Short method to evaluate $\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$? $$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$
I only know of L'hopital method but that is very long. Is there a shorter method to solve this?

Another approach is to use that $\tan x=\frac{\sin x}{\cos x}$.
Then $$\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)+\sin(x/2)}=\frac{(\cos(x/2)-\sin(x/2))^2}{\cos^2(x/2)-\sin^2(x/2)}=\frac{1-\sin(x)}{\cos(x)} .$$

So the original limit reduces to
$$\lim_{x\to \frac\pi 2} \frac{(1-\sin (x))^2}{\cos(x)(\pi-2x)^3}.$$

This is the product of the two (finite) limits 
$$\lim_{x\to \frac\pi 2} \frac{1-\sin (x)}{\cos(x)(\pi-2x)} ~~~~~~\text{  and  } ~~~~~\lim_{x\to \frac\pi 2} \frac{(1-\sin (x))}{(\pi-2x)^2}.$$
Now these limits can be calculated using l'Hospital (in second order).
