Answer
$=\frac{1}{3}$
Work Step by Step
Factor each polynomial:
$$\lim_{x\to 1}\frac{x-1}{x^3-1}\\=\lim_{x\to 1}\frac{(x-1)}{(x-1)(x^2+x+1)}.$$
Cancel the common factors: $$\lim_{x\to 1}\frac{1}{(x^2+x+1)}\\=\frac{1}{(1^2+1+1)}\\=\frac{1}{3}$$
Precalculus (10th Edition)
by Sullivan, Michael
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: 7
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