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Given the function $f(x,y)=\frac{x^3−y^3}{x^2−y^2}$. Defined by $x^2\neq y^2$ and $(x,y) \in \Bbb R$. Proof that $\lim\limits_{(x,y) \rightarrow (0,0)} f(x,y)$ exists.

I found that the limit went to $0$. And I was trieing to prove this with an $\epsilon$ $ \delta$ prove, but I'm stuck.

I got:

Let $\epsilon > 0$ be given, take $\delta= ....$. Than $|(f(x,y)-L|$=$|\frac{x^3-y^3}{x^2-y^2}|$= $|\frac{x(x+y)+y^2}{x+y}|$=$|x+\frac{y^2}{x+y}|$

I don't know how to go on.

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    I don't know, why there are so many duplicate votes.. The referenced question has a plus instead of a minus in the numerator, which is a huge difference.. Whatever, the limit in your case indeed exists (as I have shown below). btw do you have problems with my proof? Because the question has not been marked as 'answered' yet. – 2017-02-28
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    @tofurind The question is marked duplicate because with the same technique it can be proven that the limit doesn't exist,I'm not quite sure why does your method fail but take for example $y=x^2-x$ using this path you can see that the limit tends to $1$. – 2017-02-28
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    @kingW3 Ah right, I found my mistake, I've substituted in one line a $|x+y|$ with $\|(x,y)\|_1$, which is just wrong... -_- Thanks for the counter example. – 2017-03-01

2 Answers 2

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Take $y=x^2-x$ as counter example. This limit will tend to $1$, and take $y=x^2+x$ to get the limit 0, so the limit in the question doesn't exist.

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You can't prove a limit exists when it doesn't take for example $y\to - x$

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    Of course there is a definition for limits on restricted areas. Without going into much topological details, but you can restrict the limit to any subset $Y\subset X$ of a given topological $X$ space by taking the _trace topology_. Then you convergence is studied regarding the topology of $Y$. – 2017-02-28