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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.

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

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  • $\begingroup$ Thank you! (Even though that's a bit more than a hint. This is the identity I tried to use but the resulting expression seemed more complex than what I started with.) $\endgroup$ – Py42 Nov 9 '12 at 17:20
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    $\begingroup$ There are still some details that have to be filled in, though. I'm sorry if I gave you too large a slice of pie. :) $\endgroup$ – Haskell Curry Nov 9 '12 at 17:21
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    $\begingroup$ @HaskellCurry why don't you just choose $\delta=\sqrt[3] {a^2} \epsilon$ $\endgroup$ – Mathematics Nov 9 '12 at 17:32
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    $\begingroup$ 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. $\endgroup$ – Haskell Curry Nov 9 '12 at 17:50
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HINT: Use the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$:

$$\sqrt[3]x=\sqrt[3]a=\left(\sqrt[3]x-\sqrt[3]a\right)\cdot\frac{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$$

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Multiply and divide $\sqrt[3]{x}-\sqrt[3]{a}$ by conjugate: $$\dfrac{\left(\sqrt[3]{x}-\sqrt[3]{a} \right) \left(\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2} \right) }{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}=\dfrac{x-a}{\sqrt[3]{x^2} +\sqrt[3]{x}\cdot\sqrt[3]{a} +\sqrt[3]{a^2}}$$

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