Answer
$\dfrac{1}{3}$
Work Step by Step
We know
$$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$
Thus for any nonzero real number $a$, it follows that
$$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$
We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$
Now, we want to find $\underset{h\to 0^{-}}{\lim}\frac{h}{\mathrm{sin}\left(3h\right)}$ .
Note that
$$\frac{h}{\mathrm{sin}\left(3h\right)}=\frac{3h}{3\mathrm{sin}\left(3h\right)}=\frac{1}{\left(\frac{3\mathrm{sin}\left(3h\right)}{3h}\right)}.$$
Thus
$$\underset{h\to {0}^{-}}{\lim}\frac{h}{\mathrm{sin}\left(3h\right)}=\underset{h\to {0}^{-}}{\lim}\frac{1}{\left(\frac{3\mathrm{sin}\left(3h\right)}{3h}\right)}\\
=\frac{1}{3\left(1\right)}=\frac{1}{3}.$$
