Question2. $\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\ln (\sin x)}{(\pi-2 x)^{2}}\right][2 p t s]$

Studdy Solution
Evaluate the limit as $x \to \frac{\pi}{2}$:
As $x \to \frac{\pi}{2}$, $\cot x = \frac{\cos x}{\sin x} \to 0$ because $\cos(\frac{\pi}{2}) = 0$.
Thus, the entire expression approaches:
$\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-4(\pi - 2x)} = \frac{0}{0} = 0$
The limit is $\boxed{0}$.