2
$\begingroup$

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.

  • 0
    Excuse me, but I wonder why you downvote?2012-10-14
  • 0
    @AustinMohr Sorry, I forgot to add a crucial condition. That's my fault.2012-10-14
  • 0
    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
  • 0
    Could the person voted to close say why is this not a real question ? I upvoted.2012-10-14
  • 3
    @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

6

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.

  • 0
    In particular, this problem is the composition $g \circ f$, where $g(x) = \sqrt[3]{x}$.2012-10-14
  • 1
    Well, I'd like to use epsilon-delta definition to prove that. (Requirement for homework)2012-10-14
  • 0
    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.2012-10-14
  • 0
    @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. :)2012-10-14
  • 2
    @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
-3

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.

  • 0
    I'm trying to do this. But how can I bound $(f(x)A)^{\frac{1}{3}}$?2012-10-14
  • 0
    @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 $\dots$.2012-10-14
  • 2
    -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.2012-10-14
  • 0
    @did:Can you post your answer?2012-10-14
  • 7
    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.2012-10-14
  • 0
    @ymfoi: As I said you need to bound the unwanted factor.2012-10-14