# MATH 100, Section 103 Week2 Marked Homework Solutions 2009

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```					                                  MATH 100, Section 103
Week 2: Marked Homework Solutions
2009 Sep 25

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

−1 ≤ cos(1/x) ≤ 1 for all x near 0
√
and multiplying by x3/2   = 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.)

√
q
√                          1                                  5
2+ x +     1       √
2x2 +5x+1                   2x2 +5x+1                            x2        2
3. [3 each part] (a) limx→∞          4x+7
= limx→∞   x
1
(4x+7)
= limx→∞           7
4+ x
=   4
.
x
1
(b) limx→∞ (x3 − x5 ) = limx→∞ x5          x2
− 1 = limx→∞ x5 (−1) = −∞.
(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 contin-
uous at 0 (Theorem 8, p. 129).
√                        √
√1                     √1        x+√x2 −4x                x+ x2 −4x
(e) limx→+∞   x− x2 −4x
= limx→∞  x− x 2 −4x · x+ x2 −4x = limx→∞ x2 −(x2 −4x) =
√
q    √                     q
√                        1                      1+    1
x2 −4x           1+    1
(x2 −4x)
x+ x2 −4x               1+ x x2 −4x                    x2                         x2
limx→∞           =      limx→∞               = limx→∞                      = limx→∞                                  =
√ 4
4x                         4                           4                          4
1+ 1− x    1
limx→∞    4
= 2.
√                       √
2                  x− x2 −4x
(f) limx→−∞   √1
x+ x2 −4x
= limx→−∞ x+√x2 −4x · x−√x2 −4x = limx→−∞ x2 −(x2 −4x) =
1
x− x −4x q
√                          √                     q
√                     1    2 −4x            1+   1
x2 −4x           1+    1
(x2 −4x)
x− x2 −4x            1− x x                       x2                         x2
limx→−∞           = limx→−∞                = limx→−∞                  = limx→−∞                                      =
√ 4
4x                      4
√
4                          4
1+ 1− x    1                                           2 ).
limx→−∞    4
= 2 (Note: when x is negative, x = − x
√ √
f (5+h)−f (5)
4. [4] f (5) = limh→0                   = limh→0   1   √ 1         −   √1      = limh→0   1    √4− 4+h
√
h                    h    5+h−1           5−1               h     4+h · 4

2
√ √           √ √
(4)−(4+h)
= limh→0   1√4− 4+h √     √4+√4+h = limh→0 1 √          √    √ √
h  4+h · 4      4+ 4+h               h 4+h · 4 · ( 4+ 4+h)

= limh→0 √4+h · √4 −1 4+√4+h) = √4 · √4−1 4+√4) = − 16
·(
√
(
√           1

(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 (x) as x → 0± . In this question it would work, but in the next question it does not! You
can try it out.) Does limh→0 f (0+h)−f (0) exist? We need to look at the right- and left-hand
h
limits separately (why?).
f (h)−f (0)            2
The right-hand limit is limh→0+         h
= limh→0+ h = limh→0+ h = 0.
h
f (h)−f (0)                2
The left-hand limit is limh→0−        h
= limh→0− (−h ) = limh→0− (−h) =
h
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.

6. [7] (a) By Theorem 4, p. 171, the function must at least be continuous to be diﬀerentiable,
so from Question 1 we must have C = 0. To check diﬀerentiability at x = 0 (using C = 0),
we need to check: Does limh→0 f (0+h)−f (0) = limh→0 f (h)−f (0) exist? Since the formula for
h                   h
f (h) depends on whether h < 0 or h > 0, we need to consider left- and right-hand limits.
For h < 0,
f (0 + h) − f (0)        Ah2 + Bh
lim                    = lim              = B,
h→0−          h          h→0−       h
f (h)−f (0)
noting that f (0) = C = 0. On the other hand, for h > 0 we have limh→0+                h
=
3/2                √
limh→0+ h cos(1/h) = limh→0+ h cos(1/h), and we have
h
√      √          √
−1 ≤ cos(1/h) ≤ 1,   − h ≤ h cos(1/h) ≤ h       for h > 0,

3
√        √
and the limits of both − h and h are 0, as h → 0+ . Therefore by the Squeeze Theorem
for limits,
f (0 + h) − f (0)       √
lim+                   = lim+ h cos(1/h) = 0.
h→0           h          h→0

For the limit of the diﬀerence quotient to exist, the left- and right-hand limits must agree,
so
B=0
is required. The derivative f (0) = 0 is the common value of the left- and right-hand limits
of the diﬀerence quotient when B = 0. In summary, for f (0) to exist, we need

B = 0,   C = 0,    A can be any real number.

4

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