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.

  • $\begingroup$ 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

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 $k\neq 0$.


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

  • $\begingroup$ How did you draw this graph and animate? $\endgroup$ – xxx--- Nov 30 '14 at 13:28

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 $-1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.