Question

explain why the function is discontinuous at the given number a. (select all that apply.)
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 \( f(-4) \)

When \( x = -4 \), for the piece - wise function \( f(x)=

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

\), substituting \( x=-4 \) into \( \frac{1}{x + 4} \), we get \( \frac{1}{-4 + 4}=\frac{1}{0} \), which is undefined. And \( f(-4) = 1 \) is defined from the second part of the piece - wise function. But if we consider the first part (the non - constant part near \( x=-4 \)), the value of the function from the non - constant part at \( x=-4 \) is undefined.

Step2: Calculate the left - hand limit

For the left - hand limit as \( x
ightarrow - 4^{-} \), \( f(x)=\frac{1}{x + 4} \). Let \( x=-4-h \), where \( h>0 \) and \( h
ightarrow0 \). Then \( \lim_{x
ightarrow - 4^{-}}f(x)=\lim_{h
ightarrow0}\frac{1}{-4 - h+4}=\lim_{h
ightarrow0}\frac{1}{-h}=-\infty \)

Step3: Calculate the right - hand limit

For the right - hand limit as \( x
ightarrow - 4^{+} \), \( f(x)=\frac{1}{x + 4} \). Let \( x=-4 + h \), where \( h>0 \) and \( h
ightarrow0 \). Then \( \lim_{x
ightarrow - 4^{+}}f(x)=\lim_{h
ightarrow0}\frac{1}{-4 + h+4}=\lim_{h
ightarrow0}\frac{1}{h}=\infty \)

Answer:

\( f(-4) \) is undefined. \( \lim_{x
ightarrow - 4^{+}}f(x) \) and \( \lim_{x
ightarrow - 4^{-}}f(x) \) are not finite.