Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this 

$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$

Someone can help me ?
Thanks.
Since

\begin{align}
\tan(2x)&amp;=\frac{2\tan x}{1-\tan^2 x}&amp;&amp;&amp;\text{and}&amp;&amp;\tan\left(\frac{\pi}{4}-x\right)&amp;=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}
\end{align}
we have
\begin{align}
\lim_{x\to\frac{\pi}{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)&amp;=\lim_{x\to\frac{\pi}{4}}\frac{(2\tan x)(1-\tan x)}{(1-\tan x)(1+\tan x)^2}\\
&amp;=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}\\
&amp;=\frac{2}{(1+1)^2}\\
&amp;=\boxed{\color{blue}{\frac{1}{2}}}
\end{align}
hint: $\tan(2x) = \dfrac{\sin(2x)}{\cos(2x)}$ and write $P = \sin(2x)\cdot \dfrac{\tan\left(\dfrac{\pi}{4} - x\right)}{\cos(2x)}$ and use L'hospitale rule for the second factor.