How to show that $\lim_{x \rightarrow \infty } \frac{p(x)}{2^{\sqrt x}} = 0$ I want to show that the below limit is 0 for any polynomial $p(x)$ with degree $n$
$$\lim_{x \rightarrow \infty } \frac{p(x)}{2^{\sqrt x}} = 0$$
If I apply the l'Hopital's Rule, the numerator, eventually, will be zero. What about the denumerator? 

It suffices to show 
$\lim_{x \rightarrow \infty}\dfrac{x^n}{2^{√x}} =0$
(why?).
1) Let $y=√x, y >0.$
Then $F(y)= \dfrac{y^{2n}}{2^y}.$
2) Let $e^a=2,$ $a >0.$
$F(y)= \dfrac{y^{2n}}{e^{ay}}.$
$e^{ay} \gt \dfrac{(ay)^{2n+1}}{(2n+1)!}$
(Series expansion).
$F(y) \lt \dfrac{(2n+1)! y^{2n}}{(ay)^{2n+1}}=$
$\dfrac{(2n+1)!}{a^{2n+1}y}=(\dfrac{(2n+1)!}{a^{2n+1}})\dfrac{1}{y}.$
Take the limit  ${y \rightarrow \infty}.$
