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

Here is one method to evaluate 
$$\lim_{x\rightarrow\infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$
Let $Q(x)=a_{n-1}x^{n-1}+\cdots+a_{0}$ for notational convenience, and notice $\frac{Q(x)}{x^{n-1}}\rightarrow a_{n-1}$ and $\frac{Q(x)}{x^{n}}\rightarrow0$ as $x\rightarrow\infty$. The crux is the factorization 
$$y^{n}-z^{n}=(y-z)\left(y^{n-1}+y^{n-2}z+\cdots+yz^{n-2}+z^{n-1}\right).$$ 
Setting $y=\sqrt[n]{x^{n}+Q(x)}$ and $z=x$ we find 
$$\left(\sqrt[n]{x^{n}+Q(x)}-x\right)=\frac{Q(x)}{\left(\left(\sqrt[n]{x^{n}+Q(x)}\right)^{n-1}+\left(\sqrt[n]{x^{n}+Q(x)}\right)^{n-2}x+\cdots+x^{n-1}\right)}.$$
Dividing both numerator and denominator by $x^{n-1}$ yields
$$\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x=\frac{Q(x)/x^{n-1}}{\left(\left(\sqrt[n]{1+\frac{Q(x)}{x^{n}}}\right)^{n-1}+\left(\sqrt[n]{1+\frac{Q(x)}{x^{n}}}\right)^{n-2}+\cdots+1\right)}.$$ 
As $x\rightarrow\infty$, $\frac{Q(x)}{x^{n}}\rightarrow0$ so that each term in the denominator converges to $1$. Since there are $n$ terms we find $$\lim_{x\rightarrow\infty}\left(\left(\sqrt[n]{1+\frac{Q(x)}{x^{n}}}\right)^{n-1}+\left(\sqrt[n]{1+\frac{Q(x)}{x^{n}}}\right)^{n-2}+\cdots+1\right)=n$$ by the addition formula for limits. As the numerator converges to $a_{n-1}$ we see by the quotient property of limits that $$\lim_{x\rightarrow\infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x=\frac{a_{n-1}}{n}$$ and the proof is finished.
