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If we have the limit $$\displaystyle \lim_{(x,y)\to(0,0)}{\frac{xy(x-y)}{x-y}}$$ can we simply cancel $(x-y)$ from numerator and denominator and conclude $\frac{0\cdot 0\cdot 1}{1} = 0$?

user2850514
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    How is the function defined for $x=y$? – Marcin Łoś Feb 23 '14 at 00:01
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    Yes, you can cancel $(x-y)$ from the numerator and denominator, so the limit is equal to: $$\lim_{(x,y) \rightarrow (0,0)} {xy}=0$$ – Mary Star Feb 23 '14 at 00:03
  • @MarcinŁoś I believe it does not matter here, since the question does not talk about the continuity. – Math137 Feb 23 '14 at 01:25
  • @math137 Yes, but that's not about just value at the point where we compute the limit. Any neighbour contains points with $x=y$, so to say anything about the limit at $(0, 0)$ we need to know the values at such points. – Marcin Łoś Feb 23 '14 at 08:27

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You can indeed cancel $x-y$. However, we can also work with a more "comfortable" limit of a one variable if we go to polar coordinates:

$x=r\cos\theta$

$y=r\sin\theta$

Then your limit $(x,y)\rightarrow(0,0)$ corresponds to $r\rightarrow 0$ and we have

$\lim_{r\rightarrow 0}\frac{r^{3}\cos\theta\sin\theta(\cos\theta-\sin\theta)}{r(\cos\theta-\sin\theta)}=\lim_{r\rightarrow0}r^{2}\cos\theta\sin\theta$

Now, the above limit is equal to zero since $\cos\theta\sin\theta$ is bounded for all $\theta$.

GregVoit
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