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: 31


The solution is $$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{x^4+2x^3-x^2-4x-2}=4.$$

Work Step by Step

To solve this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule." $$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{x^4+2x^3-x^2-4x-2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to-1}\frac{(x^3-x^2-5x-3)'}{(x^4+2x^3-x^2-4x-2)'}=\lim_{x\to-1}\frac{3x^2-2x-5}{4x^3+6x^2-2x-4}=\left[\frac{3(-1)^2-2(-1)-5}{4(-1)^3+6(-1)^2-2(-1)-4}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to-1}\frac{(3x^2-2x-5)'}{(4x^3+6x^2-2x-4)'}=\lim_{x\to-1}\frac{6x-2}{12x^2+12x-2}=\frac{6(-1)-2}{12(-1)^2+12(-1)-2}=4$$
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.