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 - Page 616: 14


$\lim\limits_{k \to \infty} \frac{k}{\sqrt {9k^2+1}} =\frac{1}{3}$

Work Step by Step

$\lim\limits_{k \to \infty} \frac{k}{\sqrt {9k^2+1}} = \lim\limits_{k \to \infty} \frac{\frac{k}{k}}{\frac{\sqrt {9k^2+1}}{k}}= \lim\limits_{k \to \infty} \frac{1}{\sqrt {9+\frac{1}{k^2}}} = \frac{1}{\sqrt {9+0}} = \frac{1}{3}$
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