# Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$

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? –  ymfoi Oct 14 '12 at 2:59
@AustinMohr Sorry, I forgot to add a crucial condition. That's my fault. –  ymfoi Oct 14 '12 at 3:17
I don't know who down voted this question, but I will up vote for you so that it can cancel out. –  diimension Oct 14 '12 at 4:00
Could the person voted to close say why is this not a real question ? I upvoted. –  Belgi Oct 14 '12 at 4:02
@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. –  Adrián Barquero Oct 14 '12 at 4:22

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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In particular, this problem is the composition $g \circ f$, where $g(x) = \sqrt[3]{x}$. –  Austin Mohr Oct 14 '12 at 3:26
Well, I'd like to use epsilon-delta definition to prove that. (Requirement for homework) –  ymfoi Oct 14 '12 at 3:29
The links I gave are epsilon-delta proofs. Do you just want one tailored to this problem? It might be instructive to follow the links and convert their general proof into a proof specifically for this problem. –  EuYu Oct 14 '12 at 3:30
@EuYu Yes. Because this problems comes from the exercises of my textbook, I suppose that I should prove it without proving more general situation (or this specific question seems to be useless). Thanks for the material. :) –  ymfoi Oct 14 '12 at 3:42
@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. –  EuYu Oct 14 '12 at 3:48

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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I'm trying to do this. But how can I bound $(f(x)A)^{\frac{1}{3}}$? –  ymfoi Oct 14 '12 at 4:02
@ymfoi:There is no problem with this. Notice that, when $x$ close to $x_0$, ${|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}\,,$ should be close to $\dots$. –  Mhenni Benghorbal Oct 14 '12 at 4:09
-1. This is assuming that the question asked at the beginning has a trivial solution and using it to complete a proof of... the question asked at the beginning: if one knows the limit of $f(x)^{2/3}+(f(x)A)^{1/3}+A^{2/3}$, asking about the limit of $f(x)^{1/3}$ is moot. –  Did Oct 14 '12 at 9:42
@did:Can you post your answer? –  Mhenni Benghorbal Oct 14 '12 at 11:38
Yes I could (obviously), but why should I? There is already a decent answer on this page hence posting a new one would be pretty useless. In fact, this comment of yours indicates some serious misconceptions about the way the site functions. Once again (since this is not the first time you deploy similar tactics to avoid facing facts), all that matters here is that YOUR answer is wrong. THIS should be your concern (not the attributes of the messenger of the bad news) and YOU should try to do what you can to correct this sorry situation. –  Did Oct 14 '12 at 13:25