Cubic root epsilon delta proof

Question

I'm reviewing this epsilon delta proof for the continuity of the cubic root:

but I can't see why is so evident that $\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2}\ge \sqrt[3]{c^2}$. Any help please? Thanks in advance.

Answer 1

Note that $c\neq0$. So, if $x$ is close enough to $c$, $x$ and $c$ will have the same sign and therefore $xc>0$. Therefore $\sqrt[3]{x^2}+\sqrt[3]{cx}>0$, and this implies that $\sqrt[3]{x^2}+\sqrt[3]{cx}+\sqrt[3]{c^2}>\sqrt[3]{c^2}$.

Answer 2

Note that for the denominator, $\sqrt[3]{x^2}+\sqrt[3]{c^2}=x^{2/3}+c^{2/3}\geq 0$, so $|g(x)-g(c)|\geq\frac{|x-c|}{|cx|^{1/3}}$.

We can now bound $|x|$ by (if $|x-a|<\delta$, which we can do by choosing $\delta$ as the minimum of 2 numbers) $|x|<2|a|$ (prove this as a good exercise for working with limits) , so it follows that we can bound

$$|g(x)-g(c)|\leq\frac{|x-c|}{|c|^{2/3}}$$ and we can use the $\epsilon-\delta$ definition easily to prove $g$ is continuous