# 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.

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What is $tg(x)$? Is that the tangent? – icurays1 Nov 14 '12 at 16:35
@icurays1 Yes, I fixed it – o2genum Nov 14 '12 at 16:36
Your idea to take the logarithm is a good one. That will bring the exponent down front, then the limit will probably be more clear. If you're still stuck I'll submit an answer for you. – icurays1 Nov 14 '12 at 16:41
@icurays1 Still stuck. Indeed, I got an answer ($e^{-1/2}$), but it's not the one WolframAlpha gives. I made a mistake somewhere. I would like to see your calculations. – o2genum Nov 14 '12 at 16:47

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 $1$.

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

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And can it be calculated without using L'Hôpital's rule? We haven't studied it yet, and the task is supposed to solve without that rule. – o2genum Nov 14 '12 at 17:19
@o2genum I'm not sure but probably it can. Please add this restriction to your question. – Américo Tavares Nov 14 '12 at 17:22