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
 A: 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..
A: 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?)
A: 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 $r \rightarrow 0$? 
