WORKSHEET 5
MATH 220, FALL 2016
SOLUTIONS
(1) (a) We need to compute the limit
lim |x|x
x→0
and show that it equals f (0) = 1. Since the function ex is continuous,
lim |x|x = elimx→0 x ln |x| ,
x→0
provided limx→0 x ln |x| exists. To examine the latter limit, write it as
lim
x→0
ln |x|
.
x−1
L'Hopitals rule applies to this limit, which gives
ln |x|
x−1
=
−
lim
= 0.
x→0 x−1
x→0 x−2
lim
Therefore limx→0 |x|x = e0 = 1 = f (0), so our function is continuous at x = 0.
(b) For x 6= 0 we use logarithmic dierentiation,
d
d x ln |x|
|x|x =
e
= |x|x (1 + ln |x|).
dx
dx
(c) To examine whether f (x) is dierentiable at x = 0 we write down the dierence quotient
|h|h − 1
.
h→0
h
lim
We calculate this indeterminate limit using L'Hopital's rule, using our expression for f 0 (h) from (b)
when h 6= 0:
|h|h − 1
= lim |h|h (1 + ln |h|).
h→0
h→0
h
lim
Since |h|h → 1 as h → 0, and since 1 + ln |h| → −∞ as h → 0, we see that
|h|h − 1
= lim |h|h (1 + ln |h|) = −∞.
h→0
h→0
h
lim
In other words, f 0 (0) does not exist.
1
WORKSHEET 5
MATH 220, FALL 2016
SOLUTIONS
2
(2) (a) The derivative of f (x ± h) with respect to h is ±f 0 (x ± h). Therefore by L'Hopital's rule,
f (x + h) − f (x − h)
f 0 (x + h) − (−f 0 (x − h))
= lim
= f 0 (x).
h→0
h→0
2h
2
lim
(b) Let f (x) = |x|. Then (f (0 + h) − f (0 − h))/(2h) = 0 for all h, so the modied dierence quotient has
a limit. Nevertheless, f is not dierentiable at x = 0.
(3) (a) The domain of arcsin(x) is [−1, 1], and since the range of arcsin(x) (namely [−π/2, π/2]) is contained
in the domain of sin(x), the domain of f (x) is [−1, 1]. We in fact have that sin(arcsin(x)) = x for x in
[−1, 1]. Therefore f 0 (x) = 1. The graph of f (x) is the graph of y = x with x in [−1, 1].
(b) The domain of sin(x) is (−∞, ∞) and its range in [−1, 1]. Therefore the domain of arcsin(sin(x)) is
(−∞, ∞). The range is the range of arcsin(x), namely [−π/2, π/2]. Since arcsin(x) is the inverse of
sin(x) restricted to [−π/2, π/2],
g(π/4) = π/4,
g(−π/4) = −π/4.
Now sin(3π/4) = sin(π/4), we see that g(3π/4) = π/4. Similarly, sin(5π/4) = sin(−π/4), so g(5π/4) =
−π/4. The derivative of g(x) is cos(x)/| cos(x)|, which is 1 if x − 2kπ is in (−π/2, π/2) for some integer
k and −1 if x − 2kπ is in (π/2, 3π/2) for some integer k .
(c) The domain of h is (−∞, ∞) and its range is [0, π]. The derivative is − cos(x)/| cos(x)|. We can
simplify h(x) by noting that if the length of the opposite side is sin(x) and the hypotenuse has length
q
1, then the adjacent side has length 1 − sin2 (x) = | cos(x)|