Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 61



Work Step by Step

$ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$ Plug into the limit to get $\lim\limits_{x \to 0}\frac{x-ln(1+x)}{x^{2}}=\lim\limits_{x \to 0}x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$ $=\frac{1}{2}$
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