Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 21

Answer

$$-\frac{7}{3}$$

Work Step by Step

We have $$ \lim _{x \rightarrow -\infty} \frac{7x^2+4x}{9-3x^2}=-\frac{\infty}{\infty}. $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow -\infty} \frac{7x^2+4x}{9-3x^2}=\lim _{x \rightarrow -\infty} \frac{14x+4}{-6x}=-\frac{\infty}{\infty} . $$ Applying L’Hôpital’s Rule, we get $$ \lim _{x \rightarrow -\infty} \frac{14x+4}{-6x}= \lim _{x \rightarrow -\infty} \frac{14}{-6}= -\frac{7}{3}. $$
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.