Fall 2014 – MATH 151, Sections 549-551
Quiz #3 Solutions
Problem 1. Find the vertical asymptotes of the function y =
x
.
x2 −x−2
Solution. The vertical asymptotes occur at the vertical lines given by the (real) roots of
the denominator, if any exist. Factoring x2 − x − 2, we see that
y=
x
(x − 2)(x + 1)
so that the dominator has roots x = 2 and x = −1. Thus the correct answer is (b).
Problem 2. Find limx→π− csc(x).
1
Solution. By definition, csc(x) = sin(x)
. Since sin(x) → 0 and x → π − , and sin(x) > 0
for 0 < x < π, we see that csc x > 0 for 0 < x < π. Thus limx→π− csc(x) = ∞, and the
correct answer is (a).
Problem 3. Find limx→0 x2 sin(1/x)
Solution. Since −1 ≤ sin(1/x) ≤ 1, we see that −x2 ≤ x2 sin(1/x) ≤ x2 . Since
limx→0 −x2 = limx→0 x2 = 0, the Squeeze Theorem tells us that limx→0 x2 sin(1/x) = 0,
so the correct answer is (c).
Problem 4. Find limx→1.5
2x2 −3x
|2x−3|
Solution. We exame the left- and right-hand limits of (2x2 − 3x)/|2x − 3| as x → 1.5. If
x < 1.5, then 2x − 3 < 0, so |2x − 3| = −(2x − 3). In this case, we have
lim −
x→1.5
2x2 − 3x
x(2x − 3)
= lim − −
= lim − −x = −1.5.
x→1.5
x→1.5
|2x − 3|
2x − 3
On the other hand, if x > 1.5, then 2x − 3 > 0 so that |2x − 3| > 0. In this case, we have
lim +
x→1.5
2x2 − 3x
x(2x − 3)
= lim +
= lim + x = 1.5.
x→1.5
x→1.5
|2x − 3|
2x − 3
Since the left- and right-hand limits do not agree, the limit does not exists. Hence the
correct answer is (d).
1