# To prove the limit of given function does not exist.

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.

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.

• 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 $|f(x)-L|<\epsilon$ – abnry Aug 5 '13 at 14:40
• 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 $(0,0)$. – Clayton Aug 5 '13 at 14:42
• @nayrb: Rudin is referring to a situation where a domain other than an open neighborhood is explicitly specified. – Mikhail Katz Aug 5 '13 at 14:48
• @Clayton: your claim is incorrect. The expression $(x-y)/(x-y)$ is undefined when $x=y$. – Mikhail Katz Aug 5 '13 at 14:48
• It is standard practice. – Clayton Aug 5 '13 at 14:50

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.

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.

• How do you know it assumes all real values? It may, or it may not. You do know it assumes two different values. – abnry Aug 5 '13 at 15:06
• 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 – afiori Aug 6 '13 at 22:42