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: 78


Converges to $0$

Work Step by Step

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