Chapter 14, Problem 28RE

To determine

To find: $R(x)$ is continuous or not at $c=4\&-4$ .

Answer to Problem 28RE

at $x=-4$ is not continuous.

Explanation of Solution

Given information: $R(x)=\frac{x+4}{x^2-16}$

Given

  $\begin{array}{l}R(x)=\frac{x+4}{x^2-16}\\ \text{ =}\frac{x+4}{(x+4)(x-4)}\\ \text{ =}\frac{1}{x-4}\end{array}$

Now, $\underset{x\to 4^{+}}{\lim }R(x)=\underset{x\to 4^{+}}{\lim }\frac{1}{x-4}=\infty$

  $\underset{x\to 4^{-}}{\lim }R(x)=\underset{x\to 4^{-}}{\lim }\frac{1}{x-4}=-\infty$

Here, limit do’t exist

So, f is not continuous at $x=4$

And for $x=-4$

  $\begin{array}{l}\underset{x\to -4^{+}}{\lim }R(x)=\underset{x\to -4^{+}}{\lim }\frac{1}{x-4}\\ \text{ =}\frac{1}{-4-4}=-\frac{1}{8}\end{array}$

  $\begin{array}{l}\underset{x\to -4^{-}}{\lim }R(x)=\underset{x\to -4^{-}}{\lim }\frac{1}{x-4}\\ \text{ =}-\frac{1}{8}\end{array}$

Hence, $\underset{x\to -4}{\lim }R(x)=-\frac{1}{8}$

And

  $\begin{array}{l}R(-4)=\frac{-4+4}{{(-4)}^2-16}\\ \text{ =}\frac{0}{0}\end{array}$

Hence, at $x=-4$ is not continuous.