Analysis: Prove that x^(1/3) is uniformly continous but not Lipschitz

Let f(x) = x^(1/3).

Note that |f(x) - f(y)|
= |x^(1/3) - y^(1/3)|
= |x - y| / (x^(2/3) + (xy)^(1/3) + y^(2/3)).

For x, y in (-∞, 1] U [1, ∞) so that |x|, |y| ≥ 1, we can say that
|x - y| / (x^(2/3) + (xy)^(1/3) + y^(2/3))
≤ |x - y| / (1 + 1 + 1)
= |x - y|/3.

So given ε > 0, let δ = 3ε. Then for all x, y in (-∞, 1] U [1, ∞) where |x - y| < δ, we have
|f(x) - f(y)| ≤ |x - y|/3 < 3ε/3 = ε.

Thus, f(x) is uniform continuous on (-∞, 1] U [1, ∞).

However, f(x) is continuous on [-1, 1], a compact set.
Therefore, f(x) is uniformly continuous on [-1, 1] as well.
==> f(x) is uniformly continuous on R.
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f is not Lipschitz, because f '(x) = (1/3) x^(-2/3) is unbounded at x = 0.

I hope this helps!