Math 500 – Intermediate Analysis
Homework 11 Solutions
Fall 2014
29.2 Prove that | cos(x) − cos(y)| ≤ |x − y| for all x, y ∈ R.
Solution: Clearly, the result holds if x = y. Now, let x, y ∈ R be arbitrary but
distinct. Then by the Mean Value Theorem, there exists a point x∗ between x
and y such that
cos(x) − cos(y) = − sin(x∗ )(x − y).
Taking absolute values and using the fact that | sin(z)| ≤ 1 for all z ∈ R, we
have
| cos(x) − cos(y)| = | sin(x∗ )| · |x − y| ≤ |x − y|
as claimed.
29.4 Let f and g be differentiable functions on an open interval I. Suppose a, b ∈ R
satisfy a < b and f (a) = f (b) = 0. Show that f 0 (x) + f (x)g 0 (x) = 0 for some
x ∈ (a, b).
Solution: Define the function h : I → R by h(x) = f (x)eg(x) and note that g is
differentiable on (a, b) and continuous on [a, b]. In particular, h(a) = h(b) = 0
and so, by Rolle’s theorem, there exists a x ∈ (a, b) such that h0 (x) = 0. Since
h0 (x) = (f 0 (x) + f (x)g 0 (x)) eg(x) and since eg(x) 6= 0, it follows that f 0 (x) +
f (x)g 0 (x) = 0, as claimed.
29.11 Show sin(x) ≤ x for all x ≥ 0.
Solution: Define f (x) = x − sin(x) for x ∈ R and note that f is a differentiable
function on R with f 0 (x) = 1 − cos(x) ≥ 0 for all x ∈ R. It follows that f is
an increasing function on R so that, in particular, one has that for any given
a ∈ R the inequality f (x) ≥ f (a) for all x ≥ a holds. Taking a = 0 gives the
result.
29.13 Prove that if f and g are differentiable R, if f (0) = g(0) and if f 0 (x) ≤ g 0 (x)
for all x ∈ R, then f (x) ≤ g(x) for all x ≥ 0.
Solution: Define h : R → R by h(x) = g(x) − f (x) and notice that h(0) = 0
and h0 (x) ≥ 0 for all x ∈ R. For a fixed x > 0 then, the Mean Value Theorem
implies there exists a x∗ ∈ (0, x) such that
h(x) − h(0) = h0 (x∗ )x.
1
Since h(0) = 0 and h0 (x∗ ) ≥ 0 by hypothesis, it follows that h(x) ≥ 0 for all
x > 0 and hence h(x) = g(x) − f (x) ≥ 0 for all x ≥ 0 which proves the claim.
(Alternatively, you could have used Corollary 29.7 of the MVT.)
29.17 Let f and g be differentiable on an open interval I and consider a point a ∈ I.
Define h : I → R by
f (x), if x < a
h(x) =
g(x), if x ≥ a.
Prove that h is differentiable at a if and only if both f (a) = g(a) and f 0 (a) =
g 0 (a) hold.
Solution: First recall from Theorem 28.2 that in order for h to be differentiable
at a, it must be continuous at a. Taking limits as x → a and using the facts
that f and g are continuous at a, we see we must have
f (a) = lim− h(x) = lim+ h(x) = g(a).
x→a
x→a
Moreover, we have
h(x) − h(a)
=
x−a
(
f (x)−f (a)
,
x−a
g(x)−g(a)
,
x−a
if x < a
if x ≥ a
and hence taking limits as before we see that in order for h to be differentiable
we must have
f 0 (a) = lim−
x→a
h(x) − h(a)
h(x) − h(a)
= lim+
= g 0 (a).
x→a
x−a
x−a
Therefore, we conclude that if h is differentiable at a then it must be that
f (a) = g(a) and f 0 (a) = g 0 (a) as claimed. Finally, notice that if f (a) 6= g(a),
then h is not continuous at a and hence can not be differentiable, while if
f 0 (a) 6= g 0 (a) then the limit of the function
h(x) − h(a)
x−a
as x → a would not exist and hence h would not be differentiable at a. Thus h
is differentiable at a if and only if f (a) = g(a) and f 0 (a) = g 0 (a), as claimed.
3
x
30.2 (a) Determine the limit limx→0 sin(x)−x
.
Solution: As written, we have an indeterminate form 00 . By L’Hospital’s
3x2
rule, this limit would exist provided limx→0 cos(x)−1
exists, which by L’Hospital’s
6x
rule again would exist provided the limit limx→0 − sin(x)
exists, which again by
2
L’Hospital’s rule would exist provided that limx→0
6
− cos(x)
exists. This last limit
is equal to −6, and hence by L’Hospital’s rule we conclude that limx→0
exists and is equal to −6.
x3
sin(x)−x
2
(d) Determine the limit limx→0 (cos(x))1/x .
Solution: As written, this is of indeterminate form 1∞ . In order to apply
L’Hospital’s rule, we notice that
2
(cos(x))1/x = e
ln(cos(x))
x2
and hence, since the function ex is continuous on R, it is enough to analyze
the function ln(cos(x))
as x → 0. As this function is of indeterminate form 00 , by
x2
L’Hospital’s rule this limit exists provided the limit limx→0 − tan(x)
exists, which
2x
− sec(x)2
again by L’Hospital’s rule exists provided the limit limx→0
exists. Since
2
−1
this last limit does indeed exist and equal 2 , it follows by L’Hospital’s rule
that
1
ln(cos(x))
=− .
lim
2
x→0
x
2
Therefore,
2
lim (cos(x))1/x = e−1/2 .
x→0
3