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


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.