AP Calculus AB Quiz
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LIMITS AND CONTINUITY
Question 1
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Let
g
(
x
)
=
{
1
x
−
3
if
x
≠
3
5
if
x
=
3
g(x) = \begin{cases} \frac{1}{x-3} & \text{if } x \ne 3 \\ 5 & \text{if } x = 3 \end{cases}
g
(
x
)
=
{
x
−
3
1
5
if
x
=
3
if
x
=
3
To analyze the limit of the function
g
(
x
)
g(x)
g
(
x
)
as
x
x
x
approaches 3, which procedure is necessary?
A
Use direct substitution into the expression
1
x
−
3
\frac{1}{x-3}
x
−
3
1
because it defines the function near
x
=
3
x=3
x
=
3
.
B
Evaluate the one-sided limits as
x
→
3
−
x \to 3^-
x
→
3
−
and
x
→
3
+
x \to 3^+
x
→
3
+
to determine if the function approaches
∞
\infty
∞
or
−
∞
-\infty
−
∞
.
C
Conclude the limit is 5 because the function is explicitly defined as
g
(
3
)
=
5
g(3)=5
g
(
3
)
=
5
.
D
Factor the denominator and simplify the expression before applying direct substitution.
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