I’m studying multivariable calculus.
When trying to solve a limit $\lim_{(x,y)\to(x_0, y_0)}f(x,y)$ with $f:\mathbb{R}^2\to\mathbb{R}$, our professor told us that, in some cases, the limit is more easily ”computed” by changing the coordinate system to polar coordinates (using $x=ρ\cosθ$ and $y=ρ\sinθ$) and solving the limit for $\lim_{ρ\to0}f(ρ\cosθ, ρ\sinθ)$. He also tells us to consider $\theta$ as a function of $\rho$, so $\theta=\theta(\rho)$.
- note that I’ve always dealt just with limits $(x,y)\to(0,0)$, but I wrote it in general form because I don’t know if this changes something.
Now, the idea of changing system to polar coordinates is ok, but what bothers me is that I can’t find any underlying rigorous result/theorem/algebraic process or whatever that lets me say that $\theta$ is a function of $\rho$, which seems wrong to me because the “purpose” of the coordinates should be to be independent (am I wrong? maybe it’s just linearly independent?). Is it a “trick” to help us compute some limits to avoid the underlying theory? Note that, that lets me change a two-variables limit $(x, y)$ in a one-variable limit $\rho$ (which is particularly useful if all the terms with $\theta$ cancel in the limit), so it seems a crucial observation.
EDIT
Thanks to @bb_823 I saw a similar question, but still not exactly what I was asking for: here seems to be the underlying theory of doing multivariable limits in polar coordinates, and while the formulation in the link clarifies the whole question to me (“the value of the limit doesn’t depend from θ” and the uniformity), the formulation of my professor “assume that θ is function of ρ” still seems “wrong” and I don’t see how it’s equivalent to the linked answer.
Maybe this counterexample can help you:
Consider the limit $$\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}$$
Then you might think that switching to polar coordinates it results in $$\tag{*}\lim_{r\to 0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}=\lim_{r\to 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$ As if $\cos(\theta)=0$ is zero and for $r\to 0$ is $0$ as well. Thus the limit seems to be $0$.
Seems: if instead you look at the path $x=y^2$ (along a parabola) and let $y\to 0$, then: $$\lim_{y\to0}\frac{y^2y^2}{(y^2)^2+y^4}=\lim_{y\to0}\frac{y^4}{2y^4}=\frac{1}{2}.$$
So what went wrong here?
The main problem is that by switching to polar coordinates and evalutating the limit (*) you assumed that $\theta$ would be fixed, which is the same as considering only straight paths towards the origin.
As the second limit shows, using non-straight paths results in different limit, so the limit does not exist.
In conclusion $\theta$ must be able to depend on $r$ (or viceversa) in order to allow all paths to the origin.