Jtdylktuy

I want to evaluate the limit:$$lim_{x to infty}{x^aa^x} $$ where $0<a<1$

I put in Wolfram Alpha and I get the limit is zero

We see if the base is $0<a<1$ then :
$$lim_{x to infty}{a^x} =0$$

I tried to solve the limit with a teorem that says : $$ lim_{xto +infty} f(x) = 0$$ and $$g(x)$$ is a bounded .
Then $$lim_{xto +infty}f(x)g(x)=0$$

But $$lim_{x to infty}{x^a} =infty$$

and the product of $infty$ and $0$ is undefined

Here's a general result.

Theorem (Domination of exponentials and polynomials):
Let $a, b>0$.
Then
$$
lim_{xtoinfty}frac{x^a}{b^x} =
left{
begin{array}{cc}
+infty&text{if $,bleqslant 1$}\
0&text{if $,b> 1$}\
end{array}
right.
$$

Proof:
If $bleqslant 1$ then $b^xleqslant 1$ for all $xgeqslant 1$ and hence for $xgeqslant 1$ we have
$$
frac{x^a}{b^x} geqslant x^a stackrel{xtoinfty}{xrightarrow{hspace{1cm}}}+infty.
$$
If $b>1$ then $b^x to+infty$ as $xtoinfty$, so we may apply L'Hôpital's rule.
Observing that $b^x = expbig(xlog (b)big)$, we'll obtain that
$$
lim_{xtoinfty}frac{x^a}{b^x} =
lim_{xtoinfty} frac{a}{log(b)}frac{x^{a-1}}{b^x}
$$
Notice that up to the positive constant $a/log (b)$, we arrive at the same limit we began with, except the initial value of the exponent $a$ in the polynomial term has decreased by $1$.
We can successively apply L'Hôpital's rule until the exponent becomes non-positive, at which point the limit is easily seen to be $0$. $square$

For your particular question, we can apply the general result with $b=frac1a > 1$.

Let $x=1+t$ then Bernoulli's inequality shows
$$lim_{x to infty}{x^aa^x}=lim_{t to infty}{(1+t)^aa^{1+t}}>lim_{t to infty}{(at)a^{1+t}}=a^2lim_{t to infty}{ta^{t}}=0$$
the last one holds by L'Hospital.

Consider
$$y=a^x ,x^a implies log(y)=a log(x)+x log(a)=x left(afrac{ log (x)}{x}+log (a)right)$$ Since $xto infty$, $frac{ log (x)}{x}$ is negligible for any $a$. So
$$log(y) sim x log(a)$$ and, since $0 < a < 1$, $log(a) <0$. Then $log(y)to -infty$ and $y=e^{log(y)} to 0$