# continuity of natural power

###### Theorem.

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 at each point $x_{0}$.

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 at the point $x_{0}$.

Note.  Another way to prove the theorem is to use induction on $n$ and the rule 2 in limit rules of functions.

Title continuity of natural power ContinuityOfNaturalPower 2013-03-22 15:39:25 2013-03-22 15:39:25 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 26C05 msc 26A15 Exponentiation2