Find $\lim_{x \to \pi/2} \frac{1 - \sin{x}}{(2x - \pi)^2}$ without using L'Hospital? 
$$\lim_{x \to \pi/2} \frac{1 - \sin{x}}{(2x - \pi)^2}$$

I tried L'Hospital and it works like this in wolfram but if I don't use it what should I do?
 A: After multiplying top and bottom by $1 + \sin{x}$ and rearranging, this is the same as
$$\frac{1}{4} \lim_{x \to \pi/2} \frac{1}{1 + \sin{x}} \frac{\cos^2{x}}{(x - \pi/2)^2}$$
As $x \to \pi/2$, $1 + \sin{x} \to 2$. Now since $\cos{(x + \pi/2)} = -\sin{x}$, we can rewrite this as
$$\frac{1}{4} \lim_{x \to \pi/2} \left(\frac{1}{1 + \sin{x}}\right) \lim_{x \to 0} \left(\frac{\sin{x}}{x}\right)^2 = \frac{1}{4} \frac{1}{2} (1)^2$$
as desired.
A: The best thing is to put $x = \frac{\pi}{2}+h$. Now observe that as $x \to \pi/2$, $h\to 0$. Also we get $2x-\pi=2h$. Then your quantity becomes
\begin{align*}
\lim_{x\to \pi/2} \frac{1-\sin(x)}{(2x-\pi)^{2}} &=\lim_{h \to 0} \frac{1-\sin\Bigl(\frac{\pi}{2}+h\Bigr)}{4h^{2}}\\ &=\lim_{h \to 0} \frac{1-\cos(h)}{4h^{2}} \\ &= \frac{1}{4} \cdot \lim_{h \to 0} \frac{2 \cdot \sin^{2}\frac{h}{2}}{h^2} \\ &= \frac{1}{4} \cdot \lim_{h \to 0} \frac{2 \cdot \sin^{2}\frac{h}{2}}{4 \cdot \displaystyle\Bigl(\frac{h}{2}\Bigr)^{2}}
\end{align*}
A: Let $x=\frac{\pi}{2}-t$. Then we want
$$\lim_{t\to 0} \frac{1-\cos t}{4t^2}.$$
Multiply top and bottom by $1+\cos t$.
