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: 87

Answer

The solution is $$\lim_{x\to\infty}(\sqrt{x-2}-\sqrt{x-4})=0.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{x\to\infty} (\sqrt{x-2}-\sqrt{x-4})=\lim_{x\to\infty}(\sqrt{x-2}-\sqrt{x-4})\cdot\frac{\sqrt{x-2}+\sqrt{x-4}}{\sqrt{x-2}+\sqrt{x-4}}=\lim_{x\to\infty}\frac{\sqrt{x-2}^2-\sqrt{x-4}^2}{\sqrt{x-2}+\sqrt{x-4}}=\lim_{x\to\infty}\frac{x-2-(x-4)}{\sqrt{x-2}+\sqrt{x-4}}=\lim_{x\to\infty}\frac{2}{\sqrt{x-2}+\sqrt{x-4}}=\left[\frac{2}{\sqrt{\infty-2}+\sqrt{\infty-4}}\right]=\left[\frac{2}{\infty}\right]=0.$$
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.