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: 36



Work Step by Step

Consider the Taylor series for $\sin x $ as follows: $\sin x= x-\dfrac{ x^3}{3!}+\dfrac{x^5}{5!}-....$ $\lim\limits_{x \to \infty}(x+1) \sin (\dfrac{1}{x+1})=\lim\limits_{x \to \infty}(x+1) \times [-\dfrac{1}{3!(x+1)^3}+...]\\=\lim\limits_{x \to \infty} (1)- \lim\limits_{x \to \infty} \dfrac{1}{3!(x+1)^2}+\lim\limits_{x \to \infty} \dfrac{1}{5!(x+1)^4}-...)$ So, $\lim\limits_{x \to \infty}(x+1) \sin (\dfrac{1}{x+1})=1$
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