Prove $ \lim_{n\to\infty}\sqrt[n]{p(n)}=1 $ for polynomial Let $ p(x)=\sum\limits_{k=0}^{d}a_kx^k $ polynomial such that $  \forall x>0, p(x)>0 $

Prove that:
  $$ \lim_{n\to\infty}\sqrt[n]{p(n)}=1 $$

 A: Hint:
Find a lower and upper bound that converge to 1 and use the squeeze theorem. A possible upper bound would be to use
$$ \lim_{n->\infty}\sqrt[n]{p(n)} \leq \lim_{n->\infty}\sqrt[n]{c n^{k+1}}$$
because
$$ \sqrt[n]{p(n)} \leq \sqrt[n]{c n^{k+1}} $$
where c is some number takes care of the finite increase produced by $a_k$ (since it's a partial sum). The upper bound can be found with L'Hopital's limit rule.
A: From the fact that $\forall x>0$ $p(x)>0$ we get that $a_d>0$. Also $p(n)\to +\infty$ as $n\to +\infty$. So $\exists$ $N\ge 1$ such that $p(n)>1$ when $n>N$. Then we get that $\sqrt[n]{p(n)}>1$ when $n>N$.
In the rest of this answer we're assuming that $n>N$ and $n\ge d+1$.
Let $\sqrt[n]{p(n)}=1+\alpha_n$, where $\alpha_n>0$. Then by Binomial theorem:
$$p(n)=(1+\alpha_n)^n=\sum_{i=0}^n \binom{n}{i}\alpha_n^i>\binom{n}{d+1}\alpha_n^{d+1}$$
$$0<\alpha_n^{d+1}<\frac{p(n)}{\binom{n}{d+1}}=\frac{(d+1)!\left(\sum_{i=0}^d a_i n^i\right)}{\prod_{i=0}^d (n-i)}=$$
$$=\frac{(d+1)!\left(\sum_{i=0}^d a_i n^{i-d-1}\right)}{\prod_{i=0}^d (1-\frac{i}{n})}\to \frac{(d+1)!\left(\sum_{i=0}^d a_i \cdot 0\right)}{\prod_{i=0}^d (1-0)}=0$$
as $n\to +\infty$. By Squeeze theorem $\alpha_n^{d+1}\to 0$, so $\alpha_n\to 0$, so $\sqrt[n]{p(n)}\to 1$ as $n\to +\infty$.
A: That polynomials will strictly increase after some point and you will have after some $n_0$ $$a_0<p(n)<a_dn^{2d}$$
In the limit...
A: Since $p(x)>0$ for $x>0$ necessary we have $a_d>0$  then, 
$$ p(n)=\sum_{k=0}^{d}a_kn^k  =  n^d \left( a_d+  \sum_{k=0}^{d-1}a_k n^{k-d}\right) ~~~~~~and~~~~~\sum_{k=0}^{d-1}a_k n^{k-d}\to 0$$ 
Hence, since  $n^{1/n}\to1 $  it follows that, $$ \lim_{n\to\infty}\sqrt[n]{p(n)} = \lim_{n\to\infty}n^{d/n} \left( a_d+  \sum_{k=0}^{d-1}a_k n^{k-d}\right)^{1/n}=\lim_{n\to\infty}n^{d/n} \left( a_d\right)^{1/n}=1$$
A: Note that
$$\sqrt[n]{p(n)}=e^{\frac{\log p(n)}{n}}\to1$$
indeed
$$\frac{\log p(n)}{n}=\frac{\log \left(a_dn^d+a_{d-1}n^{d-1}+...a_0\right)}{n}=\frac{\log \left(n^d\right)+\log \left(a_d+a_{d-1}n^{-1}+...a_0n^{-d}\right)}{n}=\frac{d\log n}{n}+\frac{\log \left(a_d+a_{d-1}n^{-1}+...a_0n^{-n}\right)}{n}\to0+0=0$$
