## Calculus (3rd Edition)

$$\frac{1}{2}$$
We have $$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\sin 2 x}=\frac{0}{0}$$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\sin 2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\sin x}{2\cos 2 x}=\frac{1}{2}.$$