Discontinuity of $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ at $(x,y) = (0,0)$

Question

Let $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ , when $(x,y) \neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$ We are required to prove the discontinuity of this function at $(0,0)$. So, I put $y=mx$ where $x \rightarrow 0$, to get the limit to be $0$. But I can't find another instance where the limiti is different. Can you guys help? Thanks

Answer 1

Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.

Answer 2

Credit: Carmeister's solution.

I am just curious about the behavior of other similar path.

Let $y = -\frac{x^2}{2}+x^n$ where $n >0$,then we have

\begin{align}\lim_{x \to 0}\frac{2(x^3+(-\frac{x^2}{2}+x^n)^3)}{2x^n} &= \lim_{x \to 0} \frac{(x^3+(-\frac{x^2}{2}+x^n)^3)}{x^n} \\ &= \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]\end{align}

If $n \in (0,3)$, the limit is $0$.

If $n>3$, it is undefined.

If $n=3$, \begin{align} \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]=1\end{align}