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 1162: 58


To make the function $ f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ continuous $ x=9$, we define $f(9)=18$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$, For a function to be continuous at a point a, the function must satisfy the following three conditions: (a) f is defined at a. (b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ Find the value of $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}$, $\begin{align} & \underset{x\to 9}{\mathop{\lim }}\,\frac{{{x}^{2}}-81}{x-9}=\underset{x\to 9}{\mathop{\lim }}\,\frac{\left( x-9 \right)\left( x+9 \right)}{x-9} \\ & =\underset{x\to 9}{\mathop{\lim }}\,x+9 \\ & =9+9 \\ & =18 \end{align}$ For the function $ f\left( x \right)=\frac{{{x}^{2}}-81}{x-9}$ to be continuous at $ x=9$, the function must be defined at $9$, and the value of $\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}$ must be equal to $ f\left( 9 \right)$, that is $ f\left( 9 \right)=\underset{x\to 9}{\mathop{\text{ }\lim }}\,\frac{{{x}^{2}}-81}{x-9}=18$ Thus, $ f\left( 9 \right)=18$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.