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: 6


Yes, the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$ is continuous at $6$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$, First check whether the function is defined at the point $ a $ or not. Find the value of $ f\left( x \right)$ at $ a=6$, $\begin{align} & f\left( 6 \right)=\frac{{{\left( 6 \right)}^{2}}+6}{\left( 6 \right)-5} \\ & =\frac{36+6}{6-5} \\ & =42 \end{align}$ The function is defined at the point $ a=6$. Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}$, $\begin{align} & \,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5} \\ & =\frac{\,\underset{x\to 6}{\mathop{\lim }}\,\left( {{x}^{2}}+6 \right)}{\,\underset{x\to 6}{\mathop{\lim }}\,\left( x-5 \right)}\, \\ & =\frac{{{\left( 6 \right)}^{2}}+6}{6-5} \\ & =42 \end{align}$ Thus, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}=42$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 6}{\mathop{\lim }}\,\frac{{{x}^{2}}+6}{x-5}=42\text{ and }f\left( 6 \right)=42$ Therefore, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=f\left( 6 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\frac{{{x}^{2}}+6}{x-5}$ is continuous at $6$.
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