If transform it into polar coordinates we get that: $\lim_{r \rightarrow0}\frac{r^2 \cos^2(\theta)}{r\sin(\theta)}=\lim_{r \rightarrow0}r \frac{\cos^2(\theta)}{\sin(\theta)}$ which should be $0$ but then if we approach it from the curve $y=x^2$ we get $1$, but how is this the case? I thought using polar coordinates and showing the limit exists without being dependent on $\theta$ meant the limit exists but clearly it does not?
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When you use polar coordinates, you assume that $x$ and $y$ approaches $0$ with the same speed. (You take the same $r$ for both of them.) – Gary Oct 29 '21 at 21:55
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Hm, I was under the impression that polar coordinates are fool proof but I understand what you mean. Is there a way to know when polar coordinates won't work like above? – Randy Oct 29 '21 at 22:02
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I observe that along the curve $y = x^2$, as $r \rightarrow 0$, $\theta \rightarrow 0$, so $\sin(\theta) \rightarrow 0$. This means you have two zeroes in your denominator, not just the one from $r$. (In fact, we should always be on the lookout for denominators that can be zero...) – Eric Towers Oct 29 '21 at 22:16
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Problem is that $\theta$ can depend on $r$ otherwise you're basically only looking at straight lines. – kingW3 Oct 29 '21 at 22:46