$\lim_{r\to0}f(r\cos\theta,r\sin\theta)=0$ does not imply $\lim_{(x,y)\to(0,0)}f(x,y)=0$

Question

I came across this limit today:$$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^6+y^2}$$Substituting $x=r\cos\theta,y=r\sin\theta$ gave$$\lim_{r\to0}\frac{r^2\cos^3\theta\sin\theta}{r^4\cos^6\theta+\sin^2\theta}$$which is $0$. Yet the limit doesn't exist since along the path $y=x^3$ it is $1/2$. I do realize that taking $x^3=m$ in the original limit will yield$$\lim_{(m,y)\to(0,0)}\frac{my}{m^2+y^2}$$which is easily seen to be path dependent. Why did the polar substitution not work? As I see it, a lot of books make this substitution to prove that a limit exists. Are those proofs wrong? Is this not sufficient to show that a limit exists? In which case, besides the sandwich theorem and $\varepsilon-\delta$ approach, do we have no other tool to establish the existence of a limit?

Answer 1

Since $\theta$ could change as $r\to 0$, you may not conclude that $\frac{r^2\cos^3\theta\sin\theta}{r^4\cos^6\theta+\sin^2\theta}$ goes to zero!

Take $\theta=r^2$, then as $r\to 0^+$ we have $$\frac{r^2\cos^3\theta\sin\theta}{r^4\cos^6\theta+\sin^2\theta}\sim \frac{r^2\cdot r^2}{r^4+r^4}\to \frac{1}{2}.$$ Of course, if $\theta=\theta_0$, a constant angle, then as $r\to 0^+$ $$\frac{r^2\cos^3\theta_0\sin\theta_0}{r^4\cos^6\theta_0+\sin^2\theta_0} \to 0.$$ So we may conclude that, even by using polar coordinates, the given limit does not exist.

Answer 2

In order to prove that some limit does not exist you can find a substitution (like those you mentioned) such that the resulting limit does not exist. Some times polar substitution works some times it doesn't. Every time you choose the appropriate substitution depending on the expresion that you have.

Answer 3

This is more of a comment on the use of polar coordinates in problems like this.

If I were to write a proof of a limit of this type using polar coordinates, for example, $$\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}$$ I would write: $$\left|\frac{x^2y}{x^2+y^2}\right|\\ =\left|\frac{r^3\cos^2\theta\sin\theta}{r^2}\right|\\ =r|\cos^2\theta\sin\theta|\leq r$$ The last inequality is crucial for a correct proof. But your example does not allow this.