Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates.

What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$
In other words, if I am given a polynomial $P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0$, how would I find $$\lim_{x\rightarrow\infty} P(x)^{1/n}-x.$$
For example, how would I evaluate limits such as $$\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x$$ or $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$

First note that
$$
\sqrt[n]{{x^n  + a_{n - 1} x^{n - 1}  +  \cdots  + a_0 }} = \sqrt[n]{{\bigg(x + \frac{{a_{n - 1} }}{n}\bigg)^n  + O(x^{n - 2} )}}.
$$
By the mean value theorem applied to the function $f(y)=y^{1/n}$ (whose derivative is $n^{-1}y^{1/n-1}$), we have
$$
\sqrt[n]{{\bigg(x + \frac{{a_{n - 1} }}{n}\bigg)^n  + O(x^{n - 2} )}} - \sqrt[n]{{\bigg(x + \frac{{a_{n - 1} }}{n}\bigg)^n  }} = (x^n )^{1/n - 1} O(x^{n - 2} ) = O(x^{ - 1} ).
$$
Hence, 
$$
\mathop {\lim }\limits_{x \to \infty } [\sqrt[n]{{x^n  + a_{n - 1} x^{n - 1}  +  \cdots  + a_0 }} - x] = \mathop {\lim }\limits_{x \to \infty } \bigg[\bigg(x + \frac{{a_{n - 1} }}{n}\bigg) + O(x^{ - 1}) - x\bigg] = \frac{{a{}_{n - 1}}}{n}.
$$
