Is $\frac{x^3}{x^2+y^4}$ continuous at (0,0)?

Question

In an exam I was asked to determine if a function $f:\mathbb R^2\to \mathbb R$ is continuous at a point $(x_0,y_0)$, I needed to check if $$\lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0,y_0)$$

We consider the function given by $f(x,y)=\frac{x^3}{x^2+y^4}$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$.

We want to know if $f$ is continuous at $(0,0)$. What I did is the following :

We have that for $(x,y) \neq (0,0)$, $0\leq \frac{x^2}{x^2+y^4}\leq1$ and if $x>0$ $\implies 0\leq \frac{x^3}{x^2+y^4}\leq x$. And if $x<0$, $x\leq \frac{x^3}{x^2+y^4}\leq 0$

Then I used the squeeze theorem and said something very weird and stupid : $$"\lim_{(x,y) \to (0^+,0)} f(x,y)=\lim_{(x,y) \to (0^-,0)}f(x,y)=0"$$

First of all we never spoke or defined multivariable limits like that with $0^+$ and $0^-$, I am starting to realize when I am writing this that I could have been smarter and said that $-\lvert x \rvert \leq \frac{x^3}{x^2+y^4} \leq \lvert x \rvert$ and squeeze theorem to give me the continuity in $(0,0)$. The second point is that WolframAlpha tells me that the limit does not exist even though I tried many different paths such as $(t,t), (t^2,0)...$ to see if the function was discontinuous. Who is right ?

Answer 1

And Wolfram Alpha is indeed right. And so are you. In $\Bbb R^2$, that limit is indeed $0$. But in $\Bbb C^2$ that limit does not exist. For instance,\begin{align}\lim_{n\to\infty}f\left(\left(\frac1n+\frac1{n^3}\right)^2,\frac{1+i}{n\sqrt2}\right)&=\lim_{n\to\infty}\frac{\frac{1}{n^6}+\frac{6}{n^8}+\frac{15}{n^{10}}+\frac{20}{n^{12}}+\frac{15}{n^{14}}+\frac{6}{n^{16}}+\frac{1}{n^{18}}}{\frac{4}{n^6}+\frac{6}{n^8}+\frac{4}{n^{10}}+\frac{1}{n^{12}}}\\&=\frac14.\end{align}And what Wolfram Alpha says is that the value of the limit may depend on the path in the complex plane.

Answer 2

This is just an additional answer, addressing an extra question in the comments:
In fact I feel like approaching in $x$ a positive and negatives directions only and the corrector will surely tell me "what happens if I choose to zigzag between $x$ positive and negative to approach $(0,0)$ ?"

Then you would have to come up with a story, and say that your $(x,y) \to (0^+,0)$ notation meant that the point $(x,y)$ has $x$-coordinate $\ge0$, like $$\lim_{(x,y) \to (0,0)\\ x\ge0} f(x,y)=0$$ While this notation is not standard, one could argue that it makes sense and could be defined rigorously (so $(x,y)\to(0^+,0)$ means $(x,y)$ remains in the right-hand side closed half-plane $x\ge0$, not necessarily on any particular path).

Note that even if you have a function of only one variable, and say take $x_n=\frac{(-1)^n}n$, and if we knew that $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$ then we could conclude that $\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(\frac{(-1)^n}n)=0$ even if the sequence $x_n$ keeps "switching directions." You may argue that $x_n=\frac{(-1)^n}n$ does "zigzag," yet this does not present any problem.

Perhaps you didn't mean all of the above, but it is possible to formalize (and now you may start thinking that intuitively that is exactly what you meant :)

At any rate, it might be an interesting exercise to try to think of the details and whether the above makes sense, whether or not you will end up discussing it with your corrector. You could just be prepared, and try to figure it for the sake of knowing that it could be made to make sense, if you wish.

(And well, the answer by José Carlos Santos shows what happens if you use complex variables, but in my answer both $x$ and $y$ are assumed real.)