## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1161: 17

#### Answer

The function f\left( x \right)=\left\{ \begin{align} & 1-x\text{ if }x<1 \\ & 0\text{ if }x=1\text{ } \\ & {{x}^{2}}-1\text{ if }x>1 \end{align} \right. is continuous at $1$.

#### Work Step by Step

Consider the function f\left( x \right)=\left\{ \begin{align} & 1-x\text{ if }x<1 \\ & 0\text{ if }x=1\text{ } \\ & {{x}^{2}}-1\text{ if }x>1 \end{align} \right.. Find the value of $f\left( x \right)$ at $a=1$, From the definition of the function, $f\left( 1 \right)=0$ The function is defined at the point $a=1$. Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$. First find the left hand limit of $\,f\left( x \right)$. The function $f\left( x \right)=1-x$ when $x<1$. $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1-1=0$ Now find the right hand limit of $\,f\left( x \right)$. The function $f\left( x \right)={{x}^{2}}-1$ when $x>1$. $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{1}^{2}}-1=0$ Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. Thus, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=0$. From the above steps, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=0=f\left( 1 \right)$. Thus, the function satisfies all the properties of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & 1-x\text{ if }x<1 \\ & 0\text{ if }x=1\text{ } \\ & {{x}^{2}}-1\text{ if }x>1 \end{align} \right. is continuous at $1$.

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