Evaluating $\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{\pi}{2})(1-\sin x)}{(1+\tan\frac{\pi}{2})(\pi-2x)^3}$ Evaluate $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan x)(1-\sin x)}{(1+\tan x)(\pi-2x)^2}$


Question about this..
Shouldn't the answer be infinity..as I have read $x$ as $\frac{\pi}{2} + h$ where..$h$ tends to zero..
If the limit you want to solve is $\lim_{x \rightarrow \frac{\pi}{2}} \frac{(1-\tan(x))(1-\sin(x))}{(1+\tan(x))(\pi-2x)^3}$, then you could evaluate it in the following way;
$$
\lim_{x \rightarrow \frac{\pi}{2}} \frac{(1-\tan(x))(1-\sin(x))}{(1+\tan(x))(\pi-2x)^3}=\lim_{x \rightarrow 0} \frac{(1-\tan(x+\frac{\pi}{2}))(1-\sin(x+\frac{\pi}{2}))}{(1+\tan(x+\frac{\pi}{2}))(-8x^3)}
$$
using the trigonometrical identity $\sin(a+b)=\sin a \cos b+\sin b \cos a$;
$$
\lim_{x \rightarrow 0} \frac{(1-\tan(x+\frac{\pi}{2}))(1-\cos(x))}{(1+\tan(x+\frac{\pi}{2}))(-8x^3)} \sim \lim_{x \rightarrow 0} \frac{(1-\tan(+\frac{\pi}{2}))}{(1+\tan(x+\frac{\pi}{2}))(-16x)}
$$
$$
\lim_{x \rightarrow 0} \frac{\cos(x+\frac{\pi}{2})-\sin(x+\frac{\pi}{2})}{(\cos(x+\frac{\pi}{2})+\sin(x+\frac{\pi}{2}))(-16x)}=\pm \infty
$$
In fact the limit doesn't exist, since you get $+ \infty$ from the right and $- \infty$ from the left.
As we have
$$\frac{1-\tan (x)}{1+\tan (x)}=\frac{\cos (x)-\sin (x)}{\sin (x)+\cos (x)}\to -1,\text{ as } x\to\frac{\pi}{2}$$
we can write the limit as 
$$\lim_{x\to \frac{\pi }{2}} \, \frac{\sin (x)-1}{(\pi -2 x)^2}$$
using L'Hopital rule twice we get the limit that is $-\dfrac{1}{8}$