## Calculus (3rd Edition)

$$\frac{2}{3}$$
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 $\lim _{x \rightarrow \infty}\frac{2\cos \left(x\right)}{x}=\lim _{x \rightarrow \infty}\frac{\sin \left(x\right)}{x}=0$