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: 88


Converges to $-2$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}$ $\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}= \lim\limits_{n \to \infty} (\dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}) \dfrac{\sqrt {n^2-n}-\sqrt {n^2+n}}{\sqrt {n^2-n}-\sqrt {n^2+n}}$ or, $\lim\limits_{n \to \infty} \dfrac{\sqrt {n^2-1}+\sqrt {n^2+n}}{-1-n}[\dfrac{1/n}{1/n}]=-2$ Hence, $\lim\limits_{n \to \infty} a_n=-2$ and {$a_n$} is Convergent and converges to $-2$
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