$f:\mathbb{R}^{+}_0\to\mathbb{R}^{+}_0, x\mapsto\sqrt[n]{x}$ is continuous proof

Question

At the moment I would use the inverse function $x^n$.

This is invertible on whole $\mathbb{R}^{+}$.

Is there a better way?

Answer 1

You can use the equality $(a-b)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1})=a^n-b^n$, which implies that $$|x-y|=|\sqrt[n]{x}-\sqrt[n]{y}|(\sqrt[n]{x}^{(n-1)}+\sqrt[n]{x}^{(n-2)}\sqrt[n]{y}+\cdots +\sqrt[n]{y}^{(n-1)})$$

If you want to check the continuity in some $x>0$, then you can assume that $\frac{1}{2}x<y$ so that $$|x-y|\geq|\sqrt[n]{x}-\sqrt[n]{y}|n((\frac{1}{2}x)^{\frac{n-1}{n}})$$ so if $|x-y|$ is small, then so is $|\sqrt[n]{x}-\sqrt[n]{y}|$.

The continuity in $0$ follows from the fact that $y$ is small iff $\sqrt[n]{y}$ is small, or equivalently, $\varepsilon$ is small iff $\varepsilon^n$ is small.