About the cases when $N=N(\epsilon)$ is bounded or unbounded The limit of the sequence $(x_n)$ is $x$ if for each $\epsilon >0$ there exists a positive integer $N=N(\epsilon)$ such that $|x_n−x|<\epsilon$ for all $n≥N.$ 
I am asking about the cases when $N=N(\epsilon)$ is bounded or unbounded, i.e., when $\epsilon$ tending to zero. I can find examples such that $N=N(\epsilon)$ is bounded or unbounded. But I am interested on the general case.
 A: In general, if we have a sequence $(x_{n})$ converging to some $x$ and we take $N(\varepsilon)$ to be the smallest integer such that for all $n\geq N(\varepsilon)$ we have $|x_{n}-x|<\varepsilon$, then $X=\{N(\varepsilon):\varepsilon>0\}$ is bounded if and only if there exists an $N\in\mathbb{N}$ such that $x_{n}=x$ for all $n\geq N$.
First suppose that $X$ is a bounded set. Then $X$ is finite, so we can find a $\varepsilon'$ such that $N(\varepsilon')=\max X$. In particular, for all $0<\varepsilon<\varepsilon'$ we have $N(\varepsilon)=\max X$, as 
$$\max X\geq N(\varepsilon)\geq N(\varepsilon')=\max X.$$
Hence for all $n\geq\max X$ we have
$$|x_{n}-x|\leq\lim_{\varepsilon\downarrow0}\varepsilon=0,$$
so $x_{n}=x$.
Now suppose that there exists and $N\in\mathbb{N}$ such that for all $n\geq N$ we have that $x_{n}=x$. Then for all $\varepsilon>)$ we have that $N(\varepsilon)\leq N$, so $X$ is bounded.
A: In general you cannot determine the behaviour in $\epsilon$ of the sharpest $N(\epsilon)$. By sharp I mean that $N(\epsilon)$ is such that $|x-x_n|<\epsilon$ if $n\geq N(\epsilon)$ and $|x-x_n|\geq\epsilon$ if $n<N(\epsilon)$.
Put it differently, $N:(0,+\infty)\to\mathbb{N}$ defined as $N(\epsilon)=\inf\{N\in\mathbb{N}:|x-x_n|\leq\epsilon\text{ whenever }n\geq N\}$ is monotone but not necessarily unbounded. However, asking this sharp condition on $N(\epsilon)$ is a surplus with respect to the definition of limit: "it exists an $N(\epsilon)$" means that you have the freedom of the choice of it, since you are interested in an asymptotic behaviour of the sequence. Therefore, in $N(\epsilon)$ is the sharpest value for a given $\epsilon>0$, also $N(\epsilon)+1000$ is acceptable in the definition of the limit. This means that, recursively, you can always made your choice of $N$ growing faster and faster, and made it unbounded. The reverse is clearly non possible.
