Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises: 11

Answer

$0$

Work Step by Step

$\dfrac {-1}{\theta^2 }\leq \dfrac {\cos \theta }{\theta ^2}\leq \dfrac {1}{\theta^2 }$ $\lim _{\theta \rightarrow \infty }\dfrac {-1}{\theta ^2}=0$ $\lim _{\theta \rightarrow \infty }\dfrac {1}{\theta ^2}=0$ $0\leq \dfrac {\cos \theta }{\theta ^{2}}\leq 0\Rightarrow \lim _{\theta \rightarrow \infty }\dfrac {\cos \theta }{\theta ^{2}}=0$
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.