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 $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$
 A: Possibly more elementary proof based on $\frac{c^{n}-d^{n}}{c-d} = \sum_{k=0}^{n-1} c^{n-1-k} d^k$. Using this for $c = \sqrt[n]{ x^n+ \sum_{m=0}^{n-1} a_m x^m }$ and $d=x$.
$$
  c - d = \frac{c^{n}-d^{n}}{ \sum_{k=0}^{n-1} c^{n-1-k} d^k } = \frac{ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0}{ x^{n-1} \sum_{k=0}^{n-1} (\frac{c}{d})^{n-1-k}  } = \frac{ a_{n-1} + a_{n-2} x^{-1} + \ldots +  a_0 x^{1-n}}{\sum_{k=0}^{n-1} (\frac{c}{d})^{n-1-k}  }
$$
Now $\lim_{x\to \infty} \frac{c}{d} = \lim_{x \to \infty} \sqrt[n]{ 1 + \frac{a_{n-1}}{x} + \ldots + \frac{a_0}{x^n} } = 1$. This gives $\frac{a_{n-1}}{n}$.
A: Your limit can be rewritten as 
$$\lim_{x\rightarrow\infty}\left(\frac{\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1}{1 \over x}\right)$$
Or equivalently,
$$\lim_{y\rightarrow 0}\left(\frac{\sqrt[n]{1+{a_{n-1}}{y}+\cdots+{a_{0}}{y^{n}}}-1}{y}\right)$$
This, by the definition of derivative, is the derivative of the function $f(y) = {\sqrt[n]{1+{a_{n-1}}{y}+\cdots+{a_{0}}{y^{n}}}}$ at $y = 0$, which evaluates via the chain rule to ${a_{n-1} \over n}$.
A: 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  
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}.$$
A: 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.
A: 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, 
$$
\mathop {\lim }\limits_{x \to \infty } [\sqrt[n]{{x^n  + a_{n - 1} x^{n - 1}  +  \cdots  + a_0 }} - x] = \mathop {\lim }\limits_{x \to \infty } \bigg[\bigg(x + \frac{{a_{n - 1} }}{n}\bigg) + O(x^{ - 1}) - x\bigg] = \frac{{a{}_{n - 1}}}{n}.
$$
A: Let $n$ be an integer $\geqslant 2,$ and $q(y)$ a polynomial of
degree $\leqslant n - 2.$ Then $y^n + q(y) > 0$ for all sufficiently
large $y.$ For such $y,$ define $z = \sqrt[n]{y^n + q(y)}.$ Then
$z > 0,$ and
$$
y^{n-1}|z - y| \leqslant
|z^{n-1} + z^{n-2}y + \cdots + zy^{n-2} + y^{n-1}||z - y| =
|z^n - y^n| = |q(y)|,
$$
whence $z - y \to 0$ as $y \to \infty.$
Writing $x = y - a_{n-1}/{n},$ we have $y \to \infty$ as $x \to \infty,$ and so
$$
\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}} - x =
\sqrt[n]{y^n + q(y)} - y + \frac{a_{n-1}}{n} \to \frac{a_{n-1}}{n}.
$$
