Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.2 Sequences - 8.2 Exercises: 16

Answer

$\lim\limits_{n \to \infty} \sqrt {n^2 + 1} - n = 0$

Work Step by Step

$\lim\limits_{n \to \infty} \sqrt {n^2 + 1} - n = \lim\limits_{n \to \infty} [(\sqrt {n^2 + 1} - n) \times \frac{\sqrt {n^2 + 1} + n}{\sqrt {n^2 + 1} + n}] =\lim\limits_{n \to \infty} \frac{1}{\sqrt {n^2+1} + n} = \frac{1}{\infty + \infty} = 0$
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