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 10.1 - Sequences - Exercises 10.1 - Page 570: 79

Answer

converges to $1$

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})$ But $\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})=\dfrac{0}{0}$ This shows that the limit has an Indeterminate form, so we will use L-Hospital's rule. This implies that $ \lim\limits_{n \to \infty} \dfrac{\cos(\dfrac{1}{\sqrt n}) (\dfrac{-1}{2n^{3/2}})}{(1/2n^{3/2})}=\lim\limits_{n \to \infty} \cos (1/\sqrt n) $ and $\lim\limits_{n \to \infty} \cos (\dfrac{1}{\sqrt n})=1$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.
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