Proving a limit with the epsilon-delta definition I was looking to prove using the $\epsilon,\delta$ limit definition that $\lim_{x\to a}(\sqrt[3]{x})=\sqrt[3]{a}$, $(a>0)$. I'm not sure what sort of algebraic manipulation I should use on the expression $|\sqrt[3]{x}-\sqrt[3]{a}|$ (so I'll be able to continue with proving the limit).
Just a little hint would be great, thanks in advance.
 A: HINT: Use the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$: 
$$\sqrt[3]x=\sqrt[3]a=\left(\sqrt[3]x-\sqrt[3]a\right)\cdot\frac{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$$
A: You can use the identity
$$
x - a = \left( \sqrt[3]{x} - \sqrt[3]{a} \right) \left( \sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} \right),
$$
which is derived from the following identity:
$$
a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}).
$$
Then
$$
\forall x \in \mathbb{R} \setminus \{ a \}: \quad \sqrt[3]{x} - \sqrt[3]{a} = \frac{x - a}{\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}}}.
$$
Now, fix $ \epsilon > 0 $. Choose $ x \in \mathbb{R} $ so that
$$
|x - a| < \min \left( \frac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right).
$$
As $ a > 0 $, having $ |x - a| < \dfrac{a}{2} $ ensures that $ x > 0 $, and so
$$
\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} > \sqrt[3]{a^{2}}.
$$
Next, having $ |x - a| < \sqrt[3]{a^{2}} \cdot \epsilon $ yields
$$
\left| \sqrt[3]{x} - \sqrt[3]{a} \right| < \epsilon.
$$
You can therefore set $ \delta := \min \left( \dfrac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right) $.
A: Multiply and divide $\sqrt[3]{x}-\sqrt[3]{a}$ by conjugate:
$$\dfrac{\left(\sqrt[3]{x}-\sqrt[3]{a} \right) \left(\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2} \right) }{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}=\dfrac{x-a}{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}$$
