Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1161: 29

Answer

The function $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\ & 2\text{ if }x=1 \end{align} \right.$ is not discontinuous for any number.

Work Step by Step

Consider the function $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\ & 2\text{ if }x=1 \end{align} \right.$, Find the value of $ f\left( x \right)$ at $ x=1$, From the definition of the function, $ f\left( 1 \right)=2$ The function is defined at the point $ x=1$. Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$, $\begin{align} & \,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( x+1 \right)}{\left( x-1 \right)}\, \\ & =\,\underset{x\to 1}{\mathop{\lim }}\,x+1 \\ & =1+1 \\ & =2 \end{align}$ Thus, $\,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,2$ From the above two steps, $\,\,\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\,2=f\left( 1 \right)$ Thus, the function satisfies all the properties of being continuous. Thus, the function is continuous at $ x=1$. Hence, the function $ f\left( x \right)=\left\{ \begin{align} & x-2\text{ if }x\le 2 \\ & {{x}^{2}}-1\text{ if }x>2 \end{align} \right.$ is not discontinuous for any number.
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