Proof that a nonzero point is continuous for g(x)

I have come across this problem in my math book for my numerical analysis class:

Let $g(x)=\sqrt[3]{x}$.

Prove that $g$ is continuous at a point $c \neq 0$.

I start my proof off the typical way for proving a continuous function:

Given some arbitrary $\epsilon > 0$, let $\delta=$ ... and then I get lost. I cannot find what my delta should be for the life of me. I am primarily struggling with the triangle inequalities for this problem, in order to find what our delta should be. Guidance is much appreciated!

This is not an answer - please do not regard it as such. This response is too long-winded for the comments.

This response is a partial exploration of the considerations needed, under the assumption that $c$ is a fixed number, with $c > 0$. The possibility of (fixed) $c < 0$ needs to be explored separately.

These comments will guide you in a complicated setting.

• $|x - c| < \delta \iff c -\delta < x < c + \delta.$

• $|x^{1/3} - c^{1/3}| < \epsilon \iff c^{1/3} - \epsilon < x^{1/3} < c^{1/3} + \epsilon.$

Suppose (for example) that $c + \delta < \left(c^{1/3} + \epsilon\right)^3$
and that $x < c + \delta$.

Then $x^{1/3} < \left(c + \delta\right)^{1/3} < \left(c^{1/3} + \epsilon\right).$

$\left(c^{1/3} + \epsilon\right)^3 = c + 3c^{2/3}\epsilon + 3c^{1/3}\epsilon^2 + \epsilon^3.$

Can $\delta$ be specified in terms of $\epsilon$ so as to guarantee that

$(c + \delta) < c + 3c^{2/3}\epsilon + 3c^{1/3}\epsilon^2 + \epsilon^3~$?

What happens if you set $\delta = \min\left[3c^{2/3}\epsilon, ~\textit{something else}\right]$,
where (something else) will be determined by the LHS of the constraint
$\left( ~\text{i.e.} ~~c^{1/3} - \epsilon < x^{1/3} ~\right)~$?

As a potential shortcut, suppose that you demonstrate that $g(x) = \sqrt[3]{x}$ is continuous for all $c > 0$. Is there then an easy way of showing that $g(x)$ is continuous at all $d < 0~$?

Note that $\{d < 0, ~c = -d\} \implies c > 0.$

Then, $g(d) = -g(c)$.
If you then know that $g(x)$ is continuous at $(x = c)$,
can you use this fact to conclude that $g(x)$ is continuous at $(x = d = -c)~$?