AP Calculus BC Quiz
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LIMITS AND CONTINUITY
Question 1
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Let
p
(
x
)
=
2
(
x
+
1
)
2
p(x)=\dfrac{2}{(x+1)^2}
p
(
x
)
=
(
x
+
1
)
2
2
. Which statement about the limit as
x
→
−
1
x\to-1
x
→
−
1
is correct?
A
There is a horizontal asymptote at
x
=
−
1
x=-1
x
=
−
1
and
lim
x
→
−
1
p
(
x
)
=
∞
\lim_{x\to-1} p(x)=\infty
lim
x
→
−
1
p
(
x
)
=
∞
.
B
There is a vertical asymptote at
x
=
−
1
x=-1
x
=
−
1
and
lim
x
→
−
1
p
(
x
)
=
∞
\lim_{x\to-1} p(x)=\infty
lim
x
→
−
1
p
(
x
)
=
∞
.
C
There is a vertical asymptote at
y
=
−
1
y=-1
y
=
−
1
and
lim
x
→
−
1
p
(
x
)
=
−
∞
\lim_{x\to-1} p(x)=-\infty
lim
x
→
−
1
p
(
x
)
=
−
∞
.
D
There is no vertical asymptote because the exponent makes the limit finite.
E
There is a vertical asymptote at
x
=
−
1
x=-1
x
=
−
1
and
lim
x
→
−
1
p
(
x
)
=
0
\lim_{x\to-1} p(x)=0
lim
x
→
−
1
p
(
x
)
=
0
.
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