Limit of Two Variable Function using Polar Coordinate where r doesn't tend to zero

Question

I want to ask several questions related to this topic as in every case, $(x,y)$ tends to $(0,0)$ implying $r$ always tends to $0$.

1. Why in the limit using polar coordinate, the variable is only $r$, not including $\theta$?

2. In what case, $r$ approaches number besides $0$, for instance $r$ approaches $2$?

Answer 1

The change to "polar coordinates" is a tool that relies on two mathematical facts (i.e. lemmata we can prove):

• The function $\phi: \mathbb R_{\geq 0}\times [0,2\pi) \to \mathbb R^2$ given by $\phi(r,\theta)=(r\cos\theta,\,r\sin\theta)$ is surjective; that is, every point $(x,y)$ in the plane may be described as a pair $(r,\theta)$ such that $\phi(r,\theta)=(x,y)$, instead.
• If $\alpha(r)$ and $\beta(\theta)$ are functions such that $\displaystyle \lim_{r\to r_0} \alpha(r) = 0$ and $\beta(\theta)$ is bounded, then $\displaystyle \lim_{r\to r_0} \alpha(r)\beta(\theta)=0$.

Here, if $\phi(r,\theta)=(x,y)$ and $(x,y)\to(0,0)$, then necessarily $r\to 0$ as $r$ is simply the (Euclidean) norm of $(x,y)\in\mathbb R^2$. As for your questions:

1. This is only the case if the second of the previous facts applies. Given a limit $\displaystyle \lim_{(x,y)\to(0,0)} f(x,y)$, we can always write $f(x,y)=f(r\cos\theta,r\sin\theta)=g(r,\theta)$ for certain values of $r$ and $\theta$. Here $r\to 0$ as we've noted, but $\theta$ doesn't always tend to a fixed number since we only know that $(x,y)$ is approaching the origin -- we don't know the exact path it takes. (In fact, the limit exists iff any path $t\mapsto (x(t),y(t))$ such that $\displaystyle \lim_{t\to t_0} (x(t),y(t)) = (0,0)$ verifies $\displaystyle \lim_{t\to t_0} f(x(t),y(t))=L$ for a fixed number $L$.)

   If we can find functions $\alpha$ and $\beta$ such that i) $\displaystyle \lim_{r\to 0} \alpha(r)=0$, ii) $\beta$ is bounded and iii) $g(r,\theta)=\alpha(r)\beta(\theta)$, we can write our limit as $$\lim_{(x,y)\to(0,0)} f(x,y)=\lim_{\substack{r\to 0 \\ \theta\in[0,2\pi)}}g(r,\theta)=\lim_{\substack{r\to 0 \\ \theta\in[0,2\pi)}}\alpha(r)\beta(\theta)=0.$$

   It is very important that you apply this reasoning only if $g(r,\theta)$ can be written as a product of functions in this way; here's an example where that doesn't happen and thus leads to a false conclusion about the limit.

2. This is a bit more complicated; for $r$ to tend to a non-zero number, $(x,y)$ would have to tend to a point that is not the origin and then $\theta$ would have to tend to some point too. Certainly if $\displaystyle \lim_{r\to r_0} \alpha(r)=0$ and you can separate $g(r,\theta)=f(r\cos\theta,r\sin\theta)$ like before then the limit will be zero, but otherwise you'd need to evaluate in a case by case basis.

Answer 2

Because $r$ is the radius. If this shrinks to $0$ then your point $(x,y)$ also moves to the origin $(0,0)$ as wanted. However, we must require that the expression $g(r,\phi)$ always be defined for all phases $\phi$ and be bounded.

If $r$ is allowed to approach $2$ then you're homing in on so many points on a circle of radius $2,$ not the single point $(0,0)$ which you want.