# A limit which exists in polar coordinates but not in Cartesian coordinates?

Let's have look at the function $$f(x,y) \begin{cases} \frac{y(x^2+y^2)}{y^2+(x^2+y^2)^2} & (x,y)\neq(0,0) \\0 & (x,y)=(0,0)\end{cases}.$$ Switching to polar coordinates gives $$f(r,\theta)=\begin{cases} \frac{r^3 \sin \theta}{r^2\sin^2\theta+r^4} & r\neq0 \\0 & r=0\end{cases}.$$ We'd like to investigate the existence of a limit for $$f$$ at the origin. In Cartesian coordinates ($$f(x,y)$$) one can immeidately see that the limit doesn't exist because for example on the path $$y=0$$ we have $$\lim_{x\rightarrow 0,y=0} f(x,y)=0$$ and on the path $$y=x^2$$ we have $$\lim_{x\rightarrow 0,y=x^2} f(x,y)=\frac{1}{2}$$.

However, in polar coordinates we have $$f(r,\theta)=\begin{cases} \frac{r \sin \theta}{\sin^2\theta+r^2} & r\neq0 \\0 & r=0\end{cases}$$ so that $$\lim_{r\rightarrow 0}f(r,\theta) = \begin{cases} 0 & \sin\theta = 0\\ 0 & \sin\theta \neq 0 \end{cases}$$ so the limit exists and is zero regardless of $$\theta$$. Why does it look the limit doesn't exist in Cartesian coordinates but exists in polar coordinates?

Edit: Thanks to the insightful comments on this page and other similar questions in the site, the unboundedness of the expression with $$\sin^2 \theta$$ is the key to the failure of the limit existence. $$\sin\theta$$ can get arbitrarily small, making the whole expression arbitrarily large, effectively counteracting $$r\rightarrow 0$$.

The limit seems to exist in polar coordinates since you only tested the curves with fixed $$\theta$$. This is equivalent to only testing the lines $$y=mx$$ (and the line $$x=0$$) in Cartesian coordinates.

Indeed for these lines we have:

$$\lim_{(x,y) \to (0,0)}\frac {y(x^2+y^2)}{y^2 + (x^2 + y^2)^2} = \lim_{x\to 0}\frac {mx^3(1+m^2)}{m^2x^2 + x^4(1+m^2)^2}=\lim_{x\to 0}\frac {mx(1+m^2)}{m^2 + x^2(1+m^2)^2} =0$$

Suppose we take another curve: $$r = \theta$$. Then:

$$\lim_{r\to 0}\frac {r\sin\theta}{\sin^2\theta + r^2} = \lim_{r\to 0}\frac {r\sin r}{\sin^2 r+ r^2} = \lim_{r\to0}\frac {(\sin r)/r}{1 + (\sin^2 r)/r^2} = \frac1{1+1} = \frac12$$

which is not zero, so the limit cannot exist.
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