$$\lim_{x \to \frac{\pi}{2}}\frac{b(1-\sin x) }{(\pi-2x)^2}$$
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.
Best Answer
That's a good thing L'Hospital's rule has been banned. If it is not well applied, it can lead to errors, and when it works, using Taylor's formula at order $1$ is logically equivalent. Very often, using equivalents is the shortest way to compute a limit.
That said, use substitution: set $x=\dfrac\pi2-h$; $h\to 0$ if $x\to\dfrac\pi2$. Then $$\frac{b(1-\sin x)}{(\pi-2x)^2}=\frac{b(1-\cos h)}{4h^2}$$ Now it is a standard limit that $$\lim_{h\to 0}\frac{1-\cos h}{h^2}=\lim_{h\to 0}\frac{1-\cos^2 h}{h^2(1+\cos h)}=\lim_{h\to 0}\Bigl(\frac{\sin h}h\Bigr)^2\frac1{(1+\cos h)}=\frac12.$$ Thus the limit in question is equal to $\color{red}{\dfrac b8}.$