# 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.

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.

Homework Helper
Gold Member
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.

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! ♫

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Staff Emeritus
Homework Helper
Gold Member

## 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.

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