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? 
 A: $$|\sqrt[3]{x}-\sqrt[3]{a}| = |\sqrt[3]{x}-\sqrt[3]{a}| \times \frac {| x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} = \frac {|x - a|}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} \le \frac {| x - a |}{| {ax} |^{\frac 1 3}}$$
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} $
A: Let $\epsilon>0$ be given. 
Case 1:
Let $a>0$. Then we have 
\begin{align*}
|f(x)-f(a)|=\left|x^{1/3}-a^{1/3}\right|\frac{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|}{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|}=\frac{|x-a|}{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|}
\end{align*}
Now consider $x\in \mathbb R$ such that $x>\frac{a}{2}$. Then $x^{1/3}>0$ and $a^{1/3}>0$. Clearly $x^{2/3}, a^{2/3}>0$. 
$$
\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right| = x^{2/3}+a^{2/3}+x^{1/3}a^{1/3} >a^{2/3}
$$
Case 2:
Let $a<0$. Now consider $x\in \mathbb R$ such that $x<\frac{a}{2}$. Then $x^{1/3}<0$ and $a^{1/3}<0$.  Clearly $x^{2/3}, a^{2/3}>0$. 
$$
\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right| = x^{2/3}+a^{2/3}+x^{1/3}a^{1/3} > a^{2/3}
$$
Then we have 
$$
|f(x)-f(a)|< \frac{|x-a|}{a^{2/3}}
$$
Let us choose $\delta=\min \left\{a^{2/3}\epsilon, \dfrac{|a|}{2}\right\}$ and consider $x\in \mathbb R$ such that $|x-a|<\delta$. Then we have
$$
|f(x)-f(a)|<\epsilon
$$
Case 3: Suppose $a=0$, then choose $\delta=\epsilon^3$.
$$
|f(x)-f(0)|=|x^{1/3}|<\delta^{1/3}=\epsilon
$$
Hence $f$ is continuous on $\mathbb R$.
