How to prove that $x^n$ is continuous on $(0, \infty)$

Question

I want to prove that $x^n$ is continuous on the open interval $(0, \infty)$, where $n$ is a positive integer.

So I have to show:

$\forall\varepsilon>0$ there is a $\delta>0$ such that: $$|x-a|<\delta\implies|x^n - a^n|<\varepsilon.$$

I do not see how I can choose $\delta$ for which this is true.

Answer 1

It is trivial to show that $x \mapsto x$ is continuous.

Now, use that the product of 2 continuous functions is again continuous.

Answer 2

Since$$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+\cdots+a^{n-1})$$and, if $|x-a|<1$,$$|x^{n-1}+ax^{n-2}+\cdots+a^{n-1}|<(a+1)^{n-1}+a(a-1)^{n-2}+\cdots+a^{n-1},$$if, given $\varepsilon>0$, you take $\delta=\min\left\{1,\frac\varepsilon{(a+1)^{n-1}+a(a-1)^{n-2}+\cdots+a^{n-1}}\right\}$, then$$|x-a|<\delta\implies\bigl|x^n-a^n\bigr|<\varepsilon.$$

Answer 3

Let $\hat x = x - a$

$|x^n - a^n| = |(\hat x + a)^n - a^n| = |\sum^n_{m=1} \binom{n}{m} \hat x^ma^{n-m} | \leq \sum^n_{m=1} |\binom{n}{m} \hat x^ma^{n-m}| $

To progress from here let $M = {max}_{m=1..n} \binom{n}{m} a^{n-m}$.

Let $\delta < 1$ then $|x - a| = |\hat x | < \delta = 1$ and $|\hat x |^m < |\hat x |$ for all m in equation above

$|x^n - a^n| \leq M*n * |\hat x| < M*n* \delta $

So any $\delta < min(1,\frac{\epsilon}{M*n})$ will make $|x^n-a^n| < \epsilon$ This proves that they exist.