By definition of exponent operator on ordinals, we have $0^\omega=\lim_{\xi\to\omega}0^\xi$
However, Note that $0^\xi$ is not increasing, so if we still let $\lim_{\xi\to\omega}0^\xi=\sup\{0^\xi|\xi<\omega\}$ then it is followed by $0^\omega=\sup\{1,0,0,\ldots\}=1$ An incredible result.
Nevertheless if we use the definition of limit superior $\overline{\lim}_{\eta\to\beta}\alpha^\eta:=\inf_{\eta < \beta}\sup_{\eta \le\xi<\beta}\alpha^\xi$ and limit inferior $\underline{\lim}_{\eta\to\beta}\alpha^\eta:=\sup_{\eta < \beta}\inf_{\eta \le\xi<\beta}\alpha^\xi$ then we get $\lim_{\xi\to\omega}0^\xi=\underline{\lim}_{\xi\to\omega}0^\xi=\overline{\lim}_{\xi\to\omega}0^\xi=0$
Conceivable, but this is not the end of the story. Let's consider about $m^n$ with $0
So can we find a perfect definition of limit on ordinals?
Update:
By the way, if we deal the Ordinal class as a discrete topology space, then $\{\alpha\}$ is a neighborhood of $\alpha$, hence $\{\alpha\}$ must in every filterbase which converge to $\alpha$. So if we let $\overline{\lim}_{\eta\to\beta}\alpha^\eta:=\inf_{\eta \le \beta}\sup_{\eta \le\xi\le\beta}\alpha^\xi$ and $\underline{\lim}_{\eta\to\beta}\alpha^\eta:=\sup_{\eta \le \beta}\inf_{\eta \le\xi\le\beta}\alpha^\xi$
Then $\alpha^\beta=\lim_{\xi\to \beta}\alpha^\xi=\underline{\lim}_{\xi\to \beta}\alpha^\xi=\overline{\lim}_{\xi\to \beta}\alpha^\xi$ hold for every $(\alpha,\beta) \in \mathbb O^2$
But actually it cannot be a definition since $\inf_{\eta \le \beta}\sup_{\eta \le\xi\le\beta}\alpha^\xi:=\inf\{\sup\{\alpha^\xi|\eta \le\xi\le\beta\}|\eta \le \beta\}$ but $\{\alpha^\xi|\eta \le\xi\le\beta\}$ contains $\alpha^\beta$!
Update: Edited title. The eventual question is as title illustrates.