University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 79

Answer

Converges to $1$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})$ But $\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})=\dfrac{0}{0}$ Need to apply L-Hospital's rule. So, $ \lim\limits_{n \to \infty} \dfrac{\cos (1/\sqrt n) \cdot (-1/2n^{3/2})}{\dfrac{-1}{2n^{3/2}}}=\lim\limits_{n \to \infty} \cos (\dfrac{1}{\sqrt n})$ or, $=1$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is Convergent and converges to $1$
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