How to calculate $\lim\limits_{x\to \pi/4}{{\tan(x)}^{\tan(2x)}}$?

Question

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.

Answer 1

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 $1$.

Answer 2

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 $$ \lim_{x\rightarrow \pi /4}f(x)=\lim_{x\rightarrow \pi /4}e^{\log f(x)}=e^{\lim_{x\rightarrow \pi /4}\log f(x)}=e^{-1}. $$