Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 63



Work Step by Step

$sinx=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....$ Plug into the limit to get $\lim\limits_{x \to 0}\frac{sinx-x+1/6x^{3}}{x^{5}}=\lim\limits_{x \to 0}\frac{(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)-x+1/6x^{3}}{x^{5}}$ $=\frac{1}{120}$
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.