Double sequence $z_{mn}$ Converges but it doesn't imply $z_{mn}$ is bounded I have noticed an interesting thing in double sequence $z_{mn}$ and I can't see why such thing happens.
Definition: Double sequence $z_{mn}$ is a mapping from $\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$
Definition: We say that a double sequence  $\lim_{m,n\to\infty}z_{mn}=l$ where $l$ is finite if $\forall\epsilon>0$ there exists $N(\epsilon)\in \mathbb{R}$ such that $\min(m,n)>N \rightarrow |z_{mn}-l|<\epsilon$
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Let $z_{r\theta}$ be a change of variable from $z_{mn}$ to polar coordinate such that $m=r\cos\theta$ and $n=r\sin\theta$ and $r=\sqrt{m^2+n^2}$
We make $r$ and $\theta$ so that $m\rightarrow \infty$ and $n\rightarrow \infty$ as $r\rightarrow \infty$
Definition: $\lim_{r\to\infty} z_{r,\theta}=l$ if $\forall\epsilon>0$ there exists $N(\epsilon)\in \mathbb{R}$ such that $r>N \rightarrow |z_{r\theta}-l|<\epsilon$
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Theorem: $\lim_{r\to\infty} z_{r,\theta}=l$ $\rightarrow$ $\lim_{m,n\to\infty} z_{mn}=l$
Proof: Assume $\lim_{r\to\infty} z_{r,\theta}=l$. Therefore $\forall\epsilon>0$ there exists $N_1(\epsilon)\in \mathbb{R}$ such that $r>N_1 \rightarrow |z_{r\theta}-l|<\epsilon$.
Then $\exists N_2$ such that $\{z_{mn}|min(m,n)>max(N_2,N_1)\}\subset \{z_{r\theta}|r>N_1\}$ since $m\rightarrow \infty$ and $n\rightarrow \infty$ as $r\rightarrow \infty$, therefore $|z_{mn}-l|<\epsilon$ whenever $m,n>max(N_2,N_1)$. Hence $\lim_{m,n\to\infty} z_{mn}=l$.

Theorem: If $\lim_{r\to\infty} z_{r,\theta}=l$ then $z_{mn}$ is bounded.
$Proof$: Assume $\lim_{r\to\infty} z_{r,\theta}=l$. So there exists $N\in \mathbb{R}$ such that $|z_{r\theta}-l|<1$ for every $r>N$. 
Since $\{z_{mn}|\max(m,n)>N\}\subset\{z_{r\theta}|r>N\}$ hence $|z_{mn}-l|<1$ whenever $max(m,n)>N$.
Now we have
$|z_{mn}|-|l|<|z_{mn}-l|<1$ therefore $|z_{mn}|<1+l$ for every $max(m,n)>N$.
Let $M=max\{|z_{mn}||\forall m,n\le N\} $. Then clearly $|z_{mn}|\le max\{M,1+l\} \forall m,n \in \mathbb{N}$.

And Now, I noticed that for a double sequence $z_{mn}=\dfrac{m^2+n^2}{m^5+n}$ something is wrong. 
After I change the variable to polar coordinate, $z_{r\theta}=\dfrac{r}{r^4\cos^4\theta+\sin\theta} \rightarrow 0$ as $r\rightarrow \infty$. And So $z_{mn}$ is convergent to 0 BUT $z_{mn}$ isn't bounded!
Since $\forall E>0$ $\exists N$ such that $|z_{1n}|=|\dfrac{1+n^2}{1+n}|>E$ whenever $n>N$.
So, where did I do any of my proofs wrong?
$m=r\cos\theta$ and $n=r\sin\theta$. If we make $r$ and $\theta$ so that $m\rightarrow \infty$ and $n\rightarrow \infty$ as $r\rightarrow \infty$. Then first theorem should hold I think...but there should be some error in the second theorem...?
 A: Caveat: The text of the question now underwent so many nontrivial modifications, after complete answers were posted, that answers addressing previous versions, such as the one below, run the risk of being offtopic with respect to the last version. This kind of situation is so obviously dysfunctional that one could think any conscious OP would avoid it. Alas...

Both theorems are wrong. Both proofs go astray when asserting that there exists $N$ or $N_1(ϵ)$ independent on θ such that blablabla. 
A counterexample is given by the sequence $(z(m,n))$ such that $z(m,n)=0$ for every $(m,n)$ with $m\ne n^2$, and $z(n^2,n)=n$ for every $n$. Then, for every $\theta$, $z_{r,\theta}=0$ for $r$ large enough hence $\lim\limits_{r\to\infty}z_{r,\theta}=0$ for every $\theta$, but $(z(m,n))_{m,n}$ is not bounded and $\lim\limits_{m,n\to\infty}z(m,n)$ does not exist.
Edit: The revised definition does not make the convergence when $(m,n)\to\infty$ and the convergence when $r\to\infty$ equivalent, yet. One should replace the condition that $\theta\in(0,\pi/2)$ by the condition that $\theta\in[0,\pi/2]$.
With this modification of the revised definition, $z_{r,\theta}=r/(r^4\cos^4\theta+\sin\theta)$ does NOT converge to $0$ when $r\to\infty$ since $z_{r,\pi/2}\to\infty$. 
Exercise: Find $\theta(r)$ in $(0,\pi/2)$ such that $z_{r,\theta(r)}\to\infty$ when $r\to\infty$.
A: Better answers have probably been given above. Here is a layman's approach:
First theorem is wrong due to the assumption that $\{z_{m,n} |\max(m,n) > N\} \subset \{z_{m,n} | \sqrt{m^2+n^2} > N\}$ Because $\sqrt{m^2+n^2} > N$ does not  imply that the highest of them is $> N$, but the other way around (however $\min(m,n) > N \implies (\max(m,n) > N$ and $\sqrt{m^2+n^2} > N)$ and that is probably why you thought to use it in the second theorem). For any integer $N$ then $\{\min(m,n) > N\} \supset \{\max(m,n) > N \} $ 
Edit: $\{\sqrt{m^2+n^2} > N\} \cup \{\max(m,n) > N \} \neq \{\max(m,n) > N \}$; $\{\sqrt{m^2+n^2} > N\} \cap \{\max(m,n) > N \} \neq \emptyset $
(a circle is between its inscribed square and circumscribed square)
Informally, the double limit could be made to converge to a limit dependent only on one variable when the elements missed by this "projection" become of negligible size (measure) in comparison with the rest. Hence it is possible to have limits that depend just on $\sqrt{m^2+n^2}$. From a simplistic point of view, in Did's example, as $\sqrt{m^2+n^2} = r \to \infty$ there are many more $m \neq n^2$ than $m = n^2$.
