m
m
n
x→ n 0
c共冪
兲
0
m
The proof of the second part of the theorem is similar.
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Proofs of Selected Theorems
THEOREM 4.2
Summation Formulas (page 255)
n
兺 c cn,
1.
c is a constant
i1
n
n共n 1兲
2
i1
n
n共n 1兲共2n 1兲
i2 3.
6
i1
n
2
n 共n 1兲2
i3 4.
4
i1
兺i 2.
兺
兺
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof The proof of Property 1 is straightforward. By adding c to itself n times, you
obtain a sum of cn.
To prove Property 2, write the sum in increasing and decreasing order and add
corresponding terms, as follows.
n
1
2
n
→
2
1
→
→
→
→
→
共n 1兲 共n 2兲 . . . n
i1
. . . 共n 1兲 3
→
→
n
兺i →
兺i
i1
n
2
兺 i 共n 1兲 共n 1兲 共n 1兲 . . . 共n 1兲 共n 1兲
i1
n terms
So,
n
兺i
i1
n共n 1兲
.
2
To prove Property 3, use mathematical induction. First, if n 1, the result is true
because
1
兺i
2
12 1 i1
1共1 1兲共2 1兲
.
6
Now, assuming the result is true for n k, you can show that it is true for n k 1,
as follows.
k1
兺
i2 i1
k
兺i
2
共k 1兲2
i1
k共k 1兲共2k 1兲
共k 1兲2
6
k1
共2k2 k 6k 6兲
6
k1
关共2k 3兲共k 2兲兴
6
共k 1兲共k 2兲关2共k 1兲 1兴
6
Property 4 can be proved using a similar argument with mathematical induction.
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THEOREM 4.8
Proofs of Selected Theorems
A11
Preservation of Inequality (page 272)
1. If f is integrable and nonnegative on the closed interval 关a, b兴, then
冕
b
0 f 共x兲 dx.
a
2. If f and g are integrable on the closed interval 关a, b兴 and f 共x兲 ≤ g共x兲 for
every x in 关a, b兴, then
冕
b
冕
b
f 共x兲 dx a
g共x兲 dx.
a
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
冕
To prove Property 1, suppose, on the contrary, that
b
f 共x兲 dx I < 0.
a
Then, let a x0 < x1 < x2 < . . . < xn b be a partition of 关a, b兴, and let
R
n
兺 f 共c 兲 x
i
i
i1
be a Riemann sum. Because f 共x兲 0, it follows that R 0. Now, for 储储 sufficiently
small, you have R I < I兾2, which implies that
ⱍ
ⱍ
n
I
兺 f 共c 兲 x R < I 2 < 0
i
i
i1
which is not possible. From this contradiction, you can conclude that
冕
b
0 f 共x兲 dx.
a
To prove Property 2, note that f 共x兲 g共x兲 implies that g共x兲 f 共x兲 0. So, you can
apply the result of Property 1 to conclude that
冕
冕
冕
b
0 关g共x兲 f 共x兲兴 dx
a
b
0 冕
a
b
a
a
b
f 共x兲 dx a
冕
b
g共x兲 dx g共x兲 dx.
f 共x兲 dx
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Proofs of Selected Theorems
Properties of the Natural Logarithmic Function (page 319)
The natural logarithmic function is one-to-one.
lim ln x and
x→0 lim ln x x→ See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Recall from Section P.3 that a function f is one-to-one if for x1 and x2 in its
domain
f 共x1兲 f 共x2兲.
x1 x2
1
> 0 for x > 0. So f is increasing on its entire domain
x
共0, ) and therefore is strictly monotonic (see Section 3.3). Choose x1 and x2 in the
domain of f such that x1 x2. Because f is strictly monotonic, it follows that either
Let f 共x兲 ln x. Then f共x兲 f 共x1兲 < f 共x2兲 or f 共x1兲 > f 共x2兲.
In either case, f 共x1兲 f 共x2兲. So, f 共x兲 ln x is one-to-one.
1
To verify the limits, begin by showing that ln 2 2. From the Mean Value Theorem for
Integrals, you can write
冕
2
ln 2 1
1
1
1
dx 共2 1兲 x
c
c
where c is in 关1, 2兴.
This implies that
1 c 2
1
1
1 c
2
1
1 ln 2 .
2
Now, let N be any positive (large) number. Because ln x is increasing, it follows that if
x > 22N, then
ln x > ln 22N 2N ln 2.
1
However, because ln 2 2, it follows that
ln x > 2N ln 2 2N
冢12冣 N.
This verifies the second limit. To verify the first limit, let z 1兾x. Then, z → as
x → 0, and you can write
冢
冣
1
x
lim 共ln z兲
lim ln x lim ln
x→0
x→0
z→ lim ln z
z→ .
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A13
THEOREM 5.8
Continuity and Differentiability of Inverse
Functions (page 341)
Let f be a function whose domain is an interval I. If f has an inverse function,
then the following statements are true.
1.
2.
3.
4.
If f is continuous on its domain, then f 1 is continuous on its domain.
If f is increasing on its domain, then f 1 is increasing on its domain.
If f is decreasing on its domain, then f 1 is decreasing on its domain.
If f is differentiable on an interval containing c and f共c兲 0, then f 1
is differentiable at f 共c兲.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof To prove Property 1, first show that if f is continuous on I and has an inverse
function, then f is strictly monotonic on I. Suppose that f were not strictly monotonic.
Then there would exist numbers x1, x2, x3 in I such that x1 < x 2 < x3, but f 共x2兲 is not
between f 共x1兲 and f 共x3兲. Without loss of generality, assume f 共x1兲 < f 共x3兲 < f 共x2兲. By
the Intermediate Value Theorem, there exists a number x0 between x1 and x2 such that
f 共x0兲 f 共x3兲. So, f is not one-to-one and cannot have an inverse function. So, f must be
strictly monotonic.
Because f is continuous, the Intermediate Value Theorem implies that the set of
values of f
再 f 共x兲: x 僆 冎
forms an interval J. Assume that a is an interior point of J. From the previous argument,
f 1共a兲 is an interior point of I. Let > 0. There exists 0 < 1 < such that
I1 共 f 1共a兲 1, f 1共a兲 1兲 債 I.
Because f is strictly monotonic on I1, the set of values 再 f 共x兲: x 僆 I1冎 forms an interval
J1 債 J. Let > 0 such that 共a , a 兲 債 J1. Finally, if
ⱍy aⱍ < , then ⱍ f 1共 y兲 f 1共a兲ⱍ < 1 < .
So, f 1 is continuous at a. A similar proof can be given if a is an endpoint.
To prove Property 2, let y1 and y2 be in the domain of f 1, with y1 < y2. Then, there
exist x1 and x2 in the domain of f such that
f 共x1兲 y1 < y2 f 共x2兲.
Because f is increasing, f 共x1兲 < f 共x2兲 holds precisely when x1 < x2. Therefore,
f 1共 y1兲 x1 < x2 f 1共 y2兲
which implies that f 1 is increasing. (Property 3 can be proved in a similar way.)
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Proofs of Selected Theorems
Finally, to prove Property 4, consider the limit
共 f 1兲共a兲 lim
y→a
f 1共y兲 f 1共a兲
ya
where a is in the domain of f 1 and f 1共a兲 c. Because f is differentiable on an
interval containing c, f is continuous on that interval, and so is f 1 at a. So, y → a
implies that x → c, and you have
xc
f 共x兲 f 共c兲
1
lim
x→c f 共x兲 f 共c兲
xc
1
f 共x兲 f 共c兲
lim
x→c
xc
1
.
f共c兲
共 f 1兲共a兲 lim
x→c
冢
冣
So, 共 f 1兲共a兲 exists, and f 1 is differentiable at f 共c兲.
THEOREM 5.9 The Derivative of an Inverse Function (page 341)
Let f be a function that is differentiable on an interval I. If f has an inverse
function g, then g is differentiable at any x for which f共g共x兲兲 0. Moreover,
g共x兲 1
,
f共g共x兲兲
f共g共x兲兲 0.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof From the proof of Theorem 5.8, letting a x, you know that g is
differentiable. Using the Chain Rule, differentiate both sides of the equation x f 共g共x兲兲
to obtain
1 f共g共x兲兲
d
关g共x兲兴.
dx
Because f共g共x兲兲 0, you can divide by this quantity to obtain
d
1
关g共x兲兴 .
dx
f共g共x兲兲
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Proofs of Selected Theorems
A15
THEOREM 5.10
Operations with Exponential Functions
(Property 2) (page 347)
Let a and b be any real numbers.
2.
ea
e ab
eb
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
To prove Property 2, you can write
冢ee 冣 ln e
a
ln
a
b
ln eb a b ln共e a b兲
Because the natural logarithmic function is one-to-one, you can conclude that
ea
e ab.
eb
A Limit Involving e (page 360)
THEOREM 5.15
冢
lim 1 x→ 1
x
冣
x
lim
x→ 冢x x 1冣
x
e
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
冢
Let y lim 1 x→ 冤 冢
ln y ln lim 1 x→
1
x
冣
1 x
. Taking the natural logarithm of each side, you have
x
冣 冥.
x
Because the natural logarithmic function is continuous, you can write
ln y lim
x→ 冤x ln冢1 1x 冣冥 lim 冦ln 关1 1兾x共1兾x兲兴冧.
x→
1
Letting x , you have
t
ln共1 t兲
t
ln共1 t兲 ln 1
lim
t→0
t
d
ln x at x 1
dx
1
at x 1
x
1.
ln y lim
t→0
Finally, because ln y 1, you know that y e, and you can conclude that
冢
lim 1 x→ 1
x
冣
x
e.
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Proofs of Selected Theorems
THEOREM 5.16
Derivatives of Inverse Trigonometric Functions
(arcsin u and arccos u) (page 369)
Let u be a differentiable function of x.
d
u
关arcsin u兴 dx
冪1 u2
d
u
关arccos u兴 dx
冪1 u2
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof
Method 1: Apply Theorem 5.9.
Let f 共x兲 sin x and g共x兲 arcsin x. Because f is differentiable on
y 2
2
you can apply Theorem 5.9.
1
f 共g共x兲兲
1
cos共arcsin x兲
1
冪1 sin2共arcsin x兲
1
冪1 x2
g 共x兲 If u is a differentiable function of x, then you can use the Chain Rule to write
d
u
关arcsin u兴 ,
dx
冪1 u2
where u du
.
dx
Method 2: Use implicit differentiation.
Let y arccos x, 0 y . So, cos y x, and you can use implicit differentiation.
cos y x
dy
sin y
1
dx
dy
1
dx sin y
dy
1
dx 冪1 cos2 y
dy
1
dx 冪1 x2
If u is a differentiable function of x, then you can use the Chain Rule to write
d
u
关arccos u兴 ,
dx
冪1 u2
where u du
.
dx
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A17
THEOREM 8.3 The Extended Mean Value Theorem (page 558)
If f and g are differentiable on an open interval 共a, b兲 and continuous on 关a, b兴
such that g共x兲 0 for any x in 共a, b兲, then there exists a point c in 共a, b兲 such
f共c兲
f 共b兲 f 共a兲
that
.
g共c兲 g共b兲 g共a兲
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof You can assume that g共a兲 g共b兲, because otherwise, by Rolle’s Theorem, it
would follow that g共x兲 0 for some x in 共a, b兲. Now, define h共x兲 as
h共x兲 f 共x兲 冤 gf 共共bb兲兲 fg共共aa兲兲冥 g共x兲.
Then
h共a兲 f 共a兲 冤 gf 共共bb兲兲 fg共共aa兲兲冥 g共a兲
f 共a兲g共b兲 f 共b兲g共a兲
g共b兲 g共a兲
and
h共b兲 f 共b兲 冤 gf 共共bb兲兲 fg共共aa兲兲冥 g共b兲
f 共a兲g共b兲 f 共b兲g共a兲
g共b兲 g共a兲
and by Rolle’s Theorem there exists a point c in 共a, b兲 such that
h共c兲 f共c兲 f 共b兲 f 共a兲
g共c兲
g共b兲 g共a兲
0
which implies that
f 共b兲 f 共a兲
f共c兲
.
g共c兲 g共b兲 g共a兲
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Proofs of Selected Theorems
THEOREM 8.4 L’Hôpital’s Rule (page 558)
Let f and g be functions that are differentiable on an open interval 共a, b兲
containing c, except possibly at c itself. Assume that g共x兲 0 for all x in
共a, b兲, except possibly at c itself. If the limit of f 共x兲兾g共x兲 as x approaches c
produces the indeterminate form 0兾0, then
lim
x→c
f 共x兲
f共x兲
lim
x→c g共x兲
g共x兲
provided the limit on the right exists (or is infinite). This result also applies
when the limit of f 共x兲兾g共x兲 as x approaches c produces any one of the
indeterminate forms 兾, 共 兲兾, 兾共 兲, or 共 兲兾共 兲.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
You can use the Extended Mean Value Theorem to prove L’Hôpital’s Rule. Of the
several different cases of this rule, the proof of only one case is illustrated. The
remaining cases where x → c and x → c are left for you to prove.
Proof Consider the case for which lim f 共x兲 0 and lim g共x兲 0. Define the new
x→c
x→c
functions
F共x兲 冦f0,共x兲,
xc
and
xc
G共x兲 冦g0,共x兲,
xc
.
xc
For any x, c < x < b, F and G are differentiable on 共c, x兴 and continuous on 关c, x兴. You
can apply the Extended Mean Value Theorem to conclude that there exists a number z
in 共c, x兲 such that
F共z兲
F共x兲 F共c兲
F共x兲
f共z兲
f 共x兲
.
G共z兲 G共x兲 G共c兲 G共x兲 g共z兲 g共x兲
Finally, by letting x approach c from the right, x → c, you have z → c because
c < z < x, and
lim
x→c
f 共x兲
f共z兲
f共z兲
f共x兲
lim
lim
lim
.
g共x兲 x→c g共z兲 z→c g共z兲 x→c g共x兲
THEOREM 9.15 Alternating Series Remainder (page 621)
If a convergent alternating series satisfies the condition an1 an, then the
absolute value of the remainder RN involved in approximating the sum S by SN
is less than (or equal to) the first neglected term. That is,
ⱍS SNⱍ ⱍRNⱍ aN1.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof The series obtained by deleting the first N terms of the given series satisfies the
conditions of the Alternating Series Test and has a sum of RN.
RN S SN 兺 共1兲
n1
n1
ⱍRNⱍ
ⱍ
an N
兺 共1兲
n1
an
n1
1 N1
共1兲N aN1 共 兲
aN2 共1兲N2 aN3 . . .
共1兲N 共aN1 aN2 aN3 . . .兲
aN1 aN2 aN3 aN4 aN5 . . .
aN1 共aN2 aN3兲 共aN4 aN5兲 . . . aN1
ⱍ ⱍ ⱍ
Consequently, S SN RN aN1, which establishes the theorem.
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A19
THEOREM 9.19 Taylor’s Theorem (page 642)
If a function f is differentiable through order n 1 in an interval I containing
c, then, for each x in I, there exists z between x and c such that
f 共x兲 f 共c兲 f共c兲共x c兲 f 共c兲
f 共n兲共c兲
共x c兲2 . . . 共x c兲n Rn共x兲
2!
n!
where
Rn共x兲 f 共n1兲共z兲
共x c兲 n1.
共n 1兲!
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof To find Rn共x兲, fix x in I 共x c兲 and write Rn共x兲 f 共x兲 Pn共x兲, where Pn共x兲 is
the nth Taylor polynomial for f 共x兲. Then let g be a function of t defined by
f 共n兲共t兲
共x t兲n1
g共t兲 f 共x兲 f 共t兲 f共t兲共x t兲 . . . 共x t兲n Rn共x兲
.
n!
共x c兲n1
The reason for defining g in this way is that differentiation with respect to t has a
telescoping effect. For example, you have
d
关f 共t兲 f共t兲共x t兲兴 f共t兲 f共t兲 f 共t兲共x t兲 f 共t兲共x t兲.
dt
The result is that the derivative g共t兲 simplifies to
g共t兲 f 共n1兲共t兲
共x t兲n
共x t兲n 共n 1兲Rn共x兲
n!
共x c兲n1
for all t between c and x. Moreover, for a fixed x,
g共c兲 f 共x兲 关Pn共x兲 Rn共x兲兴 f 共x兲 f 共x兲 0
and
g共x兲 f 共x兲 f 共x兲 0 . . . 0 f 共x兲 f 共x兲 0.
Therefore, g satisfies the conditions of Rolle’s Theorem, and it follows that there is a
number z between c and x such that g共z兲 0. Substituting z for t in the equation for
g共t兲 and then solving for Rn共x兲, you obtain
f 共n1兲共z兲
共x z兲n
共x z兲n 共n 1兲Rn共x兲
n!
共x c兲n1
f 共n1兲共z兲
共x z兲n
0
共x z兲n 共n 1兲Rn共x兲
n!
共x c兲n1
f 共n1兲共z兲
共x c兲n1.
Rn共x兲 共n 1兲!
g共z兲 Finally, because g共c兲 0, you have
f 共n兲共c兲
0 f 共x兲 f 共c兲 f共c兲共x c兲 . . . 共x c兲n Rn共x兲
n!
f 共n兲共c兲
f 共x兲 f 共c兲 f共c兲共x c兲 . . . 共x c兲n Rn共x兲.
n!
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Proofs of Selected Theorems
THEOREM 9.20 Convergence of a Power Series (page 648)
For a power series centered at c, precisely one of the following is true.
1. The series converges only at c.
2. There exists a real number R > 0 such that the series converges absolutely
for x c < R, and diverges for x c > R.
3. The series converges absolutely for all x.
ⱍ
ⱍ
ⱍ
ⱍ
The number R is the radius of convergence of the power series. If the series
converges only at c, then the radius of convergence is R 0. If the series
converges for all x, then the radius of convergence is R . The set of all
values of x for which the power series converges is the interval of convergence
of the power series.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof In order to simplify the notation, the theorem for the power series 兺 an x n
centered at x 0 will be proved. The proof for a power series centered at x c
follows easily. A key step in this proof uses the completeness property of the set of real
numbers: If a nonempty set S of real numbers has an upper bound, then it must have a
least upper bound (see page 591).
It must be shown that if a power series 兺 an x n converges at x d, d 0, then it
converges for all b satisfying b < d . Because 兺 an x n converges, lim an d n 0. So,
ⱍⱍ ⱍⱍ
ⱍ
n→ ⱍ
there exists an integer N > 0 such that an d n < 1 for all n N. Then for n N,
ⱍ ⱍ
ⱍⱍ ⱍⱍⱍⱍ
兺ⱍ ⱍ 兺ⱍ ⱍ
ⱍan b nⱍ an b n
So, for b < d ,
bn
dn
dn
dn
ⱍ
ⱍⱍ ⱍⱍ
bn
ⱍ dn
and n
<
bn
dn
.
b
< 1, which implies that
d
b
d
n
is a convergent geometric series. By the Comparison Test, the series 兺 an b n converges.
Similarly, if the power series 兺 an x n diverges at x b, where b 0, then it diverges
for all d satisfying d > b . If 兺 an d n converged, then the argument above would
imply that 兺 an b n converged as well.
ⱍⱍ ⱍⱍ
Finally, to prove the theorem, suppose that neither Case 1 nor Case 3 is true. Then
there exist points b and d such that 兺 an x n converges at b and diverges at d. Let
S 再x: 兺 an x n converges冎. S is nonempty because b 僆 S. If x 僆 S, then x d ,
which shows that d is an upper bound for the nonempty set S. By the completeness
property, S has a least upper bound, R.
ⱍⱍ ⱍⱍ
ⱍⱍ
ⱍⱍ
ⱍⱍ
ⱍⱍ ⱍⱍ
ⱍⱍ
Now, if x > R, then xⰻS, so 兺 an x n diverges. And if x < R, then x is not an upper
bound for S, so there exists b in S satisfying b > x . Because b 僆 S, 兺 an b n
converges, which implies that 兺 an x n converges.
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A21
THEOREM 10.16
Classification of Conics by Eccentricity
(page 734)
Let F be a fixed point ( focus) and let D be a fixed line (directrix) in the plane.
Let P be another point in the plane and let e (eccentricity) be the ratio of the
distance between P and F to the distance between P and D. The collection of
all points P with a given eccentricity is a conic.
1. The conic is an ellipse for 0 < e < 1.
2. The conic is a parabola for e 1.
3. The conic is a hyperbola for e > 1.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof If e 1, then, by definition, the conic must be a parabola. If e 1, then you
can consider the focus F to lie at the origin and the directrix x d to lie to the right of
the origin, as shown in the figure.
y
P
Q
r
θ
F
x
x=d
For the point P 共r, 兲 共x, y兲, you have
ⱍPFⱍ r
Given that e and
ⱍPQⱍ d r cos .
ⱍPF ⱍ , it follows that
ⱍPQⱍ
ⱍPFⱍ ⱍPQⱍe
r e共d r cos 兲.
By converting to rectangular coordinates and squaring each side, you obtain
x 2 y 2 e2共d x兲2
e2共d 2 2dx x2兲.
Completing the square produces
冢x 1 ede 冣
2
2
2
y2
e 2d 2
.
2
1e
共1 e 2兲2
If e < 1, then this equation represents an ellipse. If e > 1, then 1 e 2 < 0, and the
equation represents a hyperbola.
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THEOREM 13.4
Sufficient Condition for Differentiability
(page 901)
If f is a function of x and y, where fx and fy are continuous in an open region
R, then f is differentiable on R.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Let S be the surface defined by z f 共x, y兲, where f, fx , and fy are continuous
at 共x, y兲. Let A, B, and C be points on surface S, as shown in the figure. From this
figure, you can see that the change in f from point A to point C is given by
z
C
B
Δz2
Δz1
Δz
A
y
x
(x + Δx, y + Δy)
(x, y)
(x + Δx, y)
z f 共x x, y y兲 f 共x, y兲
关 f 共x x, y兲 f 共x, y兲兴 关 f 共x x, y y兲 f 共x x, y兲兴
z1 z 2.
Between A and B, y is fixed and x changes. So, by the Mean Value Theorem, there is a
value x1 between x and x x such that
z1 f 共x x, y兲 f 共x, y兲 fx共x1, y兲 x.
Similarly, between B and C, x is fixed and y changes, and there is a value y1 between y
and y y such that
z2 f 共x x, y y兲 f 共x x, y兲 fy 共x x, y1兲 y.
By combining these two results, you can write
z z1 z 2 fx 共x1, y兲x fy 共x x, y1兲 y.
If you define 1 and 2 as 1 fx共x1, y兲 fx共x, y兲 and 2 fy 共x x, y1兲 fy共x, y兲, it
follows that
z z1 z 2 关1 fx共x, y兲兴 x 关2 fy 共x, y兲兴 y
关 fx 共x, y兲 x fy 共x, y兲 y兴 1x 2y.
By the continuity of fx and fy and the fact that x x1 x x and y y1 y y,
it follows that 1 → 0 and 2 → 0 as x → 0 and y → 0. Therefore, by definition, f is
differentiable.
THEOREM 13.6
Chain Rule: One Independent Variable
(page 907)
Let w f 共x, y兲, where f is a differentiable function of x and y. If x g共t兲
and y h 共t兲, where g and h are differentiable functions of t, then w is a
differentiable function of t, and
dw w dx w dy
.
dt
x dt
y dt
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proof Because g and h are differentiable functions of t, you know that both x
and y approach zero as t approaches zero. Moreover, because f is a differentiable
function of x and y, you know that w 共w兾x兲 x 共w兾y兲 y 1 x 2 y,
where both 1 and 2 → 0 as 共x, y兲 → 共0, 0兲. So, for t 0
w w x w y
x
y
1
2
t
x t
y t
t
t
from which it follows that
dw
w w dx w dy
dx
dy
w dx w dy
lim
0
0
.
t→0
dt
t
x dt
y dt
dt
dt
x dt
y dt
冢 冣
冢 冣