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

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5 Answers 5

up vote 51 down vote accepted

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 ${a_{n-1} \over n}$.

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That's the easiest way to do it and I can't see what's gained by doing it any more abstract way,Zarrax. –  Mathemagician1234 Aug 29 '11 at 19:12

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

$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}.$$

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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.

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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, $$ \mathop {\lim }\limits_{x \to \infty } [\sqrt[n]{{x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 }} - x] = \mathop {\lim }\limits_{x \to \infty } \bigg[\bigg(x + \frac{{a_{n - 1} }}{n}\bigg) + O(x^{ - 1}) - x\bigg] = \frac{{a{}_{n - 1}}}{n}. $$

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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 $\frac{a_{n-1}}{n}$.

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1  
Looks interesting, but when I read it twice, I found it quite similar to the proof of Eric Naslund's. –  user1551 Aug 30 '11 at 0:38
    
@user1551 Yes, you are right, I missed that. +1 to Eric. –  Sasha Aug 30 '11 at 4:38

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