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 =(

Write it as $\lim_{x\to\infty} x^b  e^{-ax}=\lim_{x\to\infty} e^{-ax+b\ln x}$. For large $x$ we have $|-ax|>|b\ln x|$, so only $-ax$ contributes. So we get
$$
\lim_{x\to\infty} x^b  e^{-ax} =\lim_{x\to\infty} e^{-ax} =0
$$
