The solution with l'Hôpital is quite simple.
$lim_{x \rightarrow 0} \frac{x+\sin(4x)}{2x-\sin(x)}=\frac{(x+\sin(4x))'}{(2x-\sin(x))'}=\frac{1+4\cos(4x)}{2-\cos(x)}=\frac{1+4}{2-1}=5$
But I don't know how to face the question without using l'Hôpital. Thank you for your time.
Hint: $$\lim_{x \rightarrow 0} \frac{x+\sin(4x)}{2x-\sin(x)}=\lim_{x \rightarrow 0} \frac{\frac{x}{4x}+\frac{\sin(4x)}{4x}}{\frac{2x}{4x}-\frac{1}{4}\frac{\sin(x)}{x}}=\lim_{x \rightarrow 0} \frac{\frac{1}{4}+\frac{\sin(4x)}{4x}}{\frac{1}{2}-\frac{1}{4}\frac{\sin(x)}{x}}$$ and use the fact $$\lim_{x \rightarrow 0}\frac{\sin ax}{ax}=1$$