How to calculate $\lim\limits_{x\to \pi/4}{{\tan(x)}^{\tan(2x)}}$? I guess I should apply natural logarithm here or something like that, but I can't unterstand what to do. I shouldn't apply L'Hôpital's rule as I haven't studied it yet.
 A: Recall that 
$$\tan 2x=\frac{2\tan x}{1-\tan^2 x}.$$
Put $\tan x=1+t$. Then as $x$ approaches $\pi/4$, $t$ approaches  $0$.
The expression we are taking the limit of becomes, under the substitution,
$$\left((1+t)^{-1/t}\right)^{(2+2t)/(2+t)}.$$
The function $(1+t)^{-1/t}$ approaches $e^{-1}$ as $t$ approaches $0$. The outer exponent $(2+2t)/(2+t)$ approaches $1$.
A: This evaluation uses L'Hôpital's rule. It was written prior to the edit to the question.
Let $f(x)=(\tan (x))^{\tan (2x)}$. Then
$$
\log f(x)=\tan (2x)\log \left( \tan (x)\right) =\frac{\log \left( \tan
(x)\right) }{\dfrac{1}{\tan (2x)}},
$$
and by the L'Hôpital's rule, we have
$$
\begin{eqnarray*}
\lim_{x\rightarrow \pi /4}\log f(x) &=&\dfrac{\lim_{x\rightarrow \pi /4}\dfrac{d}{dx}\left( \log \left( \tan (x)\right) \right) }{\lim_{x\rightarrow \pi /4}
\dfrac{d}{dx}\left( \dfrac{1}{\tan (2x)}\right) } \\
&=&\dfrac{\lim_{x\rightarrow \pi /4}\dfrac{1+\tan ^{2}x}{\tan x}}{
\lim_{x\rightarrow \pi /4}\left( -\dfrac{2}{\tan ^{2}2x}-2\right) } \\
&=&\frac{2}{-2}=-1.
\end{eqnarray*}
$$
So
$$
\lim_{x\rightarrow \pi /4}f(x)=\lim_{x\rightarrow \pi /4}e^{\log
f(x)}=e^{\lim_{x\rightarrow \pi /4}\log f(x)}=e^{-1}.
$$
