Limit at infinity $\lim_{x\to \infty} x^a a^x=$? $\displaystyle \lim_{x\to \infty} x^a a^x=?$; $0<a<1$
I try to use the property: $a^{\log_a x}= x$ and reescribe the expression 
$\displaystyle \lim_{x\to \infty} x^a a^x  =  \lim_{x\to \infty} \frac{a^{a\log_a x}}{a^{-x}} $ but i can't find the limit yet.
Any suggestion?

Consider $x=ay$; then your expression becomes
$$
x^aa^x=(ay)^aa^{ay}=a^a(ya^y)^a
$$
so you just need to compute
$$
\lim_{y\to\infty}ya^y
$$
For $a\ge1$ the limit is $\infty$. For $0<a<1$ rewrite it as
$$
\lim_{y\to\infty}\frac{y}{a^{-y}}=\lim_{y\to\infty}\frac{1}{-a^{-y}\log a}=0
$$
with a simple application of l'Hôpital.
