Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 14



Work Step by Step

Given: $\lim\limits_{x\to0}\dfrac{\tan 3x}{\sin 2x}$ This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{x\to0}\dfrac{\dfrac{d}{dx}(\tan 3x)}{\dfrac{d}{dx}(\sin 2x)}=\lim\limits_{x\to0}\dfrac{3\sec^2(3x)}{2\cos 2x}=\dfrac{3}{2}\dfrac{\sec^2(3\times0)}{\cos(2\times0)}=\dfrac{3}{2}$
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.