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

[Same-same]

Homework Statement

$\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$

$\frac{Lim}{x-> \pi/4}tan(2x)$ does not exist.
However, Wolfram Alpha and my TI-89 say that $\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$does exist, and that it's $\frac{1}{e}$
I submitted this answer (it's web based homework and calculators are allowed) and found it was correct. I still don't understand how though.

Homework Equations

tan(2x) = $\frac{2tan(x)}{1-tan^{2}(x)}$

The Attempt at a Solution

I attempted to split it in to
$\frac{sin(x)^{tan(2x)}}{cos(x)^{tan(2x)}}$, and then use L'Hospital's rule, but I can't seem to get the tan(2x) to go away. Both appear to be indeterminate, but neither one is 0 at the same time, so it doesn't appear that I should be using L'Hopital's rule in this case. However, I can't see any other way to proceed.

Thanks in advance.

 

[Same-same]

Thanks.

So people can't do these tasks without calculators anymore? :-)

Hey, at the very least my professor doesn't allow them on tests.

 

[jbunniii]

Have you tried using

$$\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}$$

 

[NasuSama]

Have you tried using

$$\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}$$

There is the good response here, and I particularly agree with this because this generally makes the simplification a bit simple.

Don't forget that you have the function as the exponent of the another function! Here is the hint:

Let y = lim x→π/4 (tan(x))^(tan(2x)). Then, perform logarithms, and we have...

ln(y) = lim x→π/4 tan(2x) * ln(tan(x))

Mod note: Removed intermediate steps students should work out on their own.[/color]

Also don't forget to set both sides by e. You should get the results. Let me know if this helps.

Key: ♪ Practice, practice! You will get better with limits like this! ♫

 

[SammyS]

Homework Statement

$\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$

$\frac{Lim}{x-> \pi/4}tan(2x)$ does not exist.
However, Wolfram Alpha and my TI-89 say that $\frac{Lim}{x-> \pi/4}$ tan(x)$^{tan(2x)}$does exist, and that it's $\frac{1}{e}$
I submitted this answer (it's web based homework and calculators are allowed) and found it was correct. I still don't understand how though.

Homework Equations

tan(2x) = $\frac{2tan(x)}{1-tan^{2}(x)}$

The Attempt at a Solution

I attempted to split it in to
$\frac{sin(x)^{tan(2x)}}{cos(x)^{tan(2x)}}$, and then use L'Hospital's rule, but I can't seem to get the tan(2x) to go away. Both appear to be indeterminate, but neither one is 0 at the same time, so it doesn't appear that I should be using L'Hopital's rule in this case. However, I can't see any other way to proceed.

Thanks in advance.

Yes, it's true that$\displaystyle \lim_{x\to \pi/4} \tan(2x)$ does not exist.

But $\displaystyle \lim_{x\to \pi/4} \tan(x)=1\,,$ and $\displaystyle \lim_{x\to (\pi/4)^+} \tan(2x)=+\infty\ .$


Find the limit of the log of that expression.

If $\displaystyle \lim_{x\to \pi/4} \ln\left(\left(\tan(x)\right)^{\tan(2x)}\right)=L\,,$

then $\displaystyle \lim_{x\to \pi/4} \left(\tan(x)\right)^{\tan(2x)}=e^L\ .$