# Polar coordinates in two variable function limits

Suppose that I use both $r$ and $\theta$ as variables (which means theta is not a constant it's a variable). Does switching to polar coordinates holds in any case? For example, here Calculating a limit in two variables by going to polar coordinates., in the answer of Michael E2, where he considers the function $f(x,y)=xy^2/(x^2+y^4)$, If I take $\theta = \arcsin(r \cos^2\theta)$ I can show using polar coordinates that the limit does not exist. Isn't switcing to polar coordinates equivalent to working with $x,y$? What do you think?? Thanks! Shir

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You have $\theta$ on both sides of the definition - is that intentional? – Macavity Mar 18 '13 at 12:55
oops, sorry, you are right.. I'm just saying that what you can show with x,y you can also show with $r,\theta$. So, lets do this differently. lets take $r=cos\theta/sin^2\theta$. If we take $\theta \rightarrow \pi/2$ we get $r \rightarrow 0$ and can show that the limit in this case is $1/2$ which means it does not exist.. – topsi Mar 18 '13 at 13:31

## 3 Answers

I'm just saying that what you can show with x,y you can also show with $r,\theta$. So, lets do this differently. lets take $r=cos\theta/sin^2\theta$. If we take $\theta \rightarrow \pi/2$ we get $r \rightarrow 0$ and can show that the limit in this case is $1/2$ which means it does not exist..

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This is why we can't say that $$\frac{r\cos\theta\sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}\to 0$$ as $r\to 0$. As far as we know \theta might be very near $Pi/2$ and then we have $f(x,y)\to 0/0$ – jim Mar 18 '13 at 13:35
Exactly. So, isn't its true thar working with polar coordinates is equivalent to working with x,y? isn't it's true that it is not more limited in any way? because I saw in other posts here that people tend to use them as some sort of special cases..... – topsi Mar 18 '13 at 13:39
Equivalent in that they give the same result. – jim Mar 18 '13 at 13:41
I don't get almost anyhting of what you people are talking about: if $\,\theta \sim \pi/2\,$ ,then the above expression is $$\frac{r\cos\theta\sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}\sim\frac{r\cdot 0\cdot 1}{0+r^2}$$ . I think that what you mean is that then, if $\,r\to 0\,$, then we get an expression of the form $\,1/0\,$ ? Well, then you'd be right...if we hadn't that zero in the numerator! – DonAntonio Mar 18 '13 at 17:22
$\theta \rightarrow \pi/2$ which means $\theta \neq 0$ at least thats what I ment.. and also $r=cos\theta/sin^2\theta$ – topsi Mar 18 '13 at 21:09

Why would you do that to poor $\,\theta\,$ ? We know

$$x=r\cos\theta\;,\;\;y=r\sin\theta\,\Longrightarrow$$

$$\frac{xy^2}{x^2+y^4}=\frac{r^3\cos\theta\sin^2\theta}{r^2(\cos^2\theta+r^2\sin^4\theta)}=\frac{r\cos\theta\sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}\xrightarrow[r\to 0\iff(x,y)\to(0,0)]{}0$$

The above is right for $\,\theta\ne \pi/2\;,\;3\pi/2\,$ (main angles values) (why?)

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I cannot write such this fast in Latex. The final arrow's written great. – Babak S. Mar 18 '13 at 13:14
:) Note that if we put $x=t^2, y=t$ the limit is $1/2$ so this function has no limit in (0,0). I just wanted to show that we can show this if we worl with $r,\theta$ as well.. – topsi Mar 18 '13 at 13:20
What about the $cos^{2}\theta$ in the denominator? – jim Mar 18 '13 at 13:55
@jim, are you asking me? What about it? – DonAntonio Mar 18 '13 at 16:02
@DonAntonio, you are saying that $$\frac{xy^2}{x^2+y^4}\to 0.$$ This limit does not exist. – jim Mar 18 '13 at 19:40

I think we can say that the expression goes to zero, independent of $\theta$, along any ray of constant $\theta$. The expression is equal to

$$\frac{r \cos{\theta} \sin^2{\theta}}{\cos^2{\theta} + r^2 \sin^4{\theta}}$$

Even when $\theta = \pi/2$, the expression still goes to zero because of the cosine in the numerator. The question is, does it go to zero uniformly as $r \rightarrow 0$?

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Note that if we put $x=t^2, y=t$ the limit is $1/2$ so this function has no limit in (0,0). I just wanted to show that we can show this if we worl with $r,\theta$ as well.. – topsi Mar 18 '13 at 13:20
OK, but that is a different limit altogether. When you use polars, you are weighting $x$ and $y$ the same. In this special case, you are not. – Ron Gordon Mar 18 '13 at 13:22
Why not? They are two independent variables arn't they? don't we use them here exactly in the same way as we use them in polar representation of a complex number for instance? – topsi Mar 18 '13 at 13:35
$x$ and $y$ are independent, but when you introduce polars, you assign equal weight to them. That's the whole point of a Jacobian, for example. If you want to show that a limit doesn't exist on some curve, as you do, then that is different from assigning $x$ and $y$ polar coordinates in the usual sense. – Ron Gordon Mar 18 '13 at 13:58
I can't see why you claim that the cosine in the numerator means that for $\theta=\pi/2$ the limit is $0$. It is surely the cosine in the denominator that should make us pause. – jim Mar 18 '13 at 13:59