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: 87


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

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]$ Since, $ \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]=(\infty)(\infty)$ Need to apply L-Hospital's rule. So, $\lim\limits_{n \to \infty} \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]=\lim\limits_{n \to \infty} \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}] \times \dfrac{[n+ \sqrt {n^2-n}]}{[n +\sqrt {n^2-n}]}$ or, $\lim\limits_{n \to \infty} \dfrac{n}{n +\sqrt {n^2-n}} (\dfrac{1/n}{1/n})=\lim\limits_{n \to \infty} \dfrac{1}{1+\sqrt {(1-1/n)}}=\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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