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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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    @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.2012-11-14

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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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    @o2genum I'm not sure but probably it can. Please add this restriction to your question.2012-11-14