Fdtxjr

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

$$f(x, y) = begincases dfracx^3 + y^3x - y ; ; & x neq y \
0 ; ;;;;;;;;;;; ; & x = y endcases $$

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.

For all $(x,y)in mathbb R^2$ such that $xneq y$ one has $f(x,y)=dfrac2x^3x-y-x^2-xy-y^2$, so if the limit exists, due to $lim limits_(x,y)to(0,0)left(x^2-xy-y^2right)$ existing, so does $lim limits_(x,y)to (0,0)left(dfrac2x^3x-yright)$, but this last limit can easily be seen to be $k$-dependent if $y=x-kx^3$.

So consider the paths $tmapstoleft(t,t-kt^3right)$, with $kneq 0$.

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.

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 $srightarrow 0$ 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 $-1$.