University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 34


$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}$ and $\lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}=\dfrac{\infty}{\infty}$ WenNeed to apply L-Hospital's rule: $=\lim\limits_{n \to \infty} \dfrac{3n^2}{8n}$ Again using L-Hospital's rule. This implies , we get, $= \lim\limits_{n \to \infty} \dfrac{6n}{8}$ or, $=\infty$ Thus, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
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