Homework #13
(29.3) Suppose f is differentiable on R and f (0) = 0, f (1) = 1 and f (2) = 1.
(a) Show f 0 (x) =
1
2
for some x ∈ (0, 2).
Proof. If f is differentiable on (0, 2), then, by the Mean Value Theorem, there is
a x ∈ (0, 2) such that
f 0 (x) =
1−0
1
f (2) − f (0)
=
= ,
2−0
2−0
2
which is what we wanted to show.
(b) Show f 0 (x) =
1
7
for some x ∈ (0, 2).
Proof. By the previous exercise, there is an x1 ∈ (0, 2) such that f 0 (x1 ) = 1/2.
Moreover, since f (1) = 1 and f (2) = 1, it follows from Rolle’s Theorem that
there is a x0 ∈ (1, 2) such that f 0 (x0 ) = 0. Let us assume, without loss of
generality, that x0 < x1 . Then there are x0 , x1 ∈ (0, 2) such that f 0 (x0 ) = 0 and
f 0 (x1 ) = 1/2; furthermore, f 0 (x0 ) < 1/7 < f 0 (x1 ). Then, by the Intermediate
Value Theorem for Derivatives, there is a x ∈ (x0 , x1 ) such that f 0 (x) = 1/7.
This completes the proof.
(29.16) Use Theorem 29.9 to obtain the derivative
of the inverse g = tan−1 = arctan of f
π π
where f (x) = tan(x) for x ∈ − 2 , 2 .
Solution. Let y = tan(x). By the class notes, we already know that y 0 = tan0 (x) =
sec2 (x). Therefore, by Theorem 29.9,
g 0 (y) =
1
f 0 (x)
=
1
sec2 (x)
=
1
1+tan2 (x)
=
1
.
1+y 2
(30.1) Find the following limits, if they exist.
(a) limx→0
e2x −cos x
x
Solution. Let f (x) = e2x −cos x and g(x) = x. First, we observe that limx→0 f (x) =
e0 − cos 0 = 1 − 1 = 0 and limx→0 g(x) = 0. Besides that, we have
f 0 (x)
2e2x + sin(x)
=
lim
= 2e0 + 0 = 2.
x→0 g 0 (x)
x→0
1
lim
1
Math 320: Analysis I
Spring 2015
Therefore, by L’Hospital’s Rule,
2e2x + sin(x)
e2x − cos x
= lim
= 2.
x→0
x→0
x
1
lim
(b) limx→0
1−cos x
x2
Solution. Let f (x) = 1 − cos x and g(x) = x2 . Notice that limx→0 f (x) =
1 − cos 0 = 1 − 1 = 0 and limx→0 g(x) = 0. The limit
f 0 (x)
sin x
= lim
x→0 g 0 (x)
x→0 2x
lim
(1)
still gives us an indeterminate 0/0, so we consider the limit
cos x
1
f 00 (x)
= lim
= .
00
x→0 2
x→0 g (x)
2
lim
(2)
By L’Hospital Rule and (2) above, the limit in (1) exists and equals 1/2; and by
L’Hospital Rule again and what we have just concluded for (1),
1 − cos x
cos x
1
= lim
= .
2
x→0
x→0 2
x
2
lim
(c) limx→∞
x3
e2x
Solution. Let f (x) = x3 and g(x) = e2x . Notice that limx→∞ |g(x)| = ∞. Now
we consider the limit
f 000 (x)
6
= lim
= 0.
x→∞ g 000 (x)
x→∞ 8e2x
lim
(3)
By L’Hospital Rule and (3) above, the limit
f 00 (x)
6x
= lim
x→∞ g 00 (x)
x→∞ 4e2x
lim
(4)
exists and equals 0 (notice that limx→∞ |4e2x | = ∞, thus satisfying one of the
hypotheses of L’Hospital’s Rule). Analogously, by L’Hospital Rule and (4) above,
the limit
f 0 (x)
3x2
= lim
lim 0
(5)
x→∞ 2e2x
x→∞ g (x)
exists end equals 0. Finally, by L’Hospital Rule and (5), we conclude that
x3
3x2
=
lim
= 0.
x→∞ e2x
x→∞ 2e2x
lim
2
Math 320: Analysis I
√
(d) limx→0
Spring 2015
√
1+x− 1−x
x
√
√
Solution. Let f (x) = 1 + x − 1 − x and g(x) = x. Then limx→0 f (x) =
limx→0 g(x) = 0. We have
0 (x)
limx→0 fg0 (x)
= limx→0 12 · (1 + x)−1/2 + 12 (1 − x)−1/2
i
h
1
1
+ √1−x
= 12 limx→0 √1+x
=
1
2 (1
+ 1) = 1.
Therefore, by L’Hospital’s Rule,
√
√
1
1+x− 1−x
1
−1/2
−1/2
lim
= lim
· (1 + x)
+ (1 − x)
= 1.
x→0
x→0 2
x
2
(30.2) Find the following limits, if they exist.
(a) limx→0
x3
sin x−x
Solution. Let f (x) = x3 and g(x) = sin x − x. Consider the limit
6
f 000 (x)
= lim −
= −6.
lim 000
x→0
x→0 g (x)
cos x
Since f 00 (x) = 6x, g 00 (x) = − sin x and limx→0 f 00 (x) = limx→0 g 00 (x) = 0, one of
the conditions of L’Hosptial’s Rule is satisfied, and we have
f 00 (x)
f 000 (x)
=
lim
= −6.
x→0 g 00 (x)
x→0 g 000 (x)
lim
We also have f 0 (x) = 3x2 , g 0 (x) = cos x − 1 and limx→0 f 0 (x) = limx→0 g 0 (x) = 0.
Thus L’Hospital’s Rule can be used once again, and we get
f 0 (x)
f 00 (x)
=
lim
= −6.
x→0 g 0 (x)
x→0 g 00 (x)
lim
Finally, notice that limx→0 f (x) = limx→0 g(x) = 0, so we can apply L’Hospital’s
Rule once more. This implies
f (x)
f 0 (x)
= lim 0
= −6.
x→0 g(x)
x→0 g (x)
lim
(b) limx→0
tan x−x
x3
Solution. Let f (x) = tan x − x and g(x) = x3 . We begin by calculating the
third derivative of f (x) and g(x); we get f 000 (x) = 2 sec2 x(tan2 x + sec2 x) and
g 000 (x) = 6. Thus
2 sec2 x(tan2 x + sec2 x)
f 000 (x)
2
1
=
lim
= = .
x→0 g 000 (x)
x→0
6
6
3
lim
3
Math 320: Analysis I
Spring 2015
Since f 00 (x) = 2 sec2 x tan x, g 00 (x) = 6x and limx→0 f 00 (x) = limx→0 g 00 (x) = 0,
one of the conditions of L’Hospital’s Rule is satisfied, and we have
f 00 (x)
f 000 (x)
1
=
lim
= .
00
000
x→0 g (x)
x→0 g (x)
3
lim
Now, we evaluate limx→0 [f 0 (x)/g 0 (x)]. Since f 0 (x) = sec2 x − 1, g 0 (x) = 3x2 and
limx→0 f 0 (x) = limx→0 g 0 (x) = 0, we can make use of L’Hospital’s Rule again.
Hence
f 0 (x)
f 00 (x)
1
lim 0
= lim 00
= .
x→0 g (x)
x→0 g (x)
3
Finally, observe that limx→0 f (x) = limx→0 g(x) = 0; so L’Hosptial’s Rule is
applied one last time, and we get
f 0 (x)
1
f (x)
= lim 0
= .
x→0 g (x)
x→0 g(x)
3
lim
(c) limx→0
1
sin x
−
1
x
Solution. We want to find the limit
1
1
x − sin x
lim
−
= lim
.
x→0 sin x
x→0 x sin x
x
Let f (x) = x − sin x and g(x) = x sin x; then limx→0 f (x) = limx→0 g(x) = 0.
Now, consider the limit
f 00 (x)
sin x
= lim
= 0.
00
x→0 g (x)
x→0 2 cos x − x sin x
lim
(6)
By L’Hospital’s Rule and (6) above, the limit
f 0 (x)
1 − cos x
= lim
0
x→0 g (x)
x→0 x cos x + sin x
lim
(7)
exists and is equal to 0 (notice that limx→0 (1 − cos x) = limx→0 (x cos x + sin x) =
0, thus satisfying one of the hypotheses of L’Hospital’s Rule). Therefore, by
L’Hospital’s Rule and (7), we conclude that
x − sin x
1 − cos x
= lim
= 0.
x→0 x cos x + sin x
x→0 x sin x
lim
2
(d) limx→0 (cos x)1/x
Solution. First, notice that
1/x2
2
(cos x)1/x = eln(cos x)
Now, let us evaluate the limit
ln(cos x)
.
x→0
x2
lim
4
1
= e x2 ln(cos x) .
Math 320: Analysis I
Spring 2015
Let f (x) = ln(cos x) and g(x) = x2 ; then limx→0 f (x) = limx→0 g(x) = 0. We
begin by taking the limit
− sec2 x
1
f 00 (x)
=
lim
=− .
x→0
x→0 g 00 (x)
2
2
lim
(8)
Hence, by L’Hospital’s Rule and (8) above, the limit
f 0 (x)
− tan x
= lim =
x→0 g 0 (x)
x→0
2x
lim
(9)
exists and is equal to −1/2 (notice that limx→0 − tan x = limx→0 2x = 0, and
thus one of the hypotheses of L’Hospital’s Rule applies). Hence, by L’Hospital’s
Rule and (9),
ln(cos x)
− tan x
1
lim
= lim
=− .
2
x→0
x→0
x
2x
2
This implies that
1
2
lim (cos x)1/x = lim e x2 ln(cos x) = e−1/2 .
x→0
x→0
(30.6) Let f be differentiable on the some interval (c, ∞) and suppose limx→∞ [f (x)+f 0 (x)] =
L, where L is finite. Prove limx→∞ f (x) = L and limx→∞ f 0 (x) = 0.
x
Proof. Since f (x) = f (x)e
and limx→∞ |ex | = ∞, we can apply L’Hospital’s Rule to
ex
limx→∞ f (x) by evaluating the limit
(f (x)ex )0
f 0 (x)ex + f (x)ex
=
lim
= lim [f 0 (x) + f (x)] = L;
x→∞
x→∞
x→∞
(ex )0
ex
lim
therefore,
(f (x)ex )0
= L.
x→∞
(ex )0
lim f (x) = lim
x→∞
Given that limx→∞ [f (x) + f 0 (x)] = limx→∞ f (x) + limx→∞ = L, it follows that
f 0 (x) = 0, and the statement is proven.
5