MA 421
HW #6
Fall 15
1. Evaluate the limits and prove your conclusion using the definition of limits.
(a) lim (x + 1)3
x→0
1−x
√
x→1 1 −
x
(b) lim
2. Use the definition to prove lim+
x→1
x
= ∞.
x−1
3. Prove or give a counterexample to the following statements, in which we assume that f
and g are functions defined on the indicated intervals. For your counterexamples, the
functions must be defined on the entire intervals indicated.
(a) If a function f is bounded on [a, b], then it is continuous on [a, b].
(b) If a function f is continuous on (a, b), then it is bounded on (a, b).
(c) If [f (x)]2 is continuous on (a, b), then f (x) is continuous on (a, b).
(d) If f + g and f are continuous on (a, b), then g is continuous on (a, b).
4. Prove Proposition 3.2.5 (i) and (iii).
5. Use Bolzano’s Intermediate Value Theorem to solve the given inequalities:
√
√
(a) (x − 3)(x − 3.1) > 0
2x + 1
(b)
≤3
x−5
6. Show that the equation 2x = 3x has a solution x = c for some c ∈ (0, 1).
7. If f and g are both uniformly continuous on E ⊂ R, prove that f + g is uniformly
continuous on E.
8. Determine if each function f is differentiable at the point indicated. If it is, find its
derivative at that point; if not, explain.
(
3x + 1
x<0
(a) At x = 0 where f (x) =
2
x + 3x + 1 x ≥ 0.
(
x2 x ∈ Q
(b) At x = 0 where f (x) =
0 x∈
/ Q.
(
x x∈Q
(c) At x = 0 where f (x) =
0 x∈
/ Q.
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MA 421
HW #6
Fall 15
9. Fill in the following blanks for the proof of the Quotient Rule:
Proof. Let f, g be differentiable at a with g(a) 6= 0. Then
0
f
(a) = lim
A
x→a
g
= lim
x→a
= lim
x→a
B
(Hint for part B: simplify A getting rid of double fractions)
C
(Hint for part C: add and subtract f (a)g(a) in numerator)
"
g(a) h
= lim
x→a g(x)g(a)
D
i
f (a) h
−
g(x)g(a)
E
i
#
(*)
Since g is differentiable at a, g is continuous at a by Theorem/Proposition F . And,
since g(a) 6= 0, we have 1/g is continuous at a by Theorem/Proposition G . Thus
1
=
x→a g(x)
lim
H
.
Since f and g are differentiable at a, both
g(x) − g(a)
f (x) − f (a)
and lim
x→a
x→a
x−a
x−a
lim
exist and equal f 0 (a), g 0 (a), respectively. Thus by using limit laws on (*), we get
0
f
,
(a) =
I
g
which proves the Quotient Rule.
10. (Alternative Proof of the Product Rule) Suppose that functions f and g are defined on
a neighborhood of x = a and are differentiable at x = a. Use the fact that
1
1
1
f g = (f + g)2 − f 2 − g 2
2
2
2
and the Chain Rule to prove that f g is also differentiable at x = a.
11. Recall the definitions of odd and even functions: A function f : D → R is even if
f (−x) = f (x) for all x ∈ D and is odd if f (−x) = −f (x) for all x ∈ D.
(a) Prove if f is an even function, then f 0 (x) is an odd function.
(b) Prove if f is an odd function, then f 0 (x) is an even function.
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