6

I've seen many pages on this site about the using of polar coordinates for evaluating limits but I'm really confused . I don't know when we can apply that or when it shows the correct limit . In fact , I'm looking for a theorem (with proof) that covers this problem completely . Also the references to the reliable books will be helpful .

S.H.W
  • 4,379
  • 1
    Change of variable is a strategy used to evaluate some limits. A specific change of variables would be polar coordinates (or spherical, cylindrical and etc when having three variables). I can't find a theorem that covers these problems, but I found some discussion on evaluating limits by change of variables here: link – Jonathan Perales Mar 02 '19 at 01:24
  • 1
    @JonathanPerales Thanks but the main problem here is about the function of two variables . Sometimes using the polar coordinates can be misleading . – S.H.W Mar 02 '19 at 01:30

1 Answers1

9

Theorem. Let $f:D\rightarrow \mathbb{R}$, where $D\subseteq \mathbb{R}^2$ is a suitable neighbourhood of $(0,0)$. It holds that $$ \lim_{(x,y)\rightarrow (0,0)} f(x,y) = \ell\in \mathbb{R} $$ if and only if the following two conditions hold:

(i) for all $\theta\in [0,2\pi)$ there exists the limit $\lim_{r\rightarrow 0^+} f(r\cos\theta,r\sin\theta)=\ell$;

(ii) the limit is uniform with respect to $\theta$, that is, for all $\epsilon>0$ there exists $\rho>0$ such that $$|f(r\cos\theta,r\sin\theta)-\ell|<\epsilon$$ for all $r\in (0,\rho)$ and for all $\theta\in [0,2\pi)$.

Proof. $(\Rightarrow)$ By definition of limit, for all $\epsilon>0$ there exists $\rho>0$ such that $$|f(x,y)-\ell|<\epsilon$$ for all $(x,y)\in B((0,0),\rho)$ (which is the open ball with centre $(0,0)$ and radius $\rho$). Since $(r\cos\theta,r\sin\theta)\in B((0,0),\rho)$ for all $r\in (0,\rho)$ and $\theta\in [0,2\pi)$, (i) and (ii) are both verified.

$(\Leftarrow)$ Let $\epsilon>0$. For all $(x,y)\in B((0,0),\rho)$, let $r>0$ and $\theta\in [0,2\pi)$ be such that $r\cos\theta=x$ and $r\sin\theta=y$. We have $r\in (0,\rho)$ and thus from (i) and (ii) it follows that $$ |f(x,y)-\ell|=|f(r\cos\theta,r\sin\theta)-\ell|<\epsilon. $$ Thus $f(x,y)\rightarrow \ell$ as $(x,y)\rightarrow (0,0)$. $\ \Box$

Alessio Del Vigna
  • 1,767
  • Is this theorem also true ? https://math.stackexchange.com/questions/2974062/why-exactly-limit-in-polar-coordinates-isnt-sufficient-to-find-the-limit-in-two?rq=1 – S.H.W Mar 11 '19 at 19:58
  • 1
    The statement in the question is false, and a counter example is provided. If you look at the answer of that question you can find a reformulation if the Theorem I stated above. – Alessio Del Vigna Mar 12 '19 at 00:08
  • I think applying that reformulation is easier . Is that true ? – S.H.W Mar 12 '19 at 04:14
  • The reformulation is what one uses in practice. Note that (but maybe it was already obvious) for practical purposes you actually do two steps: (i) you compute $\lim_{r\rightarrow 0+} f(r\cos \theta,r\sin\theta)$ and this must not depend on $\theta$ (othwerwise you can already conclude that the limit does not exist); (ii) if $\ell\in \mathbb{R}$ is the limit computed in (i) then you have to find $g(r)\rightarrow 0$ and such that $|f(r\cos\theta,r\sin\theta)-\ell|\leq g(r)$ for all $\theta$. This enough to prove that $\lim_{(x,y)\rightarrow (0,0)} f(x,y) = \ell$. – Alessio Del Vigna Mar 12 '19 at 08:27
  • Note that this second step is the same as point (ii) of the theorem I stated in my answer. – Alessio Del Vigna Mar 12 '19 at 08:28