Limit of $n$-th root of a polynomial as $x$ goes to infinity 
Let $P(x)$ be a monic polynomial of degree $n$, find $\lim\limits_{x_\to \infty}([P(x)]^{\frac{1}{n}}-x)$.

I'm working through some practice problems for my analysis class (quiz tomorrow!) and am stumped by this one. Can someone help me with a hint or suggestion?
I tried rationalizing but was not able to get rid of the root. Taking the log was stymied by the subtraction. This was a problem in the chapter on derivatives, so I tried taking the derivative, and tried modifying it to make L'Hospital's rule work, but they did not make it any clearer.
I have a hunch that says a series expansion at $\frac{1}{X} = 0$ could help, but the root gets in my way every time. The problem:
 A: So $$P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$
For $x\gg0$, we have a good approximation
$$ P(x)\approx\left(x+\frac{a_{n-1}}n\right)^n$$
as in: The error is a polynomial of degree $\le n-2$.
Conclude that
$$ \sqrt[n]{P(x)}-x\approx \frac{a_{n-1}}n$$
(with error bounded by some $\frac Mx$).
A: Being a monic polynomial, it looks like
$$p(x)=x^n+a_1x^{n-1}+\cdots +a_{n-1}x+a_n=x^n(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n}),$$
so
$p(x)^{1/n}-x=x((1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n}-1),$
from there we see that we are in the problematic limit $0\cdot \infty.$ As, probably, you have seen before a way to approach this is using L'Hopital rule by changing the limit to be a quotient as
$$\lim _{x\rightarrow \infty}\frac{(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n}-1}{1/x}=\lim _{x\rightarrow \infty}\frac{\frac{1}{n}(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n-1}\cdot (\frac{-a_1}{x^2}-\frac{2\cdot a_2}{x^3}\cdots -\frac{n\cdot a_n}{x^{n+1}})}{-1/x^2},$$
where the second step is the derivative in the numerator and denominator.
Notice that now everything will cancel except the $a_1$ term and the $1/n$ giving you
$$\lim _{x\rightarrow \infty}p(x)^{1/n}-x=\frac{a_1}{n}.$$
