Chapter 14, Problem 49RE

To determine

The function is continuous at the value of $x=4$

Answer to Problem 49RE

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(-2\right)=-4$ , function is continuous at that given point

Explanation of Solution

Given:

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

Concept used:

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

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

Calculation:

The function $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2},\text{if}x\ne -2\\ 4,\text{if}x=-2\end{array}\right\}\mathrm{.............}\left(1\right)$

Putting $x=-2$ in equation (1)

  $f\left(-2\right)=4$

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

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

Left continuity

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

Right continuity

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

Left continuity

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

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

Right continuity

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

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

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(a\right)=-4$

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(-2\right)=-4$

Left continuity = right continuity

So , function is continuous at that given point