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 $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$

Alternatively, rewrite this limit as 
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right).$$
Consider the Taylor expansion around $0$ of $\sqrt[n]{1+z}$. We have 
$$\sqrt[n]{1+z}=1+\frac{1}{n}z+O\left(z^{2}\right).$$ Setting $z=\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}$ we see that $z=O\left(\frac{1}{x}\right)$, and hence 
$$\sqrt[n]{1+z}=1+\frac{1}{n}\left(\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)+O\left(\frac{1}{x^{2}}\right)=1+\frac{a_{n-1}}{n}\frac{1}{x}+O\left(\frac{1}{x^{2}}\right).$$ Thus we have 
$$x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}+O\left(\frac{1}{x}\right)$$ 
and we conclude  
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}.$$
