Finding limit of $\frac{x^2y^2}{x^3+y^3}$ as $(x,y)\to (0,0)$

Question

Here, if we go along $x=-y$ then the function itself is not defined, and limit seems to exist for all other paths of approach

Putting in another way, if we use polar coords, limit is:

$$\lim_{r\to 0} \frac{r^4 \sin^2\theta\cos^2\theta}{r^3(\sin^3\theta + \cos^3\theta)}$$

which is zero only when $(\sin^3\theta + \cos^3\theta)$ is nonzero, and function itself is not defined for values where $(\sin^3\theta + \cos^3\theta) = 0$

So here, what do we say? Does the limit exist? or does it not exist

(Additional Question)

Suppose that limit is being calculated $(x,y) \to (0,0)$. Then can we choose one of the paths as, say $x=0$, although in limit we only let variables to "tend" to a value? Is it allowed to use this path?

Answer 1

We have that

• $x=y=t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4}{2t^3}=\frac t{2}\to 0$

• $x=t\to 0 \quad y=-t+t^2\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4-2t^5+t^6}{3t^4-3t^5+t^6}=\frac{1-2t+t^2}{3-3t+t^2}\to \frac13$

therefore the limit doesn't exist.

Answer 2

The limit does not exist. Consider $x\to 0$ and $y$ such that $y^3=x^9-x^3$. Then $y\to 0$ but you can see using binomial series that $$\frac{x^2 y^2}{x^3+y^3}=\frac{x^2|x^9-x^3|^{2/3}}{x^9}\approx\frac{1}{x^5}(1-\frac23 x^6) $$ tends to infinity. Choosing $y^3=-x^9-x^3$ gives you a way to approach zero where the function tends to $-\infty$. So it does not have a limit.

For your additional question, if the limit exists you may calculate it along any path of your choice. However, you can't show that the limit exists by only considering certain choices of paths.