MATH 100, Section 110 (CSP)
Week 2: Marked Homework Solutions
2010 Sep 23
1. [8] The function Ax2 + Bx + C is continuous on −∞ < x < 0 for any A, B and C (it is a
polynomial, so it is continuous for all x), and x3/2 cos(1/x) is continuous on 0 < x < ∞ (the
three functions x3/2 , 1/x and cos(x) are all continous on 0 < x < ∞; sums, products and
compositions of continuous functions are continuous; the function cos(1/x) is the composition
of the continuous functions cos(x) and 1/x so it is continuous on 0 < x < ∞ (Theorem 9, p.
130); the product of the continuous functions x3/2 and cos(1/x) is continuous).
Therefore it remains only to check continuity at x = 0. This means we have to check
that i) limx→0− f (x), ii) limx→0+ f (x) and iii) f (0) all exist and iv) are all equal to the same
number.
i) limx→0− f (x) = limx→0− (Ax2 + Bx + C) = C,
ii) The cosine of any angle is between 1 and −1, therefore
and multiplying by x3/2
−1 ≤ cos(1/x) ≤ 1 for all x near 0
√
= x x (which is positive, for x > 0), we get
−x3/2 ≤ x3/2 cos(1/x) ≤ x3/2
for all x near 0.
We easily evaluate the right-hand limits of the expressions on the left and right above:
lim+ −x3/2 = 0,
lim+ x3/2 = 0.
x→0
x→0
Then by the Squeeze Theorem for limits, we have
lim x3/2 cos(1/x) = 0.
x→0+
iii) f (0) = C by the definition of the function f at 0.
Therefore f (x) is continuous at x = 0 if and only if C = 0. Since f (x) is already known
to be continuous on −∞ < x < 0 and on 0 < x < ∞, it follows that f is continuous at all
points on the real line if and only if C = 0. A and B can be any real numbers, their values
do not affect the continuity of f .
2. [4] See the graph below. It appears that there is an intersection of the graphs of y = tan x
and y = −x between x = 3π/2 (a vertical asymptote for y = tan x) and x = 2π (where
tan 2π = 0). Note that 3π/2 ≈ 4.712388981 and 2π ≈ 6.283185308, so a = 4.72 is larger
than 3π/2 and b = 6.28 is smaller than 2π. Consider the closed interval [a, b] = [4.72, 6.28].
To prove that an intersection in [a, b] exists, we rewrite tan x = −x as
x + tan x = 0,
which is equivalent. Let f (x) = x + tan x, and look for a solution of f (x) = 0 in [a, b].
Evaluate f (a) = f (4.72) ≈ −126.7 which is negative, and f (b) = f (6.28) ≈ 6.28 which
1
is positive. Observe that f is continuous on the closed interval [4.72, 6.28], because f is
continuous on the open interval (3π/2, 5π/2), and [4.72, 6.28] is contained in (3π/2, 5π/2).
Since N = 0 is between the endpoint values f (4.72) and f (6.28), by the Intermediate Value
Theorem there is at least one number c between a = 4.72 and b = 6.28 such that f (c) = 0,
i.e., tan x = −x has at least one solution x = c between 4.72 and 6.28. (From the graph,
it looks like there is only one such number, but the Intermediate Value Theorem cannot be
used to prove that.)
√
3. [3 each part] (a) limx→∞
2x2 +5x+1
4x+7
(b) limx→∞ (x3 − x5 ) = limx→∞ x5
1
x2
1
x
√
2x2 +5x+1
1
(4x+7)
x
= limx→∞
= limx→∞
− 1 = limx→∞ x5 (−1) = −∞.
q
2+ x5 +
4+ x7
1
x2
√
=
2
.
4
(c) limt→∞ cos(t) does not exist. As t → ∞, cos(t) oscillates infinitely often between +1 and
−1 so it does not get close to a single finite number L.
(d) limt→∞ cos(1/t) = cos (limt→∞ (1/t)) = cos(0) = 1 because the cosine function is continuous at 0 (Theorem 8, p. 129).
(e) limx→+∞
√
√
x+√x2 −4x
x+ x2 −4x
√1
·
=
lim
=
x→∞
2
2
2
x −(x2 −4x)
x− x −4x
x+ x −4x q
q
√
√
1
1
2
1+
x −4x
1+
(x2 −4x)
1+ x1 x2 −4x
x2
x2
limx→∞
=
lim
=
lim
x→∞
x→∞
4
4
4
√1
x− x2 −4x
√
x+ x2 −4x
=
√4x 4
1+ 1− x
limx→∞
= 21 .
4
limx→∞
= limx→∞
=
√
√
1
x−√x2 −4x
x− x2 −4x
√1
√
=
lim
·
=
lim
x→−∞ x+ x2 −4x
x→−∞ x2 −(x2 −4x) =
x+ x2 −4x
x− x2 −4x q
q
√
√
1
1
√
1+
x2 −4x
1+
(x2 −4x)
1− x1 x2 −4x
x2
x2
x− x2 −4x
limx→−∞
=
lim
=
lim
=
lim
x→−∞
x→−∞
x→−∞
4
4
4
√4x
√
1+ 1− x4
1
2
limx→−∞
= 2 (Note: when x is negative, x = − x ).
4
(f) limx→−∞
4. [4] f 0 (5) = limh→0
f (5+h)−f (5)
h
= limh→0
1
h
√ 1
5+h−1
2
−
√1
5−1
= limh→0
1
h
√
√
√
√4− 4+h
4+h · 4
=
√ √ √
(4)−(4+h)
√
√4+√4+h = limh→0 1 √
√
√ √
√4− 4+h
h 4+h · 4 · ( 4+ 4+h)
4+h · 4
4+ 4+h
1
√ √
√ √
= √4 · √4−1
= − 16
limh→0 √4+h · √4 −1
· ( 4+ 4+h)
( 4+ 4)
= limh→0
1
h
√
=
(which agrees with use of the power rule).
5. Using the definition of |x|, rewrite the function f (x) = x · |x| as
x · (−x) = −x2 if x < 0
f (x) =
x · x = x2
if x ≥ 0.
We carefully investigate the derivative at x = 0. As instructed, we use the definition of the
derivative. (CAUTION: You don’t always get the correct answer by simply taking the limits
of f 0 (x) as x → 0± . In this question it would work, but in the next question it does not! You
(0)
can try it out.) Does limh→0 f (0+h)−f
exist? We need to look at the right- and left-hand
h
limits separately (why?).
2
f (h)−f (0)
= limh→0+ hh = limh→0+ h = 0.
h
2)
f (h)−f (0)
= limh→0− (−h
= limh→0− (−h) =
h
h
The right-hand limit is limh→0+
The left-hand limit is limh→0−
0.
Since the right- and left-hand limits both exist, and agree, the limit exists and its value is
the derivative of f (x) at x = 0: thus f 0 (0) = 0.
6.
3