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 :

$\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\frac{e^x-1}{e^x-x-1}}{-1/x^2}=\lim_{x \to 0} ~ -x^2 \cdot \frac{e^x-1}{e^x-x-1}$