Limit of (tan x)^(tan 2x) as x approaches π/4

[juantheron]

Homework Statement

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

Homework Equations

The Attempt at a Solution

let $$\frac{\pi}{4}-x = t\Leftrightarrow x=\frac{\pi}{4}-t$$ and $$t\rightarrow 0$$ $$\lim_{t\rightarrow 0}\left(\tan \left(\frac{\pi}{4}-t\right)\right)^{\tan \left(\frac{\pi}{2}-2t\right)}$$
$$\lim_{t\rightarrow 0}\left(\frac{1-\tan t}{1+\tan t}\right)^{\frac{1}{\tan 2t}}$$
as $$t\rightarrow 0,\tan t\approx t,\tan 2t\approx 2t$$
$$\lim_{t\rightarrow 0}\left(\frac{1- t}{1+ t}\right)^{\frac{1}{2t}}$$
after that how can i solve

 

[tiny-tim]

hi juantheron!

lim (1/(1 - t)) = 1 + … ?

 

[SammyS]

Use the fact that $\displaystyle \lim f(x)=e^{\lim\ln(f(x))}$.

In evaluating lim(ln(f(x)), use L'Hôpital's rule.

 

[nickalh]

SammyS's method is much closer to the most effective method I can think of.

Use
A^B = (e ^{ln A})^B = e ^{B * ln A}

Note: tan x is always positive in the vicinity we are interested in, so ln A is well defined.

Apply L'Hospital's rule to the power.

For oo * 0 forms, we need $\frac {oo} {\frac{1}{0}}$ or $\frac{0} {\frac{1}{oo}}$ before we can apply L'Hospital's rule.