Let $f ( x ) = \left\{ \begin{array} { c c }
x ^ { 2 } & - 6

Question

Let $f ( x ) = \left\{ \begin{array} { c c } 
x ^ { 2 } & - 6 < x < - 4 \
2 & x = - 4 \
6 x + 8 & - 4 < x \leq 7
\end{array} \right.$ Which of the following is true for $\lim _ { x \rightarrow - 4 } f ( x )$ ?

A)-16
B)0
C)2
D)16
E)Does not exist

Quizplus · Accepted Answer

The limit as $x$ approaches $-4$ from the left is $x^2$, which gives $(-4)^2 = 16$, and from the right is $6x + 8$, which gives $6(-4) + 8 = -24 + 8 = -16$. Since the limits from the left and right do not match, the overall limit does not exist.

Let f(x)={x2−6<x<−42x=−46x+8−4<x≤7f ( x ) = \left\{ \begin{array} { c c } x ^ { 2 } & - 6 < x < - 4 \\2 & x = - 4 \\6 x + 8 & - 4 < x \leq 7\end{array} \right.f(x)=⎩⎨⎧​x226x+8​−6<x<−4x=−4−4<x≤7​

Let $f ( x ) = \left\{ \begin{array} { c c } x ^ { 2 } & - 6 < x < - 4 \\2 & x = - 4 \\6 x + 8 & - 4 < x \leq 7\end{array} \right.$