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This is not help with homework. I just had this question on our final exam and was wondering what the answer was.

$$\lim_{x,y \to 0} \frac{x^2}{x-y}$$ I tried to use polar coordinates. So,

$$\lim_{r\to0^+} \frac{r^2\cos^2\theta}{r(\cos \theta - \sin \theta)}$$

$$= \lim_{r\to0^+} \frac{r\cos^2\theta}{\cos \theta - \sin \theta}$$

I get stuck here. Wolfram alpha says it does not exist. Could somebody provide insight as to where I went wrong?

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    $\begingroup$ Try for the directions $(x,2x)$ and $(x,x^2+x)$. limit does not exist $\endgroup$
    – Hamou
    Aug 12, 2014 at 5:33

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Hint: What happens if we approach $(0,0)$ along the path $y = x-x^3$?

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  • $\begingroup$ Strange path but effective. ;-) $\endgroup$
    – Mikasa
    Aug 12, 2014 at 5:34
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Technically, along the line $y=x$, the function doesn't even exist. So there is no way it could be called continuous at $(0,0)$ (unless you view it as a function on the complement of that set).

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Here's a hint I always used:

Try finding the limit along the path $y=mx$. Do not substitute a value for $m$. Leave it as an unknown.

If your limit is independent of $m$ (i.e. does $\textbf{NOT}$ contain $m$), then the limit is unique along all paths and thus exists and is equal to that limit, else the limit is NOT unique along all paths, since the limit would be dependent on the slope along which your path goes - hence it does not exist.

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