I'm mulling over an exercise in Abbott's textbook. We want to prove that $\forall \epsilon>0 \exists \delta>0$ such that, if $0<|x-c|<\delta$, then $|g(x)-g(x)|< \epsilon$.
A little bit of algebra confirms that
$$x-c = (x^{1/3} - c^{1/3})(x^{2/3} + c^{1/3}x^{1/3} + c^{2/3})$$
Hence, we can bound our $|g(x)-g(c)|$ appropriately,
$$|x^{1/3} - c^{1/3}| = \frac{|x-c|}{|x^{2/3} + c^{1/3}x^{1/3} + c^{2/3}|} \leq \frac{|x-c|}{c^{2/3}}$$
So choosing $\delta=:\epsilon c^{2/3}$ will do.
This is where I take issue with the answer key. For the original denominator to be greater than $c^{2/3}$, we need to know that $c^{1/3}x^{1/3} > 0$.
But this will only happen if we define $\delta < |c|$, so that transitivity guarantees that
$$|x-c| < |c| \ \Longrightarrow \ c-|c| < x<c+|c| \ \Longrightarrow \ xc>0$$
Shouldn't then $\delta :=\min\{|c|,\epsilon c^{2/3}\}$?
Yes, you are correct.
Another way to fix the proof would be to say "Without loss of generality, suppose $\epsilon < |c|^{1/3}$."