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: 77


Converges to $\dfrac{1}{2}$

Work Step by Step

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