 # How to evaluate \lim_{x\rightarrow\infty} \root(n)(x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0})-x Saige Shannon 2022-04-20 Answered
How to evaluate
$\underset{x\to \mathrm{\infty }}{lim}\sqrt[n]{{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}}-x$
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Alternatively, rewrite this limit as
$\underset{x\to \mathrm{\infty }}{lim}\sqrt[n]{{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}}-x$
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 $z=\frac{{a}_{n-1}}{x}+\dots +\frac{{a}_{0}}{{x}^{n}}$ we see that $z=O\left(\frac{1}{x}\right)$, and hence
$\sqrt[z]{1+z}=1+\frac{1}{n}\left(\frac{{a}_{n-1}}{x}+\dots +\frac{{a}_{0}}{{x}^{n}}\right)+O\left(\frac{1}{{x}^{2}}\right)=1+\frac{{a}_{n-1}}{n}\frac{1}{x}+O\left(\frac{1}{{x}^{2}}\right)$
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
$\underset{x\to \mathrm{\infty }}{lim}x\left(\sqrt[n]{1+\frac{{a}_{n-1}}{x}+\cdots +\frac{{a}_{0}}{{x}^{n}}}-1\right)=\frac{{a}_{n-1}}{n}.$