Help regarding trigonometric Limits

Question

Find the value of $$\lim_{x\to \pi/2} \frac{\cot(x) - \cos(x)}{(\pi-2x)^3}$$ The question can be solved by taking $\tan (x)$ common in the numerator. But on solving by taking $cos(x)$ common in numerator, i'm getting an extra factor $\pi/2$ with it. Where am i going wrong in the steps ? $$=\lim_{x\to \pi/2} \frac{\cos(x)(1 - \sin(x))}{\sin(x)(\pi-2x)^3}$$ $$=\lim_{x\to \pi/2} \frac{\cos(x)(1 - \sin(x))}{x\frac{\sin(x)}{x}(\pi-2x)^3}$$ $$=\lim_{x\to \pi/2} \frac{\cos(x)(1 - \sin(x))}{x(\pi-2x)^3}$$ Now, let $x-\pi/2 = h$. Then when $x \to\pi/2, h\to0$ $$=\lim_{h\to 0} \frac{\cos(\pi/2+h)(1 - \sin(\pi/2+h))}{(h+\pi/2)(\pi-2(\pi/2+h))^3}$$ $$=\lim_{h\to 0} \frac{-\sin(h)(1 - \cos(h))}{(h+\pi/2)(-2h)^3}$$ $$=\lim_{h\to 0} \frac{1}{8}\frac{\sin(h)}{h}\frac{(1 - \cos(h))}{(h^2)(h+\pi/2)}$$ $$=\lim_{h\to 0} \frac{1}{8}\frac{(1 - \cos(h))}{(h^2)(h+\pi/2)}$$ $$=\lim_{h\to 0} \frac{1}{8}\frac{(2\sin^2 (h/2))}{(h^2)(h+\pi/2)}$$ $$=\lim_{h\to 0} \frac{1}{4}\biggl(\frac{\sin (h/2)}{(h/2)}\biggr)^2\frac{1}{4(h+\pi/2)}$$ $$=\lim_{h\to 0} \frac{1}{16}\frac{1}{(h+\pi/2)}$$ On substituting $h=0,$ i get the answer $\frac{1}{8\pi}$ but the correct answer is $\frac{1}{16}$, can someone explain me why am i getting this extra term by solving in this manner ? Please forgive my formatting its my first question here.

Answer 1

Hint $$\lim_{x\to\pi/2}\dfrac{\sin x}x=?$$

Set $\pi-2x=y$