# Is polar coordinates enough to prove that a limit exists

Somewhat of a basic question but I failed to find an answer or come up with a formal one myself.

Suppose I want to find the limit $\lim_{{(x,y)} \to {(0,0)}}f(x,y)$ using spherical coordinates $x:=r\cos \theta$, $y:= r\sin\theta$. Suppose I found that $\lim_{r \to 0} f(r,\theta)$ exists and is equal to $\alpha$ regardless of $\theta$. Did I really cover every possible path? Can we say for sure that the limit is $\alpha$? maybe some other limit exists in another path that we didn't cover.

For example, take $\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2} = \lim_{r\to 0}\frac{r^3\cos^2\theta \sin \theta}{r^2} = \lim_{r \to 0}r\cos^2 \theta \sin \theta = 0$.

I agree that IF the limit exists, it has to be zero. But maybe there is some path we didn't cover and from that path the limit is something else?

• the definition of continuity at $(0,0)$ doesn't involve "every paths" but only says : for every $\epsilon > 0$ there is $\delta$ such that $|(x,y)|< \delta \implies |f(x,y)-f(0,0)| < \epsilon$ Feb 23, 2016 at 23:08

I would like to point you to the answer I gave here for an elaborate explaination on the use of $\lim\limits_{r\to0}$. To sum up what is said there; The problem is not that $\lim\limits_{r\to 0}f(r,\theta)$ only considers straight lines, the problem is that $\lim\limits_{r\to 0}f(r,\theta)$ is not what you probably think it is. $f(r,\theta)$ is a function of two variables, and letting $r\to 0$ means that we are approaching the line $r=0$ in the domain $[0,\infty)\times[0,2\pi)\subset \mathbb R^2$. And we commonly do not have many techniques for evaluating limits of this sort.

For instance, you seem to evealuate $\lim\limits_{r\to 0}r\cos^2\theta\sin\theta$ by substituting $r=0$. But this technique (limit evaluation by substitution) is something that is usually only proven to be valid for point-continous functions. It cannot easily be extended to apply to limits of the kind $\lim\limits_{r\to 0}f(r,\theta)$.

What you can do is argue that when we choose $\delta=\varepsilon$, then $$0<r<\delta\implies \vert r\cos^2\theta \sin \theta \vert<\delta \underbrace{\vert \cos^2 \theta \sin \theta\vert}_{\leq 1}<\varepsilon.$$ This does prove that $\lim\limits_{r\to 0}f(r,\theta)=0$. (Mind that you need to treat the cases where $\cos^2\theta\sin\theta=0$ with care.)

– Eric
Dec 12, 2017 at 14:57

The fine point is in the phrase "regardless of $\theta$". Let $f$ be the characteristic function of the set $$A:=\left\{{1\over k}\biggl(\cos{1\over k},\sin{1\over k}\biggr)\>\biggm|\>k\in{\mathbb N}_{\geq1}\right\}\ .$$ Then $\lim_{r\to0}\tilde f(r,\theta)=0$, regardless of $\theta$, but the $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist.

If, however, you can prove that for some $(x,y)\mapsto g(x,y)$ defined in a punctured neighborhood of $(0,0)$ you have $$|g(r\cos\theta,r\sin\theta)|\leq q(r)\quad \wedge\quad \lim_{r\to0}q(r)=0$$ then you may conclude that in fact $\lim_{(x,y)\to(0,0)} g(x,y)=0$.

You should probably read the question posted here, as well as my answer to that question. The problem is that if we write $f(r;\theta)$, then taking $$\lim_{r \to 0} f(r;\theta)$$ can be misleading, because this considers only straight-line paths to the origin unless you also permit $\theta$ to vary as a function of $r$; i.e., to consider $$f(r;\theta(r))$$ for arbitrary functions $\theta(r)$.

You are confining to the origin of coordinate system in either case. When real $r = \sqrt{x^2+y^2}=0$, both $x$ and $y$ are zero. ( $r$ is zero for all $\theta$), and so it suffices.

No, its not.

The reason is you are assuming $$\theta$$ is fixed, where it may be not.

Always, you need to check for dependency to a fixed another value such as $$m$$ before you conclude that the limit goes to $$0$$.

For example : $$lim_{(x,y)\Rightarrow (0,0)}\frac{x^3y}{x^6+y^2}$$, This limit is dependent on $$m$$ such that considering $$y = (-mx)^3$$ yields: $$\frac{-m^3x^6}{x^6-m^6x^6} = \frac{x^6(-m^3)}{x^6(1-m^6)} = \frac{-m^3}{1-m^6}$$ Thus as this limit depends on $$m$$, this limit does not exist. But if you were to use polar coordinates: $$x = rcos(\theta) \\ y = rsin(\theta) \\ \Rightarrow lim_{\:r\:\Rightarrow\: 0} \frac{r^3cos^3(\theta)rsin(\theta)}{r^6cos^6(\theta)+r^2sin^2(\theta)}$$ Simplifying this for $$r^2$$ value, we get: $$lim_{\:r\:\Rightarrow\: 0} \frac{r^2cos^3(\theta)sin(\theta)}{r^4cos^6(\theta)+sin^2(\theta)}$$ Inputting $$r = 0$$, we get: $$\frac{0}{sin^2(\theta)} = 0$$ Thus we would have found the limit result to be $$0$$, where its not.