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Given,$$f(x,y)=\begin{cases}\dfrac{x^3+y^3}{x-y}&\text{ if }x\neq y\\0&\text{ if }x=y\end{cases}$$

To prove $$\lim_{(x,y)\to (0,0)}\frac {x^3+y^3}{x-y}$$ does not exist, I used the paths $y=mx$,along x axis and along y-axis but I got $0$ in all cases.I can't figure out which path to use to get a different limit value.

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  • $\begingroup$ Wouldn't letting $x=y$ just show that it doesn't exist? Im just asking, not answering. $\endgroup$ – Erik T. Nov 16 '17 at 9:33
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    $\begingroup$ Try with m=1? Does limit still exist? $\endgroup$ – Sagar Chand Nov 16 '17 at 9:43
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Hint: For two paths $y=mx$ and $y= x-x^3$, the limit tends to zero and $2$ respectively.

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Doing the change of variable$$\left\{\begin{array}{l}x=X+Y\\y=X-Y,\end{array}\right.$$your function becomes$$f(X,Y)=\begin{cases}\dfrac{X^3}Y+3XY&\text{ if }Y\neq0\\0&\text{ otherwise.}\end{cases}$$If you take the path $Y=X^3$, then the limit is $1$, not $0$.

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