How do I show $x^\frac{1}{3}$ is uniformly continuous?

Question

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 $\epsilon-\delta$)

Answer 1

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 $[-1,1]$.