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: 30



Work Step by Step

Taylor series for $ e^x $ can be defined as: $ e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....$ Now, $\lim\limits_{x \to 0}\dfrac{e^x-e^{-x}}{x}=\lim\limits_{x \to 0} \dfrac{[(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....)-(1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-....)]}{x}$ or, $=\lim\limits_{x \to 0} \dfrac{2x+\dfrac{2x^3}{3!}+\dfrac{2x^5}{5!}+...}{x}$ or, $=\lim\limits_{x \to 0} [2+\dfrac{2x^3}{3!}+\dfrac{2x^4}{4!}+....]$ or, $=2$
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.