Research Methods in Mathematics
Homework 5 with solutions
T. PERUTZ
(1) Working carefully from the definition of a limit, prove that for any function f
(a)
(a)
defined near a, f (a+h)−f
→ L as h → 0 if and only if f (x)−f
→ L as x → a.
h
x−a
f (a+h)−f (a)
Solution. Suppose
→ L as h → 0. Then, given > 0, there exists
h
(a)
δ > 0 such that 0 < |h| < δ ⇒ | f (a+h)−f
− L| < . Let h = x − a. Then, if
h
f (x+h)−f (x)
(a)
(a)
−
L|
=
|
− L| < . Thus f (x)−f
→ L.
0 < |x − a| < δ , | f (x)−f
x−a
h
x−a
f (x)−f (a)
Conversely, if x−a → L then, given > 0, there exists δ > 0 such that
(a)
0 < |x − a| < δ ⇒ | f (x)−f
x−a − L| < . Let x = a + h. Then, if 0 < |x − a| < δ ,
(a)
(a)
(a)
| f (a+h)−f
− L| = | f (x)−f
− L| < , so f (a+h)−f
→ L.
h
h
h
(2) Starting from the definition, differentiate: (a) f (x) = (x + 1)2 ; (b) f (x) = x4 ; (c)
f (x) = x + x−1 (for x 6= 0).
Solution. (a) We have [f (x+h)−f (x)]/h = [(x+h+1)2 −(x+1)2 ]/h = 2(x+1+
h)+h = 2(x+1)+3h. In the limit as h → 0, this expression approaches 2(x+1).
Thus f 0 (x) = 2(x+1). (b) We have [f (x+h)−f (x)]/h = [(x+h)4 −x4 ]/h = 4x3 +
6x2 h + 4xh2 + h3 . In the limit as h → 0, 6x2 h + 4xh2 + h3 → 0 (because limits
are interchangeable with sums and products). Thus [f (x + h) − f (x)]/h → 4x3 .
So f 0 (x) = 4x3 . (c) To differentiate f (x) = x + x−1 , we can observe that the
derivative of g(x) := x is g0 (x) = 1, while that of h(x) = x−1 is −x−2 as proved
in class. Thus f 0 (x) = g0 (x) + h0 (x) = 1 − x−2 . Alternatively, one can use a
direct argument, similar to that used in class for h.
(3) An example of a function which is differentiable but whose derivative is not
continuous: Let s(x) = x2 sin x−1 for x 6= 0, and define s(0) = 0.
(a) Sketch the graph of s.
(b) Use the product rule and the derivative of sin to calculate the derivative of s
at x when x 6= 0.
(c) Use the definition of the derivative to show that s0 (0) = 0.
(d) Prove that the derivative s0 is not a continuous function at x = 0.
Solution
(a) See figure—the graph of s is shown sandwiched between two parabolas
y = ±x2 .
2
T. Perutz
0.3
0.2
0.1
0
-0.1
-0.2
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
-0.6
-0.3
(b) When x 6= 0, we have
s0 (x) = x2
d
(sin x−1 ) + 2x sin x−1 = − cos x−1 + 2x sin x−1
dx
using the product rule, the chain rule and the derivatives of x2 , x−1 and sin.
(c) Note that the formula derived in (b) is not valid at x = 0, nor is it valid if we
try to take its limit as x → 0. Instead note that
s0 (0) = lim h−1 (s(h) − s(0)) = lim h sin h−1 .
h→0
h→0
I claim that this limit is 0. To prove this, let > 0 be given. Let δ = . Then,
when 0 < |h| < δ ,
|h sin h−1 | ≤ |h| < δ = where the first inequality holds because sin(anything) ≤ 1. The result follows.
(d) We have s0 (x) = − cos x−1 + 2x sin x−1 for x 6= 0, while s0 (0) = 0. We
want to show this is not continuous as 0, i.e., that − cos x−1 + 2x sin x−1 6→ 0
as x → 0. By a similar argument to (c), 2x sin x−1 → 0 as x → 0. Thus if
− cos x−1 + 2x sin x−1 → 0 then − cos x−1 → 0. We will show that this is not
the case, so the premise was false.
Let xn = (2πn)−1 then − cos x−1 = −1. For any δ > 0 we may choose, there
is some n with 0 < xn < δ (namely, any integer n > (2πδ)−1 ). So there is no
δ > 0 such that | − cos x−1 − 0| < 1/2. This shows that − cos x−1 6→ 0, and
thus proves that s0 is not continuous at 0.
(4) Let p(x) = x4 + ax3 + bx2 + cx + d for constants a, b, c, d . Show that there
exists constants k > 0 and l > 0 such that
p(x) > kx4
whenever |x| > l. [Examine the argument we used to prove that every odd
degree polynomial has a root.]
Research Methods in Mathematics Homework 5 with solutions
3
This one was a tough nut! An easier exercise would have been to show that
there’s an l > 0 so that p(x) > 0 when |x| > l.
We will show that for any k in (0, 1), one can find an l so that p(x) > kx4 when
|x| > l. Note that when |x| > 1, we have x4 > |x|3 > x2 > |x| ≥ 1. Using this,
we find
|ax3 + bx2 + cx + d| ≤ |a||x|3 + |b||x|2 + |c|x + |d|
≤ (|a| + |b| + |c| + |d|)|x|3
≤ 4 max{|a|, |b|, |c|, |d|}|x|3
= C|x|3
with C = 4 max{|a|, |b|, |c|, |d|}. Thus, if |x| > max{C/(1 − k), 1} then
C < (1 − k)|x| and so
|ax3 + bx2 + cx + d| ≤ C|x|3 < (1 − k)x4 .
So: take l = max{C/(1 − k), 1}. Then, when |x| > l,
p(x) ≥ x4 − |ax3 + bx2 + cx + d| > x4 − (1 − k)|x|4 = kx4 .