Limit of goniometric function without l'Hospital's rule

Question

I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?

$$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 }$$

Answer 1

Set $t=\frac \pi2 - x,$ $$\lim_{x\to {\pi\over 2}} \frac {1-\sin x}{(\frac\pi2 -x)^2}=\lim_{t\to {0}} \frac {1-\cos t}{t^2}=\lim_{t\to {0}} \frac {2 \sin ^2(t/2)}{4(t/2)^2}={1\over 2}$$

Answer 2

We have that

$$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } =\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } \frac {1+\sin x}{1+\sin x} =\lim_{x\to \frac{\pi}2} \frac {1}{1+\sin x}\frac {\sin^2\left(\frac\pi2 -x\right)}{\left(\frac\pi2 -x\right)^2 } =\frac12$$