Answer
$$\frac{7}{3}$$
Work Step by Step
\begin{align*}
\lim _{\theta \rightarrow 0} \frac{ \sin 7\theta}{\sin 3\theta}&=\lim _{\theta \rightarrow 0} \frac{ \sin 7\theta}{\sin 3\theta}\\
&=\lim _{\theta \rightarrow 0} \frac{7}{3}\frac{ \sin 7\theta}{7\theta}\frac{ 3\theta}{\sin 3\theta}\\
&= \frac{7}{3}\lim _{7\theta \rightarrow 0}\frac{ \sin 7\theta}{7\theta}\lim _{3\theta \rightarrow 0}\frac{ 3\theta}{\sin 3\theta}\\
&= \frac{7}{3} .
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
