Answer
$$\lim_{h\to0^-}\frac{h}{\sin3h}=\frac{1}{3}$$
Work Step by Step
$$A=\lim_{h\to0^-}\frac{h}{\sin3h}=\lim_{h\to0^-}\Big(\frac{\sin3h}{h}\Big)^{-1}=\Big(\lim_{h\to0^-}\frac{\sin3h}{h}\Big)^{-1}=X^{-1}$$
Consider $X$: $$X=\lim_{h\to0^-}\frac{\sin3h}{h}$$
To be able to apply Theorem 7, we need $3h$ in the denominator, instead of $h$.
So we would multiply both numerator and denominator by $3$:
$$X=\lim_{h\to0}\frac{3\sin3h}{3\times h}=3\lim_{h\to0}\frac{\sin 3h}{3h}$$
Take $\theta=3h$. Then as $h\to0$, $\theta\to3\times 0=0$
$$X=3\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$
Apply Theorem 7: $$X=3\times1=3$$
Therefore, $$A=X^{-1}=3^{-1}=\frac{1}{3}$$
