## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}\frac{8x}{3\sin x-x}=4$$
$$A=\lim_{x\to0}\frac{8x}{3\sin x-x}$$ 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 $x$: $$A=\lim_{x\to0}\frac{8}{\frac{3\sin x}{x}-1}$$ $$A=\frac{8}{3\lim_{x\to0}(\frac{\sin x}{x})-1}$$ $$A=\frac{8}{3\times1-1}=4$$