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Suppose we've got a function $f(x,y):=\frac{y^2+2x}{y^2-x}$ defined on the real plane expect for the set $y^2=x$. Now we want to see whether $f$ can be made continuous in $(0,0)$.

My work:

I take the limit:

$$\lim_{r \to 0} \frac{r^2\sin^2 \varphi +2r \cos \varphi}{r^2 \sin^2 \varphi -r \cos \varphi}= \lim_{r \to 0} \frac{2 \cos \varphi}{- \cos \varphi}=-2$$

So the limit is $-2$ but apparently $f$ can't be made continuous in $(0,0)$ so this argument doesn't work. Where did I go wrong?

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    $\begingroup$ You went wrong in assuming that finding one sublimit would imply that all sublimits exist and are the same. $\endgroup$ – Git Gud Feb 16 '15 at 22:14
  • $\begingroup$ It's my understanding that the limit should be independent of the "approaching path" because the phis cancel out. We've used this argument in class and it was apparently correct. I.e. you can take the limit as r goes to 0 and if this limit exists then the function can be made continuous in $(0,0)$ $\endgroup$ – Mike Feb 16 '15 at 22:33
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    $\begingroup$ Just to convince you the limit doesn't exist, consider the path $t\mapsto (0,t)$. Polar coordinates can be used to prove the existence of limits sometimes, but not all the time. If it were the case that $f(r,\varphi)$ was independent of $\varphi$, then it would be correct. But it isn't. $\endgroup$ – Git Gud Feb 16 '15 at 22:43
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One problem with this argument is that it assumes that $\cos\theta\ne0$;

when $(x,y)\to(0,0)$ along the y-axis, we have $\displaystyle\lim_{y\to0}\frac{y^2}{y^2}=1\ne-2$.

More generally, $\displaystyle\bigg|\frac{r^2\sin^{2}\theta+2r\cos\theta}{r^2\sin^{2}\theta-r\cos\theta}-(-2)\bigg|=\bigg|\frac{3r\sin^{2}\theta}{r\sin^{2}\theta-\cos\theta}\bigg|$,

and this expression does not always approach 0 as $r\to0$.

For example, this does not approach 0 when $\cos\theta=0$ as noted above;

but it also does not approach 0 if $r\sin^{2}\theta=k\cos\theta$ (so $y^2=kx$) for $k\ne0$.

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