hy guys ! bagi yang bisa jawab mohon di bantu jawab yah :) thanks before

[tex]\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-\sin x}{\frac{\pi}{2}-x} \\ \lim_{x \to \frac{\pi}{2}} \frac{1-\sin x}{\frac{\pi}{2}-x}\times\frac{1+\sin x}{1+\sin x} \\ \lim_{x \to \frac{\pi}{2}} \frac{1-\sin^2 x}{(\frac{\pi}{2}-x)(1+\sin x)}= \lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{(\frac{\pi}{2}-x)(1+\sin x)} \\ \lim_{x \to \frac{\pi}{2}} \frac{\sin^2 (\frac{\pi}{2}-x)}{(\frac{\pi}{2}-x)(1+\sin x)} \\ [/tex]
[tex]\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)\times \sin (\frac{\pi}{2}-x)}{(\frac{\pi}{2}-x)(1+\sin x)} = \lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(1+\sin x)} \\ \frac{\sin(\frac{\pi}{2}-\frac{\pi}{2})}{1+\sin \frac{\pi}{2}}=\frac{0}{1+1}=\frac{0}{2}=0[/tex]