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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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    @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

2 Answers 2

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You might want to take a look at the limit of composite functions. This is a standard result in many calculus textbooks. Fishing around online immediate gives several proofs, for example here or here.

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    @ymfoi Right, I realize that. That's why I suggested you convert the general proofs into a specific proof for your problem. That way, not only will you solve your problem, but you also gain an understanding of the general case.2012-10-14
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You need to bound the term

$ \frac{1}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}\,, $

and exploit the fact that $|x-x_0|<\delta \implies |f(x)-A|<\epsilon\,.$ See here for details.

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    @ymfoi: As I said you need to bound the unwanted factor.2012-10-14