At the moment I would use the inverse function $x^n$.
This is invertible on whole $\mathbb{R}^{+}$.
Is there a better way?
$f:\mathbb{R}^{+}_0\to\mathbb{R}^{+}_0, x\mapsto\sqrt[n]{x}$ is continuous proof
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real-analysis
continuity
radicals
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0Not necessarily better, but see [kobe's answer](http://math.stackexchange.com/questions/1133795/how-do-i-prove-using-the-definition-that-the-nth-root-is-a-continuous-function) for a direct $\epsilon-\delta$ proof. – 2017-01-09
1 Answers
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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 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.