<s>[INST] 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.
 [/INST] For $x\neq \frac{\pi}{4}+k\pi/2$ and $x\neq \pi/2+k\pi$ ($k\in\mathbb{Z}$), we have 
$$\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$$
and 
$$\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\left(\frac{\pi}{4}\right)-\tan(x)}{1+\tan\left(\frac{\pi}{4}\right)\tan(x)}$$
Combining the two expressions and using the fact that $\tan(\pi/4)=1$, we get
\begin{align*}
\tan(2x)\tan\left(\frac{\pi}{4}-x\right) &=\frac{2\tan(x)}{1-\tan^{2}(x)}\cdot\frac{1-\tan(x)}{1+\tan(x)}\\
&=\frac{2\tan(x)}{(1+\tan(x))^{2}}\\
\end{align*}
Hence,
$$\lim_{x\to\frac{\pi}{4}}\tan(2x)\tan\left(\frac{\pi}{4}-x\right)=\lim_{x\to\frac{\pi}{4}}\frac{2\tan(x)}{(1+\tan(x))^{2}}=\frac{2}{(1+1)^{2}}=\frac{1}{2}$$
</s>