A question on Polar form of a two variable limit I'm new to two variable limits and i am having troubles with this:
$$ L=\lim_{||(x,y)||\rightarrow \infty } \frac {xy^2}{x^4+y^2} $$ 
I am sure this limit does not exsist becouse $ f(x,0)=f(0,y)=0 $ but $ f(x,x^2)=\frac{x}{2} $ so the limit can be $L=0$ or $ L=\infty $, but when i pass in polar coordinates i get for $ r\in [0,2\pi) $ : 
$$ \lim_{p\rightarrow \infty} p\frac{\cos(r)\sin(r)^2}{p^2\cos(r)^4+sin(r)^2}$$
The problem is i can't see how to determinate the case when $ L=\infty $ in polar form, becouse if $\cos(r)=0 $ then also numerator vanishes and we get $L=0$ , if $k=cos(r) \neq 0 $ then denominator goes to infinity as $kp^2 $ and limit is $L=0$ again, where's my mistake? I notices that the limit is not zero on $ f(x,x^2) $ wich is not a direction like $ f(x,mx) $ but it's a curve, so for every direction $ (\cos(r),\sin(r)) $ the limit goes to zero, but how i can manage to see what happens on a curve in polar coordinates? 
 A: In rectangular coordinates, $(x, x^2)$ gets you $L=\infty$. Now, we just need to convert $(x, x^2)$ to polar coordinates.
$$x=p\cos r, y=x^2=p\sin r$$
Substitute the first equation into the second equation:
$$p^2\cos^2 r=p\sin r$$
Divide both sides by $p$ and substitute $1-\sin^2 r$ for $\cos^2 r$:
$$p-p\sin^2 r=\sin r$$
Make the left side $0$:
$$0=p\sin^2 r+\sin r-p$$
Solve for $\sin r$ with quadratic formula:
$$\sin r=\frac{-1 \pm \sqrt{1+4p^2}}{2p}$$
Now, $\sqrt{4p^2+1}$ is slightly above $\lvert 2p \rvert$, so $-1+\sqrt{1+4p^2}$ is below $\lvert 2p \rvert$ and therefore the right-hand side is between -1 and 1. This means this is a valid solution for $\sin r$ even as $p \to \infty$.
Thus, the above equation will make the limit in polar coordinates converge to $\infty$ as $p \to \infty$ just like $(x, x^2)$ does in rectangular coordinates.
A: As you said, you need to be on a curve and not a straight line so you need $r$ to depend on $p$.
$$\forall p >1, \exists r \in [0,\pi] \ \text{s.t.} \ \cos(r)=\frac{1}{\sqrt{p}}$$
So when $p \to \infty$, you can follow the path so that $\cos(r) = \frac{1}{\sqrt{p}}$
And you get $$\lim_{p \to \infty}{p\frac{\cos(r)\sin^2(r)}{p^2\cos^4(r)+\sin^2(r)}}$$
$$=\lim_{p \to \infty}{\sqrt{p}\frac{(1-\frac{1}{p})}{(2-\frac{1}{p})}}=\infty$$
A: You can recover the curve $\left(x,x^2\right)$ by putting $\sin(r)=p\cos^2(r)$.
