# Thread: [SOLVED] stuck on a limit and out of ideas

1. Originally Posted by Moo
I'm sorry, I guess I was very tired yesterday, to write such a thing.

But I don't agree with Prove_it's method, since it's not exactly x^x here... but it looks like it gives the correct result

I applied twice (or three times) l'Hopital's rule :

Let's start like you did
$\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\ln(e^x-x-1)}{1/x}$
this is in the form $\displaystyle \frac{\infty}{\infty}$, because :
o $\displaystyle e^x-x-1 \to 0 \Rightarrow \ln(\dots) \to -\infty$
o $\displaystyle 1/x \to \infty$

So we can apply l'Hopital's rule :
$\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\frac{e^x-1}{e^x-x-1}}{-1/x^2}=\lim_{x \to 0} ~ -x^2 \cdot \frac{e^x-1}{e^x-x-1}$
One last thing, if everyone is not to tired of this. Part of my frustration was that for some reason I thought that to apply lHopital's rule the equation had to either be in the form $\displaystyle \frac{+\infty}{+\infty}$ or $\displaystyle \frac{-\infty}{-\infty}$. That was part of what was geting me off track. So is that not true?

2. Originally Posted by sinewave85
One last thing, if everyone is not to tired of this. Part of my frustration was that for some reason I thought that to apply lHopital's rule the equation had to either be in the form $\displaystyle \frac{+\infty}{+\infty}$ or $\displaystyle \frac{-\infty}{-\infty}$. That was part of what was geting me off track. So is that not true?
YOu can also apply the rule when you have $\displaystyle \frac 00$ (and by the way, the signs are not important, you can also apply this when you have $\displaystyle \frac{\infty}{-\infty}$)
and going from $\displaystyle \frac \infty \infty$ to $\displaystyle \frac 00$ is simple, you'll see why

assume that x goes to 0 and y goes to 0 (but different from 0 !)
$\displaystyle \frac xy$ is in the form $\displaystyle \frac 00$

now you can divide both numerator and denominator by xy :
$\displaystyle \frac xy=\frac{\frac{x}{xy}}{\frac{y}{xy}}=\frac{\frac 1y}{\frac 1x}$

and this is in the form $\displaystyle \frac \infty \infty$, since x and y go to 0 !

3. Originally Posted by Moo
YOu can also apply the rule when you have $\displaystyle \frac 00$ (and by the way, the signs are not important, you can also apply this when you have $\displaystyle \frac{\infty}{-\infty}$)
and going from $\displaystyle \frac \infty \infty$ to $\displaystyle \frac 00$ is simple, you'll see why

assume that x goes to 0 and y goes to 0 (but different from 0 !)
$\displaystyle \frac xy$ is in the form $\displaystyle \frac 00$

now you can divide both numerator and denominator by xy :
$\displaystyle \frac xy=\frac{\frac{x}{xy}}{\frac{y}{xy}}=\frac{\frac 1y}{\frac 1x}$

and this is in the form $\displaystyle \frac \infty \infty$, since x and y go to 0 !
I did know that l'Hopital's rule worked for the form $\displaystyle \frac{0}{0}$ (probably should have mentioned that) but I did not know that you could mix the signs on $\displaystyle \frac \infty \infty$. A valuable piece of information, that. Thanks, Moo.

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