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: 10


Yes, the function $ f\left( x \right)=\frac{x-7}{x+7}$ is continuous at $7$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{x-7}{x+7}$, First check whether the function is defined at the point $ a $ or not. Find the value of $ f\left( x \right)$ at $ a=7$, $\begin{align} & f\left( 7 \right)=\frac{7-7}{7+7} \\ & =0 \end{align}$ The function is defined at the point $ a=7$. Now find the value of $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}$, $\begin{align} & \underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7} \\ & =\frac{\,\underset{x\to 7}{\mathop{\lim }}\,\left( x-7 \right)}{\,\underset{x\to 7}{\mathop{\lim }}\,\left( x+7 \right)}\, \\ & =\,\frac{7-7}{7+7} \end{align}$ Thus, $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}=0$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above two steps, $\,\underset{x\to 7}{\mathop{\lim }}\,\frac{x-7}{x+7}=0\text{ and }f\left( 7 \right)=0$ Therefore, $\,\underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)=f\left( 7 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\frac{x-7}{x+7}$ is continuous at $7$.
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