Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises - Page 308: 91


The solution is $$\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)=\infty$$

Work Step by Step

To solve this limit follow the steps below: $$\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)=\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{\sqrt{x-1}}{x-1}\right)=\lim_{x\to1^+}\frac{1-\sqrt{x-1}}{x-1}=\left[\frac{1-\sqrt{1^+-1}}{1^+-1}\right]=\left[\frac{1-0^+}{0^+}\right]=\infty.$$
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.