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Is there a way to show that $x^\frac{1}{3}$ is continuous on $[-1,1]$ without resorting to arguments about compactness or continuous functions on bounded intervals being uniformly continuous? (aka manipulating only $\epsilon-\delta$)

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You could also use the Hölder condition. – dtldarek Feb 10 '14 at 11:34

Hint: the "worst" point in this example is $0$. Given $\epsilon$ find an appropriate $\delta$ that works at the point $0$. Then check that this $\delta$ works in fact for all point in $[-1,1]$.

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When I did this earlier, I got $\delta<\epsilon^3$, which should be the "worst" point. I then got that if epsilon is relatively large compared to the point in question on the real line, we get that the interval around $x=0$ isn't actually the "worst interval length"? For example, $\epsilon=.125\implies \delta<.5$. Consider the point $x=.2$. $|-.3^\frac{1}{3}-.2^\frac{1}{3}|>\epsilon$. What's going on? – user117510 Feb 10 '14 at 9:32
I'm not sure what you are asking but $0.125^3$ is not $0.5$. – Rasmus Feb 10 '14 at 10:15

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