With this $\tan 2x$ on top, I can't really think of any ways to solve it.
$ \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} $
So is it solvable and how?
What I have came up so far:
$ \begin{align} \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} \\ =\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{4x-2\pi} \end{align} $