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How to show that $\lim \limits_{(x, y) \to (0,0)} f(x, y)$ does not exist where,

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

I tried bounding the value of the function as $(x, y)$ approaches $(0,0)$ but was not successful. I graphed the function and saw that it was actually approaching $0$ but Microsoft Math failed to render some points around the $x = y$ plane. So I imagine I need to find points $(x, y)$ such that $x$ is extremely close to $y$ but $(x, y)$ is not as close to $(0,0)$ to show that the function does not approach $0$ around the origin. But was unable to think of any.

Hints or suggestions would be awesome. Any help is appreciated.

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  • $\begingroup$ Try approaching $(0,0)$ along various curves $h(t) = (x(t),y(t))$. $\endgroup$ – aes Nov 22 '14 at 3:56
  • $\begingroup$ $x$ being arbitrarily close to $y$ doesn't seem to help. Setting $x_n = \frac{1}{n+1}$ and $y_n = \frac{1}{n}$ had $f(x_n,y_n)\to 0$ as well. $\endgroup$ – JMoravitz Nov 22 '14 at 4:16
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For all $(x,y)\in \mathbb R^2$ such that $x\neq y$ one has $f(x,y)=\dfrac{2x^3}{x-y}-x^2-xy-y^2$, so if the limit exists, due to $\lim \limits_{(x,y)\to(0,0)}\left(x^2-xy-y^2\right)$ existing, so does $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^3}{x-y}\right)$, but this last limit can easily be seen to be $k$-dependent if $y=x-kx^3$.

So consider the paths $t\mapsto\left(t,t-kt^3\right)$, with $k\neq 0$.

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Hint: What happens on the curves $$(x(t), y(t)) = (1/t \pm e^{-t}, 1/t), \quad t > 0?$$

Here is a little animation of the surface in question. It was surprisingly challenging to obtain a smooth plot in the neighborhood of $(0,0)$, but I managed to find a nice parametrization. The red and green curves are the ones I described, but it is quite clear that other paths to the origin will also work.

enter image description here

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  • $\begingroup$ How did you draw this graph and animate? $\endgroup$ – pushpen.paul Nov 30 '14 at 13:28
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You can show this limit does not exist if you can show that the limits as $(x,y) \rightarrow (0,0)$ along two different paths give different answers.

So for example, try the limit as $s\rightarrow 0$ of $f(x=s, y=-s)$, which is zero, and the limit as $s \rightarrow 0$ of $f(x=s, y=\sqrt[3]{s})$ which is $-1$.

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  • $\begingroup$ The path needs to be in the domain of $f$. The second one isn't. $\endgroup$ – Git Gud Nov 22 '14 at 3:57
  • $\begingroup$ @GitGud: Second one is also in the domain. But does not work since $f$ again is constantly $0$ along that curve. $\endgroup$ – Ishfaaq Nov 22 '14 at 3:58
  • $\begingroup$ @Ishfaaq I meant in the domain of the fractional part. $\endgroup$ – Git Gud Nov 22 '14 at 3:59
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    $\begingroup$ if it's allright, can you please elaborate on the second example. It looks like $f$ is again approaching $0$??? $\endgroup$ – Ishfaaq Nov 22 '14 at 4:04

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