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I'm trying to use the $\epsilon$ - $\delta$ definition to show that $\lim_{x \rightarrow 2} \sqrt[3]{x} = \sqrt[3]{2}$ .

$\forall \epsilon>0, \exists \delta >0$ s.t. if $0< | x-2 | < \delta$ then $| \sqrt[3]{x} - \sqrt[3]{2} | < \epsilon$.

I'm not at all sure how to do this but... $\sqrt[3]{2}-\epsilon < \sqrt[3]{x} < \epsilon+\sqrt[3]{2}$

$\Rightarrow$ $(\sqrt[3]{2}-\epsilon)^3 <x< (\epsilon+\sqrt[3]{2})^3$

and $2-\delta < x < \delta + 2$. But now i'm stuck. If i say that $2-\delta = (\sqrt[3]{2}-\epsilon)$ and $\delta + 2 = (\epsilon+\sqrt[3]{2})^3$, then $\delta$ could be $2-(\epsilon+\sqrt[3]{2})^3$ or $(\epsilon+\sqrt[3]{2})^3 -2$...... How do i know which is the smaller of the two?

Also this felt like a very complicated way to prove this. Is there any way to make $| \sqrt[3]{x} - \sqrt[3]{2} |$ seem more like | x-2 |???

user125342
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  • You should find an answer to your question here or links posted here. http://math.stackexchange.com/questions/687414/prove-continuity-for-cubic-root-using-epsilon-delta/687479#687479 – Ishfaaq Mar 21 '14 at 14:50
  • Thank you so much, Ishfaaq!! – user125342 Mar 22 '14 at 13:13

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lemma:${a^3}-{b^3}=(a-b)(a^2+ab+b^2)$ now, let$a^3=x,b^3=2$,then,we have $x-2=(\sqrt[3]{x}-\sqrt[3]{2})(x^\frac{2}{3}+\sqrt[3]{2x}+2^\frac{2}{3})$, thus,$\sqrt[3]{x}-\sqrt[3]{2}=\frac{x-2}{x^\frac{2}{3}+\sqrt[3]{2x}+2^\frac{2}{3}}$. using $\epsilon-\delta$ definition as below, for all $\epsilon>0$,exists $\delta>0$ , such that if $0<|x-2|<\delta$,let$\delta=1/2$ ,$|\sqrt[3]{x}-\sqrt[3]{2}|=|\frac{x-2}{x^\frac{2}{3}+\sqrt[3]{2x}+2^\frac{2}{3}}|<\frac{|x-2|}{2^\frac{2}{3}}<\frac{\delta}{2^\frac{2}{3}}<\epsilon$. it is trivial to find that $\delta=min{\{2^\frac{-2}{3}\epsilon,1/2}\}$ is satisfied.

Wonei Lam
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