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: 120



Work Step by Step

Since \begin{align*} \lim _{x \rightarrow 0} \frac{\tanh x-\sinh x}{\sin x-x}&=\frac{0}{0} \end{align*} Then by using L’Hôpital’s Rule \begin{align*} \lim _{x \rightarrow 0} \frac{\tanh x-\sinh x}{\sin x-x}&= \lim _{x\to \:0}\left(\frac{\text{sech}^2\left(x\right)-\cosh \left(x\right)}{\cos \left(x\right)-1}\right)\\ &=\lim _{x\to \:0}\left(\frac{-2\text{sech }^2\left(x\right)\tanh \left(x\right)-\sinh \left(x\right)}{-\sin \left(x\right)}\right)\\ &= \lim _{x\to \:0}\left(\frac{2\left(-2\text{sech }^2\left(x\right)\tanh ^2\left(x\right)+\text{sech} ^4\left(x\right)\right)+\cosh \left(x\right)}{\cos \left(x\right)}\right)\\ &= 3 \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.