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: 29

Answer

$\dfrac{1}{2}$

Work Step by Step

$\lim\limits_{x \to 0}\dfrac{e^x-(1+x)}{x^2}=\lim\limits_{x \to 0} \dfrac{[(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....)-(1+x)]}{x^2}\\=\lim\limits_{x \to 0} \dfrac{\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...}{x^2}\\=\lim\limits_{x \to 0}\dfrac{1}{2}+ \lim\limits_{x \to 0} \dfrac{x}{3!}+ \lim\limits_{x \to 0} \dfrac{x^2}{4!}+.....\\=\dfrac{1}{2}$
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