ContinuityOfNaturalPower 2006-02-05 07:36:08 2006-02-05 12:21:16 Theorem continuous % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} \begin{thmplain} Let $n$ be arbitrary positive integer.\, The power function\, $x\mapsto x^n$\, from $\mathbb{R}$ to $\mathbb{R}$ (or $\mathbb{C}$ to $\mathbb{C}$) is continuous in each point $x_0$. \end{thmplain} {\em Proof.}\, Let $\varepsilon$ be any positive number.\, Denote\, $x_0+h = x$\, and\, $x^n-x_0^n = \Delta$.\, Then identically $$\Delta = (x-x_0)(x^{n-1}+x^{n-2}x_0+...+x_0^{n-1}).$$ Taking the absolute value and using the triangle inequality give $$|\Delta| = |h|\cdot|x^{n-1}+x^{n-2}x_0+...+x_0^{n-1}| \leqq |h|\cdot(|x^{n-1}|+|x^{n-2}x_0|+...+|x_0^{n-1}|).$$ But since\, $|x| = |x_0+h| \leqq |x_0|+|h|$\, and also\, $|x_0| \leqq |x_0|+|h|$,\, so each summand in the parentheses is at most equal to\, $(|x_0|+|h|)^{n-1}$,\, and since there are $n$ summands, the sum is at most equal to $n(|x_0|+|h|)^{n-1}$.\, Thus we get $$|\Delta| \leqq n|h|(|x_0|+|h|)^{n-1}.$$ We may choose\, $|h| < 1$;\, this implies $$|\Delta| \leqq n|h|(|x_0|+1)^{n-1}.$$ The right hand side of this inequality is less than $\varepsilon$ as soon as we still require $$|h| < \frac{\varepsilon}{n(|x_0|+1)^{n-1}}.$$ This means that the power function\, $x\mapsto x^n$ is continuous in the point $x_0$. \textbf{Note.}\, Another way to prove the theorem is to use induction on $n$ and the rule 2 in limit rules of functions.