<s>[INST] 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?
 [/INST] 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.
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