Answer
$1$
Work Step by Step
Let $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}f(x)=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) (\tan x)$
Re-write as: $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \tan x=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \dfrac{\sin x}{\cos x}$
But $f((\dfrac{\pi}{2})^{-})=\dfrac{0}{0}$
This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$
This implies that $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}\dfrac{-(\sin x)+(\dfrac{\pi}{2} -x) \cos x}{-(\sin x)}=\dfrac{-\sin (\dfrac{\pi}{2})+(\dfrac{\pi}{2} -\dfrac{\pi}{2}) \cos (\dfrac{\pi}{2})}{-\sin (\dfrac{\pi}{2})}$
Thus, $\dfrac{(-1)}{(-1)}=1$
