Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $\lim_{\mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \to x_0$.
For any point $c\in \Bbb C$, can we construct a free ultrafilter $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \to c$?
On
For 1, $x_0$ being an accumulation point of the sequence means that all sets $\hat{U}:=\{n \in \omega: x_n \in U\}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
On
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c \notin \overline{\{x_n\}_{n \in \mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $\omega$, then for every neighborhood $U$ of $c$ the set $\{n:x_n\in U\}$ is in $\omega$ and so in particular is infinite.