Solutions for Assignment 4
1. Suppose that f : R → R is differentiable on all of R, that f (0) = 0, and that
1 ≤ f 0 (x) ≤ 2 for all x ≥ 0. Prove that x ≤ f (x) ≤ 2x for all x ≥ 0.
Solution Fix x ≥ 0. If x = 0, then f (x) = 0 so 0 ≤ f (x) ≤ 0 and we are
done. Suppose x > 0. Since f is differentiable, f is continuous on [0, x] and
differentiable on (0, x) in particular. So the MVT gives s ∈ (0, 1) with
f (x) − f (0)
f (x)
=
.
x−0
x
Since 1 ≤ f 0 (s) ≤ 2, we must have
f 0 (s) =
1≤
f (x)
≤2
x
=⇒
x ≤ f (x) ≤ 2x
as needed.
2. Suppose that f : R → R is C 2 , that c ∈ R, that f 0 (c) = 0, and that f 00 (c) > 0.
(a) Prove that there exists δ > 0 such that |x − c| < δ =⇒ f 00 (x) > 0.
(b) Use Taylor’s theorem to prove that, with δ as in (a), we have f (x) > f (c)
for all x ∈ (c − δ, c + δ) provided x 6= c.
Solution. (a) Since f is C 2 , the second derivative f 00 is continuous everywhere,
and in particular at c. Taking = f 00 (c) in the definition of “f 00 is continuous
at c” shows that there exists δ > 0 such that
|x − c| < δ =⇒ |f 00 (x) − f 00 (c)| < f 00 (c).
But
|f 00 (x)−f 00 (c)| < f 00 (c)
⇐⇒ f 00 (x) − f 00 (c) < f 00 (c) and f 00 (c) − f 00 (x) < f 00 (c)
=⇒ f 00 (c) − f 00 (x) < f 00 (c)
=⇒ 0 < f 00 (x),
so this δ will do.
(b) Note that f is C 2 so satisfies the hypothese of Taylor’s Theorem. Fix
x ∈ (c − δ, c) ∪ (c, c + δ). Then Taylor’s theorem implies that there exists d
between x and c such that
f 00 (d)
f (x) = f (c) + f 0 (c)(x − c) +
(x − c)2 .
2
0
0
Since f (c) = 0, the term f (c)(x − c) vanishes, and can be removed. Since d is
between x and c, we have |d − c| < |x − c| < δ, and f 00 (d) > 0. Because x 6= c
we have (x − c)2 > 0, f 00 (d)(x − c)2 > 0, and
f (x) = f (c) +
f 00 (d)
(x − c)2 > f (c).
2
1
2
3. Prove that the following limit exists and find it:
x
1
lim
−
.
x→1
x − 1 ln x
Solution. We have
x
1
x ln x − x + 1
−
=
.
x − 1 ln x
(x − 1) ln x
The algebra of limits implies that x ln x − x + 1 → 0 as x → 1. We know
ln x → 0 as x → 1. So, with a view to applying l’Hôpital’s rule, we look at the
derivative of the numerator divided by the derivative of the denominator:
x(1/x) + ln x − 1
ln x
(0.1)
=
.
(x − 1)(1/x) + ln x
1 − (1/x) + ln x
The algebra of limits implies that 1 − (1/x) + ln x → 0 as x → 1 so we will
have to take derivatives again and consider:
(1/x)
x
(1/x)
x2
=
=
,
2
2
2
(1/x ) + (1/x)
(1/x ) + (1/x) x
1+x
which approaches 1/2 as x approaches 1. Since all the functions in the numerators and denominators are differentiable near a = 1 and the hypotheses of
L’Hôpital’s rule are satisfied, L’Hôpital’s rule implies first that from (0.1)
ln x
→ 1/2 as x → 1
1 − (1/x) + ln x
and second that
1
x
−
→ 1/2 as x → 1
x − 1 ln x
as well.