As in, given a regular language $V$, does there exist a deterministic Büchi automaton $\mathcal{A}$, or equivalently a regular language $W$ such that $\mathcal{L}(\mathcal{A})=\vec{W}=V^\omega$?
For clarity, if $L$ is a finite-word language, then $L^\omega$ and $\vec{L}$ are resp. the $\omega$-closure and the limit of $L$; see here.
Edit: My guess is that $V^\omega=\vec{V^\ast}$, but I'm unable to (dis)prove it. see comments.
I assumed that $\vec{R}$ is the language of infinite words $w$ that have infinitely many prefixes in $R$, a.k.a $lim(R)$.
Consider a deterministic Büchi automaton $\mathcal{A} = (\Sigma, Q, q_0, \delta, \alpha)$, and let $\mathcal{A}_f$ denote $\mathcal{A}$ when viewed as an automaton over finite words. As both automata are deterministic, it is not hard to show that $L(\mathcal{A}) = lim(L(\mathcal{A}_f))$. Indeed, if we consider two runs of $r_x$ and $r_y$ where $x$ is a prefix of $y$, then the run $r_y$ extends the run $r_x$. In particular, if an infinite word $w$ has infinitely many prefixes $x_1 < x_2 < x_3 < \ldots$ (by $x_i < x_j$, we mean that $x_i$ is a proper prefix of $x_j$) in $L(\mathcal{A}_f)$, then their corresponding runs induce a unique accepting run of $\mathcal{A}$ on $w$.
It is also not hard to prove the converse similarly. Meaning, for every DFA $\mathcal{A}$, let $\mathcal{A}_B$ denote $\mathcal{A}$ when viewed as a Büchi automaton, then it holds that $L(\mathcal{A}_B) = lim(L(\mathcal{A}))$. Thus, we can conclude with the following:
Claim[reference]: A language $L$ over infinite words is recognizable by a deterministic Büchi automaton iff there is a regular language $R$ such that $L(\mathcal{A}) = lim(R)$.
Now consider a regular language $R$. Regarding whether the language $R^\omega$ is always deterministic-Büchi-recognizable, the answer is negative and a counter-example was recently shown here. So the answer to your question is negative, although the limit of every regular language is deterministic-Büchi-recognizable.
Edit: The specific counter-example is given by the regular expression $$R = $ \cup (0 \cdot (0\cup 1 \cup $)^*\cdot1)$$
Note that the only way to have only finitely many 1’s in a
word in $R^\omega$ is to have an infinite tail of $\\\$$’s. Therefore,
it follows that for every word $w\in R^\omega$, if $w$ contains infinitely many 1’s, then $w$ contains infinitely many 0’s. The latter implies that the word $$ w=011\text{$\\\$$}1\text{$\\\$$}\text{$\\\$$}1\text{$\\\$$}\text{$\\\$$}\text{$\\\$$}1\text{$\\\$$}\text{$\\\$$}\text{$\\\$$}\text{$\\\$$}\cdots = 0\cdot \prod_{i=0}^\infty 1\text{$\\\$$}^i$$
is not in $R^\omega$. It is not hard to see however that $w \in R^\# = \{w\in \Sigma^\omega: \text{ for all $i\geq 1$ there exists a prefix of $w$ in $R^i$}\}$, and this implies that $R^\omega$ is not deterministic-Büchi-recognizable as the authors prove the following claim:
Claim: Consider a regular language $R$. Then, $R^\omega$ is deterministic-Büchi-recognizable iff $R^\omega = R^\#$.
