Computing: $ \lim_{x\rightarrow \frac{\pi}{4}} (\tan{x})^{\tan{2x}} $ without L'Hopital - tan and exponential

Question

I've encountered the following limit:

$$ \lim_{x\rightarrow \frac{\pi}{4}} (\tan{x})^{\tan{2x}} $$

How to find this limit?

I tried the following formula:

$$ \tan{x} = \frac{\tan{2x}-\tan{x}}{1+\tan{2x}\tan{2}}$$

But I still haven't figure it out yet. Still I hope it is helpful.

Thanks!

Answer 1

$$ \lim_{x\rightarrow \frac{\pi}{4}} (\tan{x})^{\tan{2x}} = \lim_{x\rightarrow \frac{\pi}{4}} (1+\tan{x}-1)^{\frac{1}{\tan{x}-1}\cdot(\tan{x}-1)\tan{2x}}=$$ $$=e^{- \lim\limits_{x\rightarrow \frac{\pi}{4}}\frac{2\tan{x}}{1+\tan{x}}}=\frac{1}{e}$$

Answer 2

To make life easier, let $x=y+\frac \pi 4$ which makes $$A=\left(\tan{(x)}\right)^{\tan{(2x)}}=\left(\tan \left(y+\frac{\pi }{4}\right)\right)^{-\cot (2 y)}$$ Take logarithms $$\log(A)={-\cot (2 y)}\log \left(\tan \left(y+\frac{\pi }{4}\right)\right)$$ Now, using Taylor expansions around $y=0$ $$\tan \left(y+\frac{\pi }{4}\right)=1+2 y+2 y^2+\frac{8 y^3}{3}+O\left(y^4\right)$$ $$\log \left(\tan \left(y+\frac{\pi }{4}\right)\right)=2 y+\frac{4 y^3}{3}+O\left(y^4\right)$$ $$\cot (2 y)=\frac{1}{2 y}-\frac{2 y}{3}-\frac{8 y^3}{45}+O\left(y^4\right)$$ All of that makes $$\log(A)=-1+\frac{2 y^2}{3}+O\left(y^3\right)$$ Now, Taylor again $$A=e^{\log(A)}=\frac{1}{e}+\frac{2 y^2}{3 e}+O\left(y^3\right)$$ which shows not only the limit but also how it is approached.

Edit

Since we know the exact values of trigonometric functions for multiples of $\frac \pi {24}$ (see here and here) let us make the computations for $y=\pm \frac \pi {24}$. The "exact" value is $\approx 0.372170$ while the small expansion leads to $\approx 0.372082$.