# 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.

• Remember $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$ – Nameless Nov 9 '12 at 17:02

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 $\delta := \min \left( \dfrac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right)$.
• @HaskellCurry why don't you just choose $\delta=\sqrt[3] {a^2} \epsilon$ – Mathematics Nov 9 '12 at 17:32
• I would like to do that, but I want $x > 0$ so as to be able to exploit the inequality $\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} > \sqrt[3]{a^{2}}$. If I choose $\delta = \sqrt[3]{a^{2}} \cdot \epsilon$, then this $\delta$ may be too big to be able to force $x$ close enough to $a$ so that it stays positive. Also, $\epsilon$ is arbitrary here, so it can be as large as you please. You can try to see if there is a simpler $\delta$ that takes care of everything though. I believe it is possible. – Haskell Curry Nov 9 '12 at 17:50
HINT: Use the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$:
$$\sqrt[3]x=\sqrt[3]a=\left(\sqrt[3]x-\sqrt[3]a\right)\cdot\frac{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$$
Multiply and divide $\sqrt[3]{x}-\sqrt[3]{a}$ by conjugate: $$\dfrac{\left(\sqrt[3]{x}-\sqrt[3]{a} \right) \left(\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2} \right) }{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}=\dfrac{x-a}{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}$$