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: 9


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

Work Step by Step

Consider the function $ f\left( x \right)=\frac{x-5}{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=5$, $\begin{align} & f\left( 5 \right)=\frac{5-5}{5+5} \\ & =0 \end{align}$ The function is defined at the point $ a=5$. Now find the value of $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}$, $\begin{align} & \underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)\,=\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5} \\ & =\frac{\,\underset{x\to 5}{\mathop{\lim }}\,\left( x-5 \right)}{\,\underset{x\to 5}{\mathop{\lim }}\,\left( x+5 \right)}\, \\ & =\frac{5-5}{5+5} \\ & =0 \end{align}$ Thus, $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}=0$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x+5}=0\text{ and }f\left( 5 \right)=0$ Therefore, $\,\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=f\left( 5 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\frac{x-5}{x+5}$ is continuous at $5$.
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