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: 9

Answer

$\lim\limits_{n \to \infty}\frac{n^3}{n^4 + 1} = 0$

Work Step by Step

$\lim\limits_{n \to \infty}\frac{n^3}{n^4 + 1} = \lim\limits_{n \to \infty}\frac{\frac{n^3}{n^4}}{\frac{n^4}{n^4} + \frac{1}{n^4}} = \frac{\frac{1}{\infty}}{1+\frac{1}{\infty}}= \frac{0}{1+0} = 0$
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