# What is the limit of $\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{y}?$

I am somewhat rusty with the limits when changing polar coordinates and the following question has arisen.

What is the limit of $$\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{y}?$$

In Wolfram, $$\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{y}=0$$ but, in polar coordinates, with $$\theta=r$$ $$\lim_{r\to 0, \theta=r} \frac{r}{\sin(\theta)}=\lim_{r\to 0} \frac{r}{\sin(r)}=1$$.

The limit not exists? Thanks.

• $x=r\cos \theta$ not $r\sin \theta$ Aug 26, 2022 at 5:57
• sorry I put x wrong and it was y. I fixed it. Aug 26, 2022 at 6:01
• – user
Aug 26, 2022 at 6:06

Approach the limit along two different paths; $$y=x$$ and $$y=x^2.$$ Can you complete now? Even in polar version also the limit value varies with the different values of $$\theta!$$
• Thanks. I have just understood. Yes, I was confused by the fact that in general $r\cos(\theta)\to 0$ if $r\to 0$, but that's true since $\cos(\theta)$ is bounded (just because of that) Aug 26, 2022 at 6:12