Solving limits without using L'Hôpital's rule $$\lim_{x \to \frac{\pi}{2}}\frac{b(1-\sin x) }{(\pi-2x)^2}$$
I had been solving questions like these using L'Hôpital's rule since weeks. But today, a day before the examination, I got to know that its usage has been 'banned', since we were never officially taught L'Hôpital's rule. 
Now I am at a loss how to solve questions I breezed through previously. While it is not possible to learn in 18 hours methods that would cover all kinds of problems, I am hoping I can pick up enough of them to salvage the examination tomorrow.
It has been hinted to me that the above could be solved using trigonometric techniques, but I'm at a loss as to how. 

If $b = 0$, then there is nothing to work out; let $b \neq 0$. But
$$
\lim_{x \to \pi/2} \frac{b(1 - \sin x)}{(\pi - 2x)^{2}} = b\lim_{h \to 0}\frac{1 - \sin (h + \pi/2)}{4h^{2}} = b\lim_{h \to 0}\frac{1 - \sin h \cos (\pi /2) - \cos h \sin (\pi/2)}{4h^{2}}\\ = b\lim_{h \to 0}\frac{1 - \cos h}{4h^{2}} = b\lim_{h \to 0}\frac{\frac{h^{2}}{2} + o(h^{2})}{4h^{2}} = \frac{b}{8}.
$$
