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: 10


$\lim\limits_{n \to \infty} \frac{n^{12}}{3n^{12}+4}=\frac{1}{3}$

Work Step by Step

$\lim\limits_{n \to \infty} \frac{n^{12}}{3n^{12}+4} = \lim\limits_{n \to \infty} \frac{\frac{n^{12}}{n^{12}}}{\frac{3n^{12}}{n^{12}}+\frac{4}{n^{12}}} = \lim\limits_{n \to \infty}\frac{1}{3 + \frac{4}{n^{12}}} = \frac{1}{3+0} = \frac{1}{3}$
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.