Expert Answer to "How to evaluatelimx→∞xn+an−1xn−1+⋯+a0n−x"

Saige Shannon

Answered question

2022-04-20

How to evaluate

lim_{x \to \infty} \sqrt[n]{x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}} - x

Answer & Explanation

Salvador Ayala

Beginner2022-04-21Added 10 answers

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

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