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It's known that $\lim\limits_{x \rightarrow x_0}f(x) = A$, how to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$?

Here's what I've got now:

When $A = 0$, to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = 0$: Since we have $\lim\limits_{x \rightarrow x_0}f(x) = A = 0$, so $|f(x)| < \epsilon$. => $|\sqrt[3]{f(x)}| < \epsilon_0^3 < \epsilon$

When $A \ne 0$, $|\sqrt[3]{f(x)} - \sqrt[3]{A}| = \frac{|f(x) - A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}$...

How can I deal with $(f(x)A)^{\frac{1}{3}}$? Thanks.

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    Excuse me, but I wonder why you downvote?2012-10-14
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    @AustinMohr Sorry, I forgot to add a crucial condition. That's my fault.2012-10-14
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    I don't know who down voted this question, but I will up vote for you so that it can cancel out.2012-10-14
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    Could the person voted to close say why is this not a real question ? I upvoted.2012-10-14
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    @Belgi I didn't vote to close, but if you check the edit history of this question, you will see why someone could have voted to close the first version as not being a real question.2012-10-14

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