University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 107: 12


$$\lim_{r\to\infty}\frac{r+\sin r}{2r+7-5\sin r}=\frac{1}{2}$$

Work Step by Step

$$A=\lim_{r\to\infty}\frac{r+\sin r}{2r+7-5\sin r}$$ Here we can divide both numerator and denominator by $r$, the highest power of $r$ in the denominator: $$A=\lim_{r\to\infty}\frac{1+\frac{\sin r}{r}}{2+\frac{7}{r}-\frac{5\sin r}{r}}$$ - We already know $\lim_{r\to\infty}(7/r)=0$ - For $\lim_{r\to\infty}(\sin r)/r$: As $-1\le\sin r\le1$, we have $(-1/r)\le(\sin r/r)\le(1/r)$. But we know that $\lim_{r\to\infty}(-1/r)=\lim_{r\to\infty}(1/r)=0$ Therefore, following Squeeze Theorem: $\lim_{r\to\infty}(\sin r)/r=0$ That means, $$A=\frac{1+0}{2+0-5\times0}=\frac{1}{2}$$
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.