Question

explain why the function is discontinuous at the given number a. (select all that ap
f(x)=left{\begin{array}{ll}\frac{1}{x + 4} & \text { if } x
eq-4 \\ 1 & \text { if } x=-4end{array} quad a=-4
ight.
( f(-4) ) is undefined.
( lim _{x
ightarrow-4} f(x) ) is not finite.
( f(-4) ) is defined and ( lim _{x
ightarrow-4} f(x) ) is finite, but they are not equal.
( lim _{x
ightarrow-4^{+}} f(x) ) and ( lim _{x
ightarrow-4^{-}} f(x) ) are finite, but are not equal.
none of the above
sketch the graph of the function.

Explanation:

Step1: Analyze \( f(-4) \)

Given \( f(x)=

$$\begin{cases}\frac{1}{x + 4}&x eq - 4\\1&x=-4\end{cases}$$

\), \( f(-4) = 1\), so \( f(-4) \) is defined.

Step2: Calculate \(\lim_{x

ightarrow - 4}f(x)\)
For \( x
eq - 4\), \( f(x)=\frac{1}{x + 4}\).
As \( x
ightarrow - 4^{+}\), let \( x=-4 + h\), \( h
ightarrow0^{+}\), then \( f(x)=\frac{1}{h}
ightarrow+\infty\).
As \( x
ightarrow - 4^{-}\), let \( x=-4 - h\), \( h
ightarrow0^{+}\), then \( f(x)=\frac{1}{-h}
ightarrow-\infty\).
So \(\lim_{x
ightarrow - 4}f(x)\) does not exist (is not finite).

Answer:

\(\lim_{x
ightarrow - 4}f(x)\) is not finite.