Chapter 14, Problem 14RE

To determine

To find: Whether $f$ is continuous at $c$ .

Answer to Problem 14RE

The function $f$ is not continuous at $c=-2$ .

Explanation of Solution

Given information:

  $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2};\text{ if }x\ne -2\\ 4;\text{ if }x=-2\end{array}\right\}\\ c=-2;\end{array}$

Solution:

  $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2};\text{ if }x\ne -2\\ 4;\text{ if }x=-2\end{array}\right\}$

Let’s check whether $f$ is continuous at $c=-2;$ ,

  $\begin{array}{l}\underset{x\to -2^{-}}{\lim }f\left(x\right)\\ =\underset{x\to -2^{-}}{\lim }\frac{x^2-4}{x+2}\\ =\underset{x\to -2^{-}}{\lim }\frac{\left(x+2\right)\left(x-2\right)}{x+2}\\ =\underset{x\to -2^{-}}{\lim }\left(x-2\right)\\ =-2-2\\ =-4\end{array}$

  $\begin{array}{l}\underset{x\to -2^{+}}{\lim }f\left(x\right)\\ =\underset{x\to -2^{+}}{\lim }\frac{x^2-4}{x+2}\\ =\underset{x\to -2^{+}}{\lim }\frac{\left(x+2\right)\left(x-2\right)}{x+2}\\ =\underset{x\to -2^{+}}{\lim }\left(x-2\right)\\ =-2-2\\ =-4\end{array}$

Now,

  $f\left(-2\right)=4$

Therefore, $\underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }f\left(x\right)\ne f\left(-2\right)$

Hence, the function $f$ is not continuous at $c=-2$ .