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? 

Base on the nice hint;
$$\lim_{x \rightarrow \infty } \frac{p(x)}{2^\sqrt{x}} = \lim_{u^2 \rightarrow \infty } \frac{p(u^2)}{2^u} = \lim_{u^2 \rightarrow \infty } \frac{
  \frac{d^{2n}}
       {du^{2n}} p(u^2)} 
  {\frac{d^{2n}}
       {du^{2n}} 2^u}
   = \lim_{u^2 \rightarrow \infty } \frac{c}{2^{u} \log^{2n} 2}=0,$$ for some $c\in \mathbb{Z},$
