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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.

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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 $\theta!$

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    $\begingroup$ 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) $\endgroup$
    – eraldcoil
    Aug 26, 2022 at 6:12
  • $\begingroup$ Welcome. That's great. $\endgroup$
    – user159888
    Aug 26, 2022 at 6:14

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