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

Question

$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$

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.)

Answer 1

Hint: $$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))=\lim_{x \to 0} {\cos (2x)\over \sin 2x}\tan (x) =\lim_{x \to 0} {\cos (2x)\over 2\sin x\cos x}{\sin x\over \cos x}$$ $$=\lim_{x \to 0} {\cos (2x)\over 2\cos^2 x} = {1\over 2}$$

Answer 2

Hint: $$\lim_{x \to 0} \cot (2x)\cot (\frac{\pi }{2}-x)=\lim_{x \to 0} \cot (2x)\tan x=\lim_{x \to 0} \dfrac{\cos2x}{\sin2x}\dfrac{\sin x}{\cos x}=\lim_{x \to 0} \dfrac{\cos2x}{1}\dfrac{2x}{\sin2x}\dfrac{\sin x}{x}\dfrac{1}{\cos x}\dfrac12$$

Answer 3

HINT

We have that

$$\cot (2x)\cot\left(\frac{\pi }{2}-x\right)=\frac{\cos(2x)}{\sin(2x)}\frac{\sin x}{\cos x}$$

Answer 4

$\displaystyle \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\cot (2x)\tan (x)=\dfrac{\cos 2x\cdot \sin x}{2\sin x\cos^2 x}=\dfrac{\cos 2x}{2\cos^2x}$

$\displaystyle\lim_{x\to0} \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\displaystyle\lim_{x\to0}\dfrac{\cos 2x}{2\cos^2x}=\dfrac12 $

Answer 5

$$F(x)=\cot2x\cot\left(\dfrac\pi2-x\right)=\dfrac{\tan x}{\tan2x}=\dfrac{\tan x(1-\tan^2x)}{2\tan x}$$

For $\tan x\ne0,F(x)=\dfrac{1-\tan^2x}2$

As $x\to0,x\ne0\implies\lim_{x\to0}F(x)=\lim_{x\to0}\dfrac{1-\tan^2x}2=?$