Chapter 14, Problem 13RE

To determine

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

Answer to Problem 13RE

The function $f$ is continuous at $c=-2$ .

Explanation of Solution

Given information:

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

Solution:

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

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

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

Now,

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

  $\begin{array}{l}\therefore f\left(-2\right)=-2-2\\ \Rightarrow f\left(-2\right)=-4\end{array}$

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

Hence, the function $f$ is continuous at $c=-2$ .