2022 WTW114 Worksheet 5 Selected answers

• Remember that in general it is not enough to simply arrive at the correct answer, you need a solid argument
to support your conclusions. So even if your final answer matches with one here, it does not necessarily mean
that you mastered the problem.
• Please make use of the padlet, discord, signal message group, tutorials and consultions to get more feedback
should you need it.

Selected answers.


3 if/as x < 0,

2 if/as x = 0,
1(a) One possible answer is g(x) = and the sketch is


3 if/as 0 < x ≤ 4,
4 if/as x < 4,


1
1(b) One possible answer is g(x) = (x−1)2 and the sketch is

2. g is continuous at x = 1. Use the Composition Law to prove that lim− g(x) = g(1) = lim− g(x).
x→1 x→1
(
−1 if x ≤ 0
3. False. f (x) = .
1 if x > 0
4. Now f continuous everywhere means that f is continuous at 2 and 3. Continuity at 2 means that f (2) =
lim f (x) and continuity at 3 means f (3) = lim f (x). This gives
x→2− x→3+

7 = 2p + p
7 = 9q − p

56 21
Solving this you get p = 19 and q = 19 .

5. Show that −x4 ≤ f (x) ≤ x4 for all x ∈ R. How apply the Squeeze Theorem to prove that lim f (x) = f (0).
x→0

Page 1/3
 f f
6. Apply the Quotient Law for limits to prove that lim (x) = (a). Carefully explain why the Quotient Law
x→a g g
applies, and where you use the continuity of f and g.
7. Let L = f (a) = g(a). By continuity of f and g at a, lim f (x) = L = lim g(x). Because f (x) ≤ h(x) ≤ g(x) for
x→a x→a
all x ∈ R, it follows from the Squeeze Theorem that lim h(x) = L. But L = f (a) ≤ h(a) ≤ g(a) = L so that
x→a
L = h(a). Therefore lim h(x) = h(a) so that h is continuous at a.
x→a

8. He used the Product Law when he was not allowed to, since lim sin(1/x) does not exist.
x→0

3
9(a) Let f (x) = x − x − 1. Explain why f is continuous on [1, 2]. Verify that 0 is between f (1) and f (2). Apply the
IVT.
√
9(b) Let f (x) = x2 − 1 + x. Explain why f is continuous on [1, 2]. Verify that 0 is between f (1) and f (2). Apply
the IVT.

10. Apply the IVT to the function h(x) = f (x) − g(x). Carefully explain why the conditions of the IVT are satisfied.
11. The function f is a polynomial, hence continuous on [0, 1]. We have f (0) = −3 and f (1) = a + b − 3. But it is
given that a + b = 4. Therefore f (1) = 4 − 3 = 1. Hence f (0) < 0 < f (1). By the Intermediate Value Theorem,
there exists at least one number c ∈ (0, 1) so that f (c) = 0.
2
12(a) lim = 0.
x→∞ x
12(b) lim cos x DNE. [Look at the graph]
x→∞
p
5
12(c) For x > 5/2,
2x5 + x x5 (2 + 14 2 + 14
5
= 5 10 x = 10 x .
10 − 4x x ( x5 − 4) x5 − 4
We have    
1 10
lim 2+ 4 = 2 and lim − 4 = −4 ̸= 0.
x→∞ x x→∞ x5
Hence by the Quotient Law,
2x5 + x 2 + x14 1
lim 5
= lim 10 =− .
x→∞ 10 − 4x x→∞ 5 − 4 2
x

2x2 − 6x
12(d) lim = 0.
x→∞ x3 + 1
√
4x2 + 5
12(e) lim = −2. See L16, Example (i) on p 8 of the PDF.
x→−∞ x+5
p 1
12(f) lim ( x2 + x − x) = . See L16, Example (iii).
x→∞ 2
sin(x2 )
12(g) lim = 0, see L16, Example (iv) on p 8 of the PDF.
x→∞ x2
1
13(a) f (x) = x2 and g(x) = . Explain why this is a counterexample. Note: there are many more valid examples.
x
1
13(b) f (x) = x2 and g(x) = . Explain why this is a counterexample. Note: there are many more valid examples.
x2
13(c) f (x) = x2 and g(x) = x. Explain why this is a counterexample. Note: there are many more valid examples.
13(d) f (x) = 2x and g(x) = x. Explain why this is a counterexample. Note: there are many more valid examples.
14. Assume that y = 5 is a horizontal asymptote of y = f (x). Then

mx3 + 3x + 6 mx3 + 3x + 6
lim = 5 or lim = 5.
x→∞ m − 2x − 4x2 − 7x3 x→−∞ m − 2x − 4x2 − 7x3

Compute the limits to show that m = −35.

Page 2/3
 f (x)g(x)
15. For all x ∈ R so that g(x) ̸= 0, f (x) = . Because lim [f (x)g(x)] = 4 and lim g(x) = 1 ̸= 0 it follows
g(x) x→∞ x→∞
from the Quotient Law that

f (x)g(x) lim [f (x)g(x)] 4

lim f (x) = lim = x→∞ = = 4.
x→∞ x→∞ g(x) lim g(x) 1
x→∞

16. (a) f ′ (x) = 3x2 + 2x and Df = R

−1
(b) g ′ (x) = and Dg = R \ {−2}
(x + 2)2
−1
(c) h′ (x) = √ and Dh = (−1, ∞)
2(x + 1) x + 1
17. Use the definition of the derivative to show that:
1
• when x < 0, f ′ (x) = − ;
x2
• when 0 < x < 1, f ′ (x) = 2x;
• when x > 1, f ′ (x) = 2.
f is not differentiable at 0 because it is not continuous there.
Since
f (1 + h) − f (1) f (1 + h) − f (1)
lim = 2 = lim+
h→0− h h→0 h
f (1 + h) − f (1)
then f ′ (1) = lim exists hence f is differentiable at 1 and f ′ (1) = 2.
h→0 h
Hence f : R \ {0} → R is given by

1
 − 2 when x < 0


x
f ′ (x) =

 2x when 0 < x < 1
 2 when x ⩾ 1

and Df = R \ {0}.
18.

f (x) − f (a) x100 − a100

19. f ′ (a) = lim = lim =
x→a x−a x→a x−a
(x − a)(x99 + x98 a + x97 a2 + · · · + xa98 + a99 )
lim = . . . = 100a99
x→a x−a

Page 3/3