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

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?

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.

• Isn't $\theta$ arbitrary but fixed? – Shubham Johri Sep 4 '19 at 11:23
• $\theta$ is arbitrary and generally NOT fixed. It is fixed (an arbitrary constant), if we are taking the limit along a line. – Robert Z Sep 4 '19 at 11:35
• This is something novel for me, taking $\theta$ non-fixed, that is. This little detail was never discussed in my book or class. Yet it is intuitive, necessary for the polar substitution to be able to describe arbitrary paths. Thank you for your time and effort! – Shubham Johri Sep 4 '19 at 11:42
• @ShubhamJohri Glad to be of help. – Robert Z Sep 4 '19 at 11:50

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.

• This is as good as sandwich theorem/epsilon-delta, wouldn't you say? – Shubham Johri Sep 4 '19 at 11:44
• Pretty much. If the expression can be bounded by another one that tends to $0$ as $r\to0$ and is independent of $\theta$, then you can conclude the original one goes to $0$ as well. Although it may not be necessary, most of the times if you want to do it using polar coordinates then you have to do it this way. – trisct Sep 4 '19 at 11:47

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.