# Limit with Tan()

1. Jul 21, 2011

### juantheron

1. The problem statement, all variables and given/known data
$$\displaystyle \lim_{x\rightarrow \frac{\pi}{4}}\left(\tan x\right)^{\tan 2x}$$

2. Relevant equations

3. 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

2. Jul 21, 2011

### tiny-tim

hi juantheron!

lim (1/(1 - t)) = 1 + … ?

3. Jul 21, 2011

### SammyS

Staff Emeritus
Re: limit

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

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

4. Jul 22, 2011

### nickalh

Re: limit

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

Use
AB = (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.