Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 14 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 906: 9

Answer

$0$

Work Step by Step

Factor each polynomial: $$\lim_{x\to -1^-}\frac{x^2-1}{x^3-1}\\=\lim_{x\to -1^-}\frac{(x+1)(x-1)}{(x-1)(x^2+x+1)}.$$ Cancel the common factors: $$\lim_{x\to -1^-}\frac{x+1}{x^2+x+1}\\=\frac{(-1)+1}{(-1)^2+(-1)+1}\\=\frac{0}{1}=0$$
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.