Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 910: 73

Answer

not continuous at $x=-1$ or $x=1$.

Work Step by Step

1. See graph for $R(x)=\frac{x-1}{x^2-1}=\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}, x\ne1$ with V.A. $x=-1$ and a hole at $(1, \frac{1}{2})$. 2. Although $\lim_{x\to1}R(x)=\frac{1}{2}$, the function is not continuous at $x=1$ because of the hole. 3. We have $\lim_{x\to -1^-}R(x)=-\infty$ and $\lim_{x\to -1^+}R(x)=\infty$, thus the function is not continuous at $x=-1$.
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.