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Thread: Epsilon-Delta proof in polar coordinates

  1. #1
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    Post Epsilon-Delta proof in polar coordinates

    EDIT: Problem solved; taking $\displaystyle \delta =\frac{12}{\epsilon}$ is (of course) just a mistake. However, I guess this thread still serves as a good exercise in proving multivariable limits by using polar coordinates. By the way, the original function in x and y is given by $\displaystyle f(x,y)=\frac{3(x^4+y^6-x^3y^4)}{x^4+y^6}$.

    Consider the following solution to an exercise in which one is to give a proof using polar coordinates to prove the limit of a function at $\displaystyle (x,y)=(0,0)$

    Given
    $\displaystyle
    f(rcos(\theta),rsin(\theta))={\frac {3\,{r}^{4} \left( \cos \left( \theta \right) \right) ^{4}+3\,
    {r}^{6} \left( \sin \left( \theta \right) \right) ^{6}-3\,{r}^{7}
    \left( \cos \left( \theta \right) \right) ^{3} \left( \sin \left(
    \theta \right) \right) ^{4}}{{r}^{4} \left( \cos \left( \theta
    \right) \right) ^{4}+{r}^{6} \left( \sin \left( \theta \right)
    \right) ^{6}}}
    $
    we want to show that the function $\displaystyle f$ has limit $\displaystyle L=3$ as $\displaystyle r$ approaches zero.

    Assuming $\displaystyle r\leq 1$, we can establish the following,
    $\displaystyle
    \left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta
    \right) \right) -3 \right|\leq 3\,{\frac {{r}^{7}}{{r}^{4} \left( \cos \left( \theta \right)
    \right) ^{4}+{r}^{6} \left( \sin \left( \theta \right) \right) ^{6}}
    }
    $
    $\displaystyle
    \Rightarrow \left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta
    \right) \right) -3 \right|\leq 3\,{\frac {{r}^{7}}{{r}^{6} \left( \cos \left( \theta \right) \right) ^{6}+{r}^{6} \left( \sin \left( \theta \right) \right) ^{6}}}

    $
    Further, noting that $\displaystyle \left( \cos \left( \theta \right) \right) ^{6}+ \left( \sin \left(
    \theta \right) \right) ^{6}\geq \frac {1}{4}
    $, we have finally

    $\displaystyle
    \left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta
    \right) \right) -3 \right|\leq 12r
    $

    Thus we may define $\displaystyle \delta =\frac{12}{\epsilon}$ so that,

    $\displaystyle
    0 < r < \delta =\frac{12}{\epsilon} \Rightarrow \left| f(rcos(\theta),rsin(\theta)) - 3\right| < \epsilon
    $

    Now, here's my problem; I'm perfectly fine with the reasoning here up until the conclusion where $\displaystyle \delta$ is taken to be $\displaystyle \frac{12}{\epsilon}$. To me it is fairly obvious that, instead, letting $\displaystyle \delta =\frac{\epsilon}{12}$ is a valid choice in establishing this proof, and that's what I've been teaching when instructing a few of the students who take a basic course in multivariable calculus at my university up until I was emailed this solution from a professor who teach the subject.

    I don't see why taking $\displaystyle \delta =\frac{12}{\epsilon}$ could be correct, since with $\displaystyle r \leq 1$ we'd be forced to take $\displaystyle \epsilon > 12$, but according to the definition we should be able to choose an arbitrary $\displaystyle \epsilon > 0$.

    So, am I missing something here? Is there any reason why the first choice of $\displaystyle \epsilon$ should be valid or is it simply a typo?

    Would be nice if someone could comment on this so that I can know if I'm really running around lying to the students around here or not.
    Last edited by bkoal; Mar 16th 2009 at 11:55 AM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    an opinion

    I think you mean delta = epslion/12 not 12/epsilon

    I believe you are right that it is a typo--as with all epsilon-delta proofs they start "Let epsilon be given......." I am pretty sure he is talking about delta.



    You are assuming r << 1 which is obviously a good assumption since we are interested in r near 0.

    However to cross all the ts take delata = min (e/12,1)
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