Answer
$$\lim_{x\to0}\frac{8x}{3\sin x-x}=4$$
Work Step by Step
$$A=\lim_{x\to0}\frac{8x}{3\sin x-x}$$
In these exercises where both function $x$ and function $\sin$ are present, we would think of applying the identity $\lim_{x\to0}\frac{\sin x}{x}=1$.
To do so here, divide both numerator and denominator by $x$:
$$A=\lim_{x\to0}\frac{8}{\frac{3\sin x}{x}-1}$$
$$A=\frac{8}{3\lim_{x\to0}(\frac{\sin x}{x})-1}$$
$$A=\frac{8}{3\times1-1}=4$$
