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: 34



Work Step by Step

$\lim\limits_{y \to 0}\dfrac{\tan^{-1} y -\sin y}{y^3 \cos y}=\lim\limits_{y \to 0} \dfrac{(y-\dfrac{y^3}{3!}+\dfrac{y^5}{5!}-....)-(y-\dfrac{y^3}{3}+\dfrac{y^5}{5}-....)}{y^3}\\=\lim\limits_{y \to 0} \dfrac{\dfrac{-y^3}{6}+ \dfrac{23y^5}{5!}+...}{y^3 \cos y}\\=\dfrac{\lim\limits_{y \to 0} (-\dfrac{1}{6})+ \lim\limits_{y \to 0} \dfrac{23y^2}{5!}-...}{ \lim\limits_{y \to 0} \cos y}$ So, $\lim\limits_{y \to 0}\dfrac{\tan^{-1} y -\sin y}{y^3 \cos y}=-\dfrac{1}{6}$
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