so we have I have the limit $\lim\limits_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}}$ Now i know that this limit does not exist. I even saw the proof on this website. However, if you substitute x and y with polar coordinates it seems like the limit exist, to me at least. $x=rcos(L)$ $y=rsin(L)$ $\lim\limits_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}}=\lim\limits_{(x,y) \to (0,0)} \frac{{rcos(L)r^2{sin(L)^2}}}{{{r^2cos(L)^2} + {r^4sin(L)^4}}}=\lim\limits_{(x,y) \to (0,0)} \frac{rcos(L)}{cos(L)^2+r^2sin(L)^4}=\frac {0}{cos(L)^2}=0$
I am clearly doing something not allowed. Is it becouse $cos(L)=0$ for some L?
Note that $L$ need not be constant. If we arrange things so that on our path we have $\cos(L)=r$, then the limit will not be $0$.
Remark: An argument like yours can be used to show that if $\cos(L)$ and $\sin(L)$ stay bounded away from $0$ as we approach $(0,0)$ along some path, then the limit along that path is $0$.