Two-variable limit via change of variables It often happens that by considering polar coordinates we can transform a limit of the form
$$\lim_{(x,y) \to (0,0)} f(x,y) \quad \quad \quad (1)$$
into one of the form
$$\lim_{(r,\theta),\, r \to 0} F(r)G(\theta). \quad \quad (2)$$
I am assuming that $f(x,y)$ is continuous in a punctured ball centered at the origin of $\mathbb{R}^{2}$.
I have two questions regarding this process:
a) Given that $x^{2}+y^{2} = r^{2}$ and $(x,y) \to (0,0)$ iff $r \to 0$, is it OK to omit the $\theta$ in $(2)$?
b) In case $F(r)\to 0$ as $r \to 0$ and $G$ is a bounded function, it does follow that $\lim_{(x,y) \to (0,0)} f(x,y)=0$, right?
Can you recommend a calculus text wherein an "observation" similar to the one in b is stated and proved in detail?
Thanks in advance for you replies...
 A: Recall
We write a function $f(x,y) = F(r)G(\theta)$ with $r=\sqrt{x^2+y^2}$ and $\theta = \bar{\text{atan}}(x,y)$, where $$\bar{\text{atan}}(x,y) = \begin{cases}
\arctan(\frac{y}{x})&\text{if } x\neq0\\
\frac{\pi}{2}&\text{if }x=0 \land y>0\\
\frac{3\pi}{2}&\text{if }x=0 \land y<0\\
\text{undefined}&\text{if }x=0 \land y=0\\
\end{cases}$$
(You can change the definition above as long as the $\theta$ is discussable.)
Then we can easily prove (by the $\epsilon-\delta$ definition) that
$$ \lim_{(x,y)\to(0,0)} f(x,y) = L$$
implies
$$ \lim_{r\to0} F(r)G(\theta) = L, \; \forall \theta$$
For $f(x,y)$ whose limit does not exist, the polor coordinates expression of the limit might not exists itself or depends on the value $\theta$. Nothing further is guaranteed.
Noted that the converse statement normally does not hold. The polar coordinate one limit exists and equals to $L$ for all $\theta$ does not imply the original limit exist. See b. for further discussion.
The Questions

a. Given that $r^2={x^2+y^2}$ and $(x,y)→(0,0)$ iff $r→0$, is it OK to omit the $θ$ in (2)?

By "omit", you might think $\theta$ does not affect the limit. Since nothing further is guaranteed in your statement(e.g. the original limit exists), I'd say No-the second limit might depends on $\theta$.
For instance
$$ f(x,y) = \bar{\text{atan}}(x,y)$$
with
$$ F(r)=1, G(\theta)=\theta$$
Beside that, how will you replace $G(\theta)$ if you omit the $\theta$ generally? You may still find ways to omit $\theta$ with more restrictions(for example, the restriction in b.), but generally it is a No.

b. In case $F(r)→0$ as $r→0$ and $G$ is a bounded function, it does follow that $\lim_{(x,y) \to (0,0)}{f(x,y)=0}$, right?

The answer is Yes. Here's a proof.
We know that since $G$ is bounded, we may let $G(\theta) < M$, and $F(r) \to 0$ as $r \to 0$. Then given $\epsilon$ there exists $\delta$ such that for all $r<\delta$ and $\theta$
$$ |F(r)G(\theta)| < M|F(r)| < \epsilon \;$$
Noted that $\delta$ is only revelant to $\epsilon$, and is irrevelant to $\theta$.
Now for $(x,y)$, we may write $\lim\limits_{(x,y) \to (0,0)}{f(x,y)}$
$$=\lim_{(x,y) \to (0,0)}{F(\sqrt{x^2+y^2})G(\bar{\text{atan}}(x,y))}$$
Let $r=\sqrt{x^2+y^2}$ and take $\delta$ in the above statement. Then for all   $|(x,y)|<\delta \to r<\delta$
$$|F(\sqrt{x^2+y^2})G(\bar{\text{atan}}(x,y))|<\epsilon$$
The choice of $\delta$ does not rely $\theta$, which, is $\bar{\text{atan}}(x,y)$ in this case. And hence proved.
Note
Noted that even if this statement is correct, you should still carefully use polar coordinate to prove an existence of multi-variable limits. Here is an example of mis-using polar coordinates to examine the limit. Similar proof for that example does not work because $\delta$ is revelant to $\theta$, unlike question b. Which cause there does not exist a common $\delta$ for different angle. That is why $G(\theta)$ being bounded is neccessary here.
tl;dr
Without further restrictions, you should not omit $\theta$ in the second limit. The hypo in b. does imply the original limit exist, but carefully use this theorem- especially be sure that $G(\theta)$ is bounded.
A: Well since r is defined as $\sqrt(x^2+y^2)$
As x --> 0 and y -->0, it becomes apparent that r --> 0.
Which is why we can use this. However this doesn't mean we can omit the $\theta$. As it is an independent variable.
