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I have to prove that the limit $$\lim\limits_{(x,y) \to (0,0)}\dfrac{x^2}{x+y}$$ does not converge.

This is fairly 'easy' to do, but while I was doing it I came across some doubts. I took the limit $(x,y)\to(0,0)$ when $y=mx$. In this case, the function approches zero as x approaches zero.

Then I tried to approximate it with a curve. But none of the curves $y=x^{n}$ worked, so I did $y=x^2-x$ and then took the limit, which is $1$.

My question is, can I do that? I mean, the only condition that, as far as I know, I have to check, is that the path have to include the point (0,0), but is that the only condition? I know it may be a silly question, but I've never find any theorem regarding the path method and I want to be sure that I'm not messing things up.

I'm sorry if the redaction is a bit messy, It's been a while since I write something in English, please, if you don't understand something I wrote, ask me. Thanks!

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$m=-1$, i.e. $y=-x$, for your first condition shows the limit does not exist – Adam Hughes Jul 28 '14 at 16:48
@AdamHughes That path is not within the domain of the function... – Lessa121 Jul 28 '14 at 16:51
I mean, how do you procede when you set $y=-x$? – Lessa121 Jul 28 '14 at 16:53
It doesn't particularly matter if you're not allowing that, I suppose. Your choice of $y=x^2-x$ is a good one, that will show the limit does not exist because you get different results from different paths. I've also included a general counterexample in my answer, below, in case you are still a bit leery about the "two different limits" approach. – Adam Hughes Jul 28 '14 at 16:56
@LunaSage What you have done in the original post is perfectly acceptable. – apnorton Jul 28 '14 at 17:01
up vote 7 down vote accepted

Aside from showing that two limits differ as you do for $y=x$ and $y=x^2-x$, you can also do this in one fell swoop:

Let $y=x^3-x$, then you get

$$\lim_{(x,y)\to (0,0)}{x^2\over x^3}$$

which clearly diverges to infinity.

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