Chapter 14, Problem 24RE

To determine

The function $f\left(x\right)=\frac{x^2-9}{x+10}c=2$ is continuous at the value of $c$

Answer to Problem 24RE

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(a\right)=-\frac{5}{12}$ , function is continuous at that given point

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{x^2-9}{x+10}c=2$

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 $f\left(x\right)=\frac{x^2-9}{x+10}c=\mathrm{2.............}\left(1\right)$

Left continuity

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

Right continuity

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

  $\begin{array}{l}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{x^2-9}{x+10}\\ \text{apply}\text{limit}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\frac{2^2-9}{2+10}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=-\frac{5}{12}\end{array}$

  $\begin{array}{l}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{x^2-9}{x+10}\\ \text{apply}\text{limit}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\frac{2^2-9}{2+10}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=-\frac{5}{12}\end{array}$

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(a\right)=-\frac{5}{12}$

Left continuity = right continuity

So , function is continuous at that given point