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?
 A: Set $\pi-2x=4y$ to find $$\lim_{y\to0}\dfrac{\tan y(1-\cos2y)}{(2y)^3}=\lim{...}\left(\dfrac{\sin y}y\right)^3\dfrac1{4\cos y}=?$$
A: Note that $$\frac{1-\text{tan}\frac{x}{2}}{1+\text{tan}\frac{x}{2}}=\text{tan}(\frac{\pi}{4}-\frac{x}{2})$$
So, given limit is same as
$$\text{lim}_{x\to\pi/2}\frac{\text{tan}(\pi/4-x/2)(1-\text{sin}x)}{(\pi-2x)^3}$$
Can you do it now using $\lim_{t\to0}\frac{\text{sin}t}{t}=1$ and $\lim_{t\to0}\frac{\text{tan}t}{t}=1$?
You'll finally get the limit to be $1/32$.
A: Another trick is to multiply both the numerator and denominator by 
$(1+\tan(x/2))(1+\sin x)$ and use that
\begin{align}
1-\sin^2x&=\cos^2x,\\
1-\tan^2(x/2)&=\frac{\cos^2(x/2)-\sin^2(x/2)}{\cos^2(x/2)}=\frac{\cos x}{\cos^2(x/2)}.
\end{align}
Then you get after simplification
$$
\frac{1}{(1+\tan(x/2))^2(1+\sin x)\cos^2(x/2)}\frac{1}{2^3}\color{red}{\left(\frac{\cos x}{\pi/2-x}\right)^3}
$$
where the only uncertainty  when $x\to\pi/2$ is in the red part. The limit of $\frac{\cos x}{\pi/2-x}$ can be calculated by L'Hopital.
A: 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).
A: Here's yet another approach. First of all, notice that the factor of $\left(1+\tan\dfrac{x}{2}\right)$ in the denominator is the only one that is not equal to zero. So it has nothing to do with the $\dfrac{0}{0}$ indeterminate form in this limit; it's only there (informally speaking) as an extra detractor. And we might as well take it out:
$$\begin{split}
\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(1+\tan\frac{x}{2})(\pi-2x)^3}&=\lim_{x\to\frac{\pi}{2}}\frac{1}{1+\tan\frac{x}{2}}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}\\
&=\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}.\end{split}$$
We could apply L'Hôpital's Rule now, but… Although the denominator looks much more manageable now, the numerator is still a product, and taking its derivatives will be difficult. Instead, let's break this limit further apart:
$$\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}=\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{1-\tan\frac{x}{2}}{\pi-2x}\times\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{(\pi-2x)^2}.$$
Of course, this is only valid when both limits in the product exist. But as we will see in a moment, they do, and so the desired answer is their product.
In fact, now both limits can be found by applying L'Hôpital's Rule — once for the first limit and twice for the second. Also, for the second limit we can use the Taylor series expansion of the numerator to find this limit — either expand in powers of $\left(x-\dfrac{\pi}{2}\right)$, which is quite doable, albeit a bit ugly; or make the change of variables $y=x-\dfrac{\pi}{2}$ first. (Note: of course, we could use the Taylor series for $\tan$ for the first limit too; but it's a much lesser known and more advanced expansion involving Bernulli's numbers, and it doesn't appear in standard undergraduate Calculus courses.)
