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 1160: 14

Answer

No, the function $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right.$ is not continuous at $6$.

Work Step by Step

Consider the function $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right.$, First check whether the function is defined at the point $ a $ or not. Find the value of $ f\left( x \right)$ at $ a=6$, Here use the second line of the piecewise function where $ x=6$ because the first line of the piecewise function is not defined at 6. $ f\left( 6 \right)=13$ The function is defined at the point $ a=2$. Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}$, $\begin{align} & \underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6} \\ & =\,\underset{x\to 6}{\mathop{\lim }}\,\frac{\left( x-6 \right)\left( x+6 \right)}{\left( x-6 \right)}\, \\ & =\,\underset{x\to 6}{\mathop{\lim }}\,x+6 \\ & =6+6 \end{align}$ Further solve the above expression, $\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=12$ Thus, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}=12$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}-36}{x-6}=12\text{ and }f\left( 6 \right)=13$ Therefore, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 6 \right)$ Thus, the function does not satisfy the third property of being continuous. Hence, the function $ f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-36}{x-6}\text{ if }x\ne 6 \\ & \text{13 if }x=6 \end{align} \right.$ is not continuous at $6$.
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