continuity of natural power

Theorem.

Let $n$ be arbitrary positive integer.  The power function  $x\mapsto x^n$  from $ℝ$ to $ℝ$ (or $ℂ$ to $ℂ$) is continuous at each point $x_0$.

Proof.  Let $\epsilon$ be any positive number.  Denote  $x_0+h=x$  and  $x^n-x_0^n=\mathrm{\Delta }$.  Then identically

$$\mathrm{\Delta }=(x-x_0)(x^{n-1}+x^{n-2}{x}_0+\mathrm{\dots }+x_0^{n-1}).$$

Taking the absolute value and using the triangle inequality give

$$|\mathrm{\Delta }|=|h|\cdot |x^{n-1}+x^{n-2}{x}_0+\mathrm{\dots }+x_0^{n-1}|\leqq |h|\cdot (|x^{n-1}|+|x^{n-2}{x}_0|+\mathrm{\dots }+|x_0^{n-1}|).$$

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

$$|\mathrm{\Delta }|\leqq n|h|{(|x_0|+|h|)}^{n-1}.$$

We may choose  ;  this implies

$$|\mathrm{\Delta }|\leqq n|h|{(|x_0|+1)}^{n-1}.$$

The right hand side of this inequality is less than $\epsilon$ as soon as we still require

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.