## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1161: 32

#### Answer

The given function is discontinuous at $x=3$.

#### Work Step by Step

Consider the given function. First check the discontinuity of the function at $x=0$. Find the value of $f\left( x \right)$ at $x=0$, From the definition of the function, for $x=0$, $f\left( x \right)=x+7$ Then the value of $f\left( x \right)$ at $x=0$ is, $f\left( 0 \right)=0+7=7$ The function is defined at the point $x=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0+7=7$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=7$ Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=7=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. Thus $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=7$. From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=7=f\left( 0 \right)$. Thus, the function satisfies all the properties of being continuous. Thus, the given function is continuous at $x=0$. Now check the discontinuity of the function at $x=3$ Find the value of $f\left( x \right)$ at $x=3$, From the definition of the function, $f\left( 3 \right)=7$ The function is defined at the point $x=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=7$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{3}^{2}}-1=8$ Since the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$ does not exist. Thus, the function does not satisfy the second property of being continuous. Hence, the given function is discontinuous at $x=3$.

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