Fall 2016, Math 409, Section 502
Solutions to some problems from the ninth assignment
Exercise 1. Let f : (a, b) → R be a differentiable function, x0 ∈ (a, b), and
λ ∈ R. If lim f 0 (x) = λ, prove that f 0 is continuous at x0 and f 0 (x0 ) = λ.
x→x0
Hint: use the intermediate value theorem for derivatives.
Solution. By the limit characterization of continuity, it is sufficient to show
that f 0 (x0 ) = λ. Towards a contradiction, assume that f 0 (x0 ) 6= λ. We
shall assume f 0 (x0 ) > λ, as the case f 0 (x0 ) < λ is treated identically. Define
ε = f 0 (x0 ) − λ, which is strictly positive. By the definition of the limit of
a function, there exists δ > 0 so that for all x ∈ (a, b) with x 6= x0 and
|x − x0 | < δ, we have |f 0 (x) − λ| < ε/2.
Fix an arbitrary y0 in (a, b) ∩ (x0 − δ, x0 ). The lower triangle inequality
yields f 0 (y0 ) 6 λ + |f 0 (y0 ) − λ| < λ + ε/2 < f 0 (x0 ). Setting c = λ + ε/2,
the intermediate value theorem for derivatives yields that there exists z0 ∈
(y0 , x0 ) so that f 0 (z0 ) = c = λ + ε/2. However, z0 must be in (a, b) ∩
(x0 − δ, x0 ), therefore |f 0 (z0 ) − λ| < ε/2, which implies λ + ε/2 = f 0 (z0 ) 6
λ + |f 0 (z0 ) − λ| < λ + ε/2. In conclusion, λ + ε/2 < λ + ε/2, which is
absurd.
Comment: if the derivative f 0 of a function f is discontinuous at some point
x0 in the interior of the domain of f , then lim f 0 (x) cannot exist.
x→x0
Exercise 2. Prove that for all x ∈ [0, +∞) we have
0 6 x − sin(x) 6
x3
.
6
You may use that if f (x) = sin(x), then f 0 (x) = cos(x), f 00 (x) = − sin(x)
and −1 6 cos(x) 6 1 for all x ∈ R.
Hint: use Taylor’s formula and 3! = 6.
Solution. If f (x) = sin(x), then f is (at least) three times differentiable on R.
We start by observing that if g(x) = x − sin(x), then g 0 (x) = 1 − cos(x) > 0
for all x ∈ [0, ∞). Therefore g is increasing. For any x ∈ [0, ∞), x − sin(x) =
g(x) > g(0) = 0, i.e. the left part of the inequality holds.
We now show the right part of the inequality. If x = 0, then 0−sin(0) = 0,
hence the conclusion trivially holds. Otherwise, x > 0 and by Taylor’s
formula, there exists c ∈ (0, x) so that
f 00 (0) 2 f 000 (c) 3
x +
x , or in other words
2
6
− sin(0) 2 − cos(c) 3
x3
sin(x) = sin(0) + cos(0)x +
x +
x = x − cos(c) .
2
6
6
f (x) = f (0) + f 0 (0)x +
We deduce x − sin(x) 6 |x − sin(x)| = | cos(c)|x3 /6 6 x3 /6.
Exercise 3. Let f, g : R → R be differentiable functions so that g 0 (x) 6= 0
for all x ∈ R and and the function h : R → R with h(x) = f 0 (x)/g 0 (x) is
bounded. If g is uniformly continuous, show that f is uniformly continuous
as well.
Hint: use the generalized mean value theorem.
Solution. Define C = sup{|f 0 (x)/g 0 (x)| : x ∈ R}. By assumption, C is a
(non-negative) real number. We distinguish two cases, namely whether C =
0 or C > 0. In the first case, for all x ∈ R we have |f 0 (x)| 6 C|g 0 (x)| = 0,
hence f 0 (x) = 0 for all x ∈ R. We conclude that f is constant, hence it is
uniformly continuous.
Otherwise, C > 0. Fix ε > 0. Since g is uniformly continuous, there exists
δ > 0, so that for all x, y ∈ R with |x − y| < δ, we have |g(x) − g(y)| < ε/C.
Fix x, y ∈ R with |x − y| < δ. We will show that |f (x) − f (y)| < ε. If
x = y, then f (x) = f (y) and the conclusion holds. Otherwise, x 6= y and
we may assume that x < y. By the generalized mean value theorem there
exists c ∈ (x, y) so that f 0 (c)(g(x) − g(y)) = g 0 (c)(f (x) − f (y)) and hence
|f (x) − f (y)| = |f 0 (c)/g 0 (c)||g(x) − g(y)| < C(ε/C) = ε.