Chapter 14, Problem 38RE

To determine

To find:

The function is continuous at the value of $c=-2$

Answer to Problem 38RE

  $\underset{x\to -4^{+}}{\lim }f\left(x\right)=-6$ , function is continuous at that given point

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{x^2-4}{x+2}c=-2$

Concept used:

The left continuity and right continuity of a function $f$ at a point $x=c$ are , equal

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

Calculation:

The function $f\left(x\right)=\frac{x^2-4}{x+2}c=-\mathrm{2.............}\left(1\right)$

Left continuity

  $\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)$

Right continuity

  $\underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$

  $\begin{array}{l}\underset{x\to -4^{+}}{\lim }f\left(x\right)=\underset{x\to -4^{+}}{\lim }\frac{x^2-4}{x+2}\\ \underset{x\to -4^{+}}{\lim }f\left(x\right)=\underset{x\to -4^{+}}{\lim }\frac{\left(x-2\right)\left(x+2\right)}{x+2}\\ \underset{x\to -4^{+}}{\lim }f\left(x\right)=\underset{x\to -4^{+}}{\lim }\left(x-2\right)\\ \text{apply}\text{limit}\\ \underset{x\to -4^{+}}{\lim }f\left(x\right)=\left(-4-2\right)\\ \underset{x\to -4^{+}}{\lim }f\left(x\right)=-6\end{array}$

  $\underset{x\to -4^{+}}{\lim }f\left(x\right)=-6$

So , function is continuous at that given point