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.
 A: Since
\begin{align}
\tan(2x)&=\frac{2\tan x}{1-\tan^2 x}&&&\text{and}&&\tan\left(\frac{\pi}{4}-x\right)&=\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)&=\lim_{x\to\frac{\pi}{4}}\frac{(2\tan x)(1-\tan x)}{(1-\tan x)(1+\tan x)^2}\\
&=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}\\
&=\frac{2}{(1+1)^2}\\
&=\boxed{\color{blue}{\frac{1}{2}}}
\end{align}
A: 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,
$$\lim_{x\to\frac{\pi}{4}}\tan(2x)\tan\left(\frac{\pi}{4}-x\right)=\lim_{x\to\frac{\pi}{4}}\frac{2\tan(x)}{(1+\tan(x))^{2}}=\frac{2}{(1+1)^{2}}=\frac{1}{2}$$
A: Notice, let $\frac{\pi}{4}-x=t\implies t\to 0$ as $x\to \pi/4$ 
$$\lim_{x\to \pi/4}\tan 2x\tan\left(\frac{\pi}{4}-x\right)$$
$$=\lim_{t\to 0}\tan\left(\frac{\pi}{2}-2t\right)\tan (t)$$
$$=\lim_{t\to 0}\cot\left(2t\right)\tan (t)$$
$$=\lim_{t\to 0}\frac{\cos\left(2t\right)}{\sin (2t)}\tan (t)$$
$$=\frac 12\lim_{t\to 0}\frac{\frac{\tan (t)}{t}}{\frac{\sin (2t)}{2t}}\cdot \cos\left(2t\right)$$
$$=\frac 12\frac{\lim_{t\to 0}\left(\frac{\tan (t)}{t}\right)}{\lim_{t\to 0}\left(\frac{\sin (2t)}{2t}\right)}\cdot \lim_{t\to 0}\cos\left(2t\right)$$
$$=\frac 12\cdot \frac{1}{1}\cdot 1=\color{red}{\frac 12}$$
A: Take $y=x-\frac{\pi}{4}$ You have  $$f(x)=\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)=-\frac{\sin(2y+\frac{\pi}{2})}{\cos(2y+\frac{\pi}{2})}\frac{\sin y}{\cos y}=\frac{\cos 2y}{\sin 2y}\frac{\sin y}{\cos y}=\frac{\cos 2y}{2\cos^2y}$$ Hence $$\lim\limits_{x \to \frac{\pi}{4}} f(x)=\frac{1}{2}$$
A: In THIS ANSWER, I showed that the tangent function satisfies the inequalities
$$\left|z\right| \le \left|\tan(z)\right|\le \left|\frac{z}{\cos(z)}\right| \tag 1$$
for $|z|<\pi/2$.
For the problem of interest, we note first that 
$$\tan(2x)\tan(\pi/4-x)=\frac{\tan(x-\pi/4)}{\tan(2(x-\pi/4))}.$$
Then, using the inequalities in $(1)$, we obtain the upper bound for $0<x-\pi/4<\pi/4$
$$\begin{align}
|\tan(2x)\tan(\pi/4-x)|&=\tan(2x)\tan(\pi/4-x)\\\\
&\le \left|\frac{\frac{(x-\pi/4)}{\cos(x-\pi/4)}}{2(x-\pi/4)}\right|\\\\
&=\frac{1}{2\cos(x-\pi/4)} \tag 2
\end{align}$$
and the lower bound
$$\begin{align}
|\tan(2x)\tan(\pi/4-x)|&=\tan(2x)\tan(\pi/4-x)\\\\
&\ge \left|\frac{x-\pi/4}{{\frac{2(x-\pi/4)}{\cos(2(x-\pi/4))}}}\right|\\\\
&=\frac{\cos(x-\pi/4)}{2} \tag 3
\end{align}$$
Putting together $(2)$ and $(3)$ we obtain
$$ \frac{\cos(x-\pi/4)}{2} \le \tan(2x)\tan(\pi/4-x)\le \frac{1}{2\cos(x-\pi/4)}$$
whereupon applying the Squeeze Theorem results in the limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \pi/4}\tan(2x)\tan(\pi/4-x)=\frac12}$$
A: 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.
