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: 14


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

Work Step by Step

We will evaluate this limit using L'Hopital's rule. $\text{LR}$ will stand for "Apply L'Hopital's rule" $$\lim_{x\to -1}\frac{x^4+x^3+2x+2}{x+1}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to -1}\frac{(x^4+x^3+2x+2)'}{(x+1)'}=\lim_{x\to -1}\frac{4x^3+3x^2+2}{1}=\left[\frac{4\cdot(-1)^3+3\cdot(-1)^2+2}{1}\right] =1.$$
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.