Originally Posted by

**lvleph** There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.

I need to determine the convergence of

$\displaystyle \lim_{n \to \infty} x^n e^{-nx}$

for $\displaystyle x \in [0,1]$ and $\displaystyle x \in [0, +\infty)$

Trying to use l'Hopital's Rule I get

$\displaystyle \lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}} $

$\displaystyle \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}$

Using l'Hopital n time results in

$\displaystyle \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}$

However, I am not sure this even makes sense, because $\displaystyle f_n$ is not continuous with respect to n. So, what am I suppose to do?