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Ques: I want to show that a limit of a function $$f(x,y)=\frac{x^{3}+y^{3}}{x-y}$$ does not exist at point $(0,0)$.

My try: I am just taking path $y=x-x^{3}$ then $$\lim _{(x,y)\rightarrow(0,0)}\frac{x^{3}+y^{3}}{x-y}=\lim_{x\rightarrow 0}\frac{x^{3}+(x-x^{3})^{3}}{x^{3}}=2$$ Then i am taking path $y=2(x-x^{3})$ and i got limit $0$. So, in the both case the limit does not remain same. it means limit of given function does not exist.

Am i right? please give your valuable suggestions!

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You don't have to work so hard. For the limit to exist at a point, the function has to be defined in a punctured neighborhood of this point. But your function is undefined whenever $x=y$. Hence the limit does not exist.

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    $\begingroup$ This depends on your definition of a limit. You can find in "little Rudin" that the definition for a limit requires points in the domain within $\delta$ to imply $|f(x)-L|<\epsilon$ $\endgroup$ – abnry Aug 5 '13 at 14:40
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    $\begingroup$ I think the neighborhood has to be a subset of the domain. In this case, the domain doesn't include the set $y=x$. For example, $(x-y)/(x-y)$ clearly exists at $(0,0)$. $\endgroup$ – Clayton Aug 5 '13 at 14:42
  • $\begingroup$ @nayrb: Rudin is referring to a situation where a domain other than an open neighborhood is explicitly specified. $\endgroup$ – Mikhail Katz Aug 5 '13 at 14:48
  • $\begingroup$ @Clayton: your claim is incorrect. The expression $(x-y)/(x-y)$ is undefined when $x=y$. $\endgroup$ – Mikhail Katz Aug 5 '13 at 14:48
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    $\begingroup$ It is standard practice. $\endgroup$ – Clayton Aug 5 '13 at 14:50
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I think you're right, but I get limit 2 for the first one, not 1. For the second limit, I get 9/2, not zero. As @Clayton points out, $y = 0$ will suffice, and as @user72694 points out, there is a much easier way.

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That is perfectly right. You could also say that in every square of edge l around the (0,0) the function assume all real value, so it clearly can not have a unique limit.

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    $\begingroup$ How do you know it assumes all real values? It may, or it may not. You do know it assumes two different values. $\endgroup$ – abnry Aug 5 '13 at 15:06
  • $\begingroup$ Using polar coordinates. (x,y) = (rcos(t), rsin(t)) if we leave r fixed and send t to pi / 4 from above it will go to - inf and if we send t to pi*5/4 from below then it will go to +inf and if we restrict the function to x-y < 0 it will be continuos; thus intermediate value Theorem applies $\endgroup$ – afiori Aug 6 '13 at 22:42

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