Limit for sequence with $e$-function I need some help :)
$a,b \in \mathbb R$ and $a,b > 0$.
What is? $\lim_{x\to\infty} x^b  e^{-ax} $
It should be $0$ but I'm not sure how to prove it... - EDIT: Without L-Hospital =(

$$\lim_{x\rightarrow\infty}x^be^{-ax}=\lim_{x\rightarrow\infty}\frac{x^b}{e^{ax}}$$
Now use L'Hopital $\lfloor b+1 \rfloor$ times...
