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