# Math Help - Epsilon-Delta proof in polar coordinates

1. ## 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 $(x,y)=(0,0)$

Given
$
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 $f$ has limit $L=3$ as $r$ approaches zero.

Assuming $r\leq 1$, we can establish the following,
$
\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}}
}
$

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

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

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

$
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 $\delta$ 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.

2. ## 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)