Chapter 14, Problem 52RE

To determine

To define:

The function $f\left(x\right)=\frac{3x^2+6x}{x^2-2}c=-2andc=2$ is continuous at the value of $c$

Answer to Problem 52RE

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right)=0,\underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }f\left(x\right)=f\left(-2\right)=\frac{3}{2}$ , function is continuous at that given point

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{3x^2+6x}{x^2-2}c=-2andc=2$

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 $f\left(x\right)=\frac{3x^2+6x}{x^2-2}c=-2andc=\mathrm{2.............}\left(1\right)$

Left continuity

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

Right continuity

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

  $\begin{array}{l}\underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{-}}{\lim }\frac{3x^2+6x}{x^2-2}c=-2\\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{-}}{\lim }\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{-}}{\lim }\frac{3x}{\left(x-2\right)}\\ \text{apply}\text{limit}\\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=\frac{3\left(-2\right)}{\left(-2-2\right)}\\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=\frac{-6}{-4}\\ \underset{x\to -2^{-}}{\lim }f\left(x\right)=\frac{3}{2}\end{array}$

  $\begin{array}{l}\underset{x\to -2^{+}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }\frac{3x^2+6x}{x^2-2}c=-2\\ \underset{x\to -2^{+}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to -2^{+}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }\frac{3x}{\left(x-2\right)}\\ \text{apply}\text{limit}\\ \underset{x\to -2^{+}}{\lim }f\left(x\right)=\frac{3\left(-2\right)}{\left(-2-2\right)}\\ \underset{x\to -2^{+}}{\lim }f\left(x\right)=\frac{-6}{-4}\\ \underset{x\to -2^{+}}{\lim }f\left(x\right)=\frac{3}{2}\end{array}$

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(-2\right)=\frac{3}{2}$

Left continuity = right continuity at the point $c=-2$

Left continuity

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

  $\begin{array}{l}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{3x^2+6x}{x^2-2}c=2\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{3x}{\left(x-2\right)}\\ \text{apply}\text{limit}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\frac{3\left(2\right)}{\left(2-2\right)}\\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=0\end{array}$

  $\begin{array}{l}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\frac{3x^2+6x}{x^2-2}c=2\\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\frac{3x}{\left(x-2\right)}\\ \text{apply}\text{limit}\\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=\frac{3\left(2\right)}{\left(2-2\right)}\\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=0\end{array}$

  $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right)=0$

Left continuity = right continuity at the point $c=2$

So , function is continuous at that given point