Answer
$3$
Work Step by Step
Since we have
$$\lim _{\theta \rightarrow 0} \frac{2 \sin \theta-\sin 2 \theta}{\sin \theta-\theta \cos \theta}=\frac{0}{0}.$$
We can apply L’Hôpital’s Rule as follows
$$\lim _{\theta \rightarrow 0} \frac{2 \sin \theta-\sin 2 \theta}{\sin \theta-\theta \cos \theta}=\lim _{\theta \rightarrow 0} \frac{2 \cos\theta-2\cos2 \theta}{\cos\theta- \cos \theta+\theta\sin \theta}=\frac{0}{0}.$$
Again, we can apply L’Hôpital’s Rule as follows
$$ \lim _{\theta \rightarrow 0} \frac{2 \cos\theta-2\cos2 \theta}{\theta\sin \theta}= \lim _{\theta \rightarrow 0} \frac{-2 \sin\theta+4\sin\theta}{\sin \theta+\theta\cos\theta}=\frac{0}{0}.$$
Again, we can apply L’Hôpital’s Rule as follows
$$ \lim _{\theta \rightarrow 0} \frac{-2 \sin\theta+4\sin2\theta}{\sin \theta+\theta\cos\theta}= \lim _{\theta \rightarrow 0} \frac{-2 \cos\theta+8\cos2\theta}{\cos\theta+\cos\theta-\theta\sin\theta}=\frac{6}{2}=3.$$
Hence,
$$\lim _{\theta \rightarrow 0} \frac{2 \sin \theta-\sin 2 \theta}{\sin \theta-\theta \cos \theta}=3.$$
