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: 18

Answer

The solution is $$\lim_{x\to 1}\frac{4\tan^{-1}x-\pi}{x-1}=2.$$

Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to 1}\frac{4\tan^{-1}x-\pi}{x-1}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 1}\frac{(4\tan^{-1}x-\pi)'}{(x-1)'}=\lim_{x\to 1}\frac{4\frac{1}{1+x^2}}{1}=\lim_{x\to 1}\frac{4}{1+x^2}=\left[\frac{4}{1+1^2}\right]=2.$$
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.