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Thread: Epsilon-Delta proof in polar coordinates

  1. #1
    Mar 2009

    Post Epsilon-Delta proof in polar coordinates

    EDIT: Problem solved; taking \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 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 (x,y)=(0,0)

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

    Assuming r\leq 1, we can establish the following,
    <br />
\left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta<br />
\right) \right) -3 \right|\leq 3\,{\frac {{r}^{7}}{{r}^{4} \left( \cos \left( \theta \right)<br />
\right) ^{4}+{r}^{6} \left( \sin \left( \theta \right) \right) ^{6}}<br />
}<br />
    <br />
\Rightarrow \left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta<br />
\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}}}<br /> <br />
    Further, noting that  \left( \cos \left( \theta \right) \right) ^{6}+ \left( \sin \left(<br />
\theta \right) \right) ^{6}\geq \frac {1}{4}<br />
, we have finally

    <br />
\left| f \left( r\cos \left( \theta \right) ,r\sin \left( \theta<br />
\right) \right) -3 \right|\leq 12r<br />

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

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

    Now, here's my problem; I'm perfectly fine with the reasoning here up until the conclusion where \delta is taken to be \frac{12}{\epsilon}. To me it is fairly obvious that, instead, letting \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 \delta =\frac{12}{\epsilon} could be correct, since with r \leq 1 we'd be forced to take \epsilon > 12, but according to the definition we should be able to choose an arbitrary \epsilon > 0.

    So, am I missing something here? Is there any reason why the first choice of \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 12:55 PM.
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009

    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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