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: 32

Answer

$\dfrac{1}{120}$

Work Step by Step

$\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\lim\limits_{\theta \to 0} \dfrac{-\theta+\dfrac{\theta^3}{6}+\sin \theta}{\theta^5}\\=\lim\limits_{\theta \to 0} \dfrac{\dfrac{\theta^5}{5!}- \dfrac{\theta^7}{7}+...}{\theta^5}\\=\lim\limits_{\theta \to 0} \dfrac{1}{5!}- \lim\limits_{\theta \to 0} \dfrac{\theta^2}{7!}+\lim\limits_{\theta \to 0} \dfrac{\theta^4}{9!}-...$ So, $\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\dfrac{1}{120}$
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