Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 73: 57



Work Step by Step

Since $ \frac{1}{x-1}-\frac{c}{x^{3}-1}=\frac{x^{2}+x+1-c}{(x-1)\left(x^{2}+x+1\right)} $ In order for the limit to exist as $x \rightarrow 1$, the numerator must evaluate to 0 at $x=1 .$ Thus, we must have $3-c=0,$ which implies that $c=3$.
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.