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 $x_0$ .

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