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: 11

Answer

$\lim\limits_{n \to \infty} \frac{3n^3-1}{2n^3+1} = \frac{3}{2}$

Work Step by Step

$\lim\limits_{n \to \infty} \frac{3n^3-1}{2n^3+1} = \lim\limits_{n \to \infty} \frac{\frac{3n^3}{n^3}-\frac{1}{n^3}}{\frac{2n^3}{n^3}+\frac{1}{n^3}} = \lim\limits_{n \to \infty} \frac{3-\frac{1}{n^3}}{2+\frac{1}{n^3}} = \frac{3-0}{2+0} = \frac{3}{2}$
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