Polar coordinates for the evaluating limits

Question

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 .

Answer 1

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 $(x,y)\rightarrow (0,0)$. $\ \Box$