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 307: 15


The solution is $$\lim_{x\to 1}\frac{\ln x}{4x-x^3+3}=\frac{1}{2}.$$

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". $$\lim_{x\to 1}\frac{\ln x}{4x-x^3+3}=\left[\frac{0}{0}\right][\text{LH}]=\lim_{x\to 1}\frac{(\ln x)'}{(4x-x^2-3)'}=\lim_{x\to1}\frac{\frac{1}{x}}{4-2x}=\lim_{x\to1}\frac{1}{4x-2x^2}=\left[\frac{1}{4\cdot1-2\cdot1^2}\right]=\frac{1}{2}.$$
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.