University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 549: 32



Work Step by Step

Taylor series for $\sin \theta $ can be defined as: $\sin \theta= \theta-\dfrac{ \theta^3}{3!}+\dfrac{ \theta^5}{5!}-....$ Now, $\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\lim\limits_{\theta \to 0} \dfrac{-\theta+\dfrac{\theta^3}{6}+\sin \theta}{\theta^5}$ or, $=\lim\limits_{\theta \to 0} \dfrac{\dfrac{\theta^5}{5!}- \dfrac{\theta^7}{7}+...}{\theta^5}$ or, $=\lim\limits_{\theta \to 0} [\dfrac{1}{5!}-\dfrac{\theta^2}{7!}+\dfrac{\theta^4}{9!}-....]$ or, $=\dfrac{1}{120}$
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.