Answer
$$\frac{2}{3}$$
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 $
