Prove continuity for cubic root using epsilon-delta I am trying to prove that a function is continuous at a point a using the $\epsilon$-$\delta$ theorem. I managed to find a $\delta$ in this case $|2x^2+1 - (2a^2+1)| < \epsilon$. But I have a hard time when the function under consideration is $f(x) = \sqrt[3]{x}$. That is, I want to have if $|x-a|<\delta$, then $|\sqrt[3]{x}-\sqrt[3]{a}| < \epsilon$. Any suggestions? 

$$|\sqrt[3]{x}-\sqrt[3]{a}| = |\sqrt[3]{x}-\sqrt[3]{a}| \times \frac {| x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} = \frac {|x - a|}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} \le \frac {| x - a |}{| {ax} |^{\frac 1 3}}$$
The final inequality is due to the fact that $x^{\frac 2 3} + a^{\frac 2 3} \ge 0$. 
Let us assume $a \neq 0$. Then We can bound $|x|$ as follows. 
Say $|x - a| \lt |a| $ then $ |x| \lt 2|a| \implies |ax| \lt 2|a|^2 \implies \frac {1}{2|a|^2} \lt \frac {1} {|ax|}$.  
Therefore $|\sqrt[3]{x}-\sqrt[3]{a}| \lt \frac {| x - a |}{| {a} |^{\frac 2 3}}$ as long as $|x - a| \lt |a|$ and $a \neq 0$. So if we pick $\delta = \text {Min} \{ |a|, \epsilon|a|^{\frac 2 3} \}$ then $|x - a| \lt \delta \implies |\sqrt[3]{x}-\sqrt[3]{a}| \lt \epsilon$. This proves the function is continuous everywhere except at $0$. To get rid of the case when $a = 0$ just pick $\delta = \epsilon^{3} $
