Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 45


Converges to $1$

Work Step by Step

Given: $a_n=nsin(\frac {1} {n})$ It can be re-written as $a_n=\frac{sin(\frac {1} {n})}{\frac {1}{n}}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{sin(\frac {1} {n})}{\frac {1}{n}}$ Consider $\frac {1}{n}=x$ and $x \to 0$ as $n \to \infty $ Thus, $=\lim\limits_{x \to 0} \frac {sin(x)}{x}$ Note that the limit is of the form $\frac{0}{0}$ Therefore, we can apply the L-Hospital's Rule. $=\lim\limits_{x \to 0} \frac {cos(x)}{1}$ $=1$ Therefore, the given sequence converges to $1$.
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