Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 31

Answer

$-\dfrac{1}{24}$

Work Step by Step

$\lim\limits_{t \to 0}\dfrac{1-\cos t-\dfrac{t^2}{2}}{t^4}=\lim\limits_{t \to 0} \dfrac{(1-\dfrac{t^2}{2})-(1-\dfrac{t^2}{2}+\dfrac{t^4}{4}-\dfrac{t^6}{6}+....)}{t^4}\\=\lim\limits_{t \to 0} \dfrac{-\dfrac{t^4}{4!}+\dfrac{t^6}{6!}+...}{t^4}\\=\lim\limits_{x \to 0} -\dfrac{1}{4!}+ \lim\limits_{x \to 0} \dfrac{t^2}{6!}+......\\=-\dfrac{1}{24}$
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.