Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital) 
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$

I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
 A: HINT
We have that
$$\cot (2x)\cot\left(\frac{\pi }{2}-x\right)=\frac{\cos(2x)}{\sin(2x)}\frac{\sin x}{\cos x}$$
A: Hint: $$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))=\lim_{x \to 0} {\cos (2x)\over \sin 2x}\tan (x) =\lim_{x \to 0} {\cos (2x)\over 2\sin x\cos x}{\sin x\over \cos x}$$
$$=\lim_{x \to 0} {\cos (2x)\over 2\cos^2 x} =  {1\over 2}$$
A: Hint:
$$\lim_{x \to 0} \cot (2x)\cot (\frac{\pi }{2}-x)=\lim_{x \to 0} \cot (2x)\tan x=\lim_{x \to 0} \dfrac{\cos2x}{\sin2x}\dfrac{\sin x}{\cos x}=\lim_{x \to 0} \dfrac{\cos2x}{1}\dfrac{2x}{\sin2x}\dfrac{\sin x}{x}\dfrac{1}{\cos x}\dfrac12$$
A: $\displaystyle  \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\cot (2x)\tan (x)=\dfrac{\cos 2x\cdot \sin x}{2\sin x\cos^2 x}=\dfrac{\cos 2x}{2\cos^2x}$ 
$\displaystyle\lim_{x\to0} \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\displaystyle\lim_{x\to0}\dfrac{\cos 2x}{2\cos^2x}=\dfrac12 $
A: $$F(x)=\cot2x\cot\left(\dfrac\pi2-x\right)=\dfrac{\tan x}{\tan2x}=\dfrac{\tan x(1-\tan^2x)}{2\tan x}$$
For $\tan x\ne0,F(x)=\dfrac{1-\tan^2x}2$
As $x\to0,x\ne0\implies\lim_{x\to0}F(x)=\lim_{x\to0}\dfrac{1-\tan^2x}2=?$
