Answer
$$-4$$
Work Step by Step
Since $$\sin 3\theta =3\sin \theta -4\sin^3\theta,$$
then we have
\begin{align*}
\lim _{\theta \rightarrow 0} \frac{ \sin 3\theta -3\sin \theta }{ \theta^3}&= \lim _{\theta \rightarrow 0} \frac{ -4\sin^3 \theta }{ \theta^3}\\
&=-4 \lim _{\theta \rightarrow 0} \frac{ \sin^3 \theta }{ \theta^3}\\
&= -4 \left(\lim _{\theta \rightarrow 0} \frac{ \sin \theta }{ \theta}\right)^3\\
&= -4.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$.
