Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 121



Work Step by Step

Given $$ \lim _{x \rightarrow \infty} \frac{2 x-\sin x}{3 x+\cos 2 x} $$ Since as $x\to \infty $, then $2 x-\sin x,\ 3 x+\cos 2 x\to \infty $ L'Hôpital's Rule applies to $$\lim _{x \rightarrow \infty} \frac{2 x-\sin x}{3 x+\cos 2 x} $$However, the resulting limit $$\lim _{x \rightarrow \infty} \frac{2-\cos x}{3-2 \sin 2 x}$$ does not exist due to the oscillation of $ \sin x$ and $ \cos x$. Thus \begin{align*} \lim _{x \rightarrow \infty} \frac{2 x-\sin x}{3 x+\cos 2 x}&= \lim _{x\to \infty \:}\left(\frac{2-\frac{\sin \left(x\right)}{x}}{3+\frac{2\cos \left(x\right)}{x}}\right)\\ &= \frac{\lim _{x \rightarrow \infty}2-\lim _{x \rightarrow \infty}\frac{\sin \left(x\right)}{x}}{\lim _{x \rightarrow \infty}3+\lim _{x \rightarrow \infty}\frac{2\cos \left(x\right)}{x}}\\ &= \frac{2-0}{3+0}=\frac{2}{3} \end{align*} where by using the squeeze theorem we get $\lim _{x \rightarrow \infty}\frac{2\cos \left(x\right)}{x}=\lim _{x \rightarrow \infty}\frac{\sin \left(x\right)}{x}=0 $
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