91Ó°ÊÓ

We have $R\left(x\right)=\frac{x+4}{x^2-16}$

As we know, for a function to be continuous at $x=c$, then the function must be defined at $x=c$and $\underset{x→c^{-}}{\lim }f\left(x\right)=\underset{x→c^{+}}{\lim }f\left(x\right)=f\left(c\right)$

First we have to find the limit of $R\left(x\right)=\frac{x+4}{x^2-16}$at $c=-4$.

$$\begin{gathered} \underset{x→c^{-}}{\lim }R\left(x\right)=\underset{x→-4}{\lim }\frac{x+4}{x^2-16} \\ =\underset{x→-4}{\lim }\frac{x+4}{(x-4)(x+4)} \\ =\underset{x→-4}{\lim }\frac{1}{(x-4)} \\ =\frac{1}{-4-4} \\ =-\frac{1}{8} \end{gathered}$$

Therefore, here the limit of $R\left(x\right)$when$c=-4$is$-\frac{1}{8}$, but the function is not itself is not defined at role="math" localid="1647192157375" $x=-4$.

That is, there is a hole and discontinuity at the point$(-4,-\frac{1}{8})$.

Hence $R\left(x\right)$is not continuous at $x=-4$.

Now, we have to find the limit of $R\left(x\right)=\frac{x+4}{x^2-16}$at $c=4$.

$$\begin{gathered} \underset{x→c^{+}}{\lim }R\left(x\right)=\underset{x→4}{\lim }\frac{x+4}{x^2-16} \\ =\underset{x→-4}{\lim }\frac{x+4}{(x-4)(x+4)} \\ =\underset{x→-4}{\lim }\frac{1}{(x-4)} \\ =\frac{1}{4-4} \\ =\frac{1}{0} \end{gathered}$$

Therefore, here near $c=4$, the graph $R\left(x\right)=\frac{x+4}{x^2-16}$approaches infinity.

That is, there is an asymptote at $x=4$.

Hence $R\left(x\right)$is not continuous when$c=4$

Let us graph $R\left(x\right)$with the help of a graphing utility.

Here, we could see that the graph has a hole or discontinuity at $(-4,-\frac{1}{8})$.

From the left side of$x=4$ , the graph approaches $-∞$and from the right side the graph approaches $+∞$. This shows that $R\left(x\right)$is not continuous and has an asymptote at$x=4$.