Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 119



Work Step by Step

Since \begin{align*} \lim _{x \rightarrow 0} \frac{\sinh \left(x^{2}\right)}{\cosh x-1}&=\frac{0}{0} \end{align*} Then by using L’Hôpital’s Rule \begin{align*} \lim _{x \rightarrow 0} \frac{\sinh \left(x^{2}\right)}{\cosh x-1}&= \lim _{x\to \:0}\left(\frac{\cosh \left(x^2\right)\cdot \:2x}{\sinh \left(x\right)}\right)\\ &= \lim _{x\to \:0}\left(\frac{2\left(2x^2\sinh \left(x^2\right)+\cosh \left(x^2\right)\right)}{\cosh \left(x\right)}\right)\\ &= 2 \end{align*}
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.