AP Calculus AB Quiz
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LIMITS AND CONTINUITY
Question 1
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Define
t
(
x
)
=
{
(
x
−
4
)
2
,
x
≠
4
7
,
x
=
4
t(x)=\begin{cases}(x-4)^2,&x\ne4\\7,&x=4\end{cases}
t
(
x
)
=
{
(
x
−
4
)
2
,
7
,
x
=
4
x
=
4
. Is
t
t
t
continuous at
x
=
4
x=4
x
=
4
, and why?
A
Yes;
t
(
4
)
=
7
t(4)=7
t
(
4
)
=
7
, so
t
t
t
is continuous at
4
4
4
.
B
No;
lim
x
→
4
t
(
x
)
=
7
\lim_{x\to4}t(x)=7
lim
x
→
4
t
(
x
)
=
7
but
t
(
4
)
=
0
t(4)=0
t
(
4
)
=
0
.
C
No;
lim
x
→
4
t
(
x
)
=
0
\lim_{x\to4}t(x)=0
lim
x
→
4
t
(
x
)
=
0
but
t
(
4
)
=
7
t(4)=7
t
(
4
)
=
7
.
D
Yes;
lim
x
→
4
t
(
x
)
=
7
\lim_{x\to4}t(x)=7
lim
x
→
4
t
(
x
)
=
7
equals
t
(
4
)
t(4)
t
(
4
)
.
E
No;
lim
x
→
4
t
(
x
)
\lim_{x\to4}t(x)
lim
x
→
4
t
(
x
)
does not exist since squaring creates a corner.
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