$$
\forall\varepsilon > 0\ \exists \delta > 0\ \forall x \Big( 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - \ell| < \varepsilon\Big)
$$
I would phrase this differently:
$$
\forall\varepsilon > 0\ \exists \delta \in(0,3)\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)
$$
Now the question is whether that is equivalent to this:
$$
\forall\varepsilon > 0\ \exists \delta>0\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)
$$
If it is true that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$ then it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, simply because every number in $(0,3)$ is $>0$.
Now the question is whether the converse holds. If it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, does it necessarily follow that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots\,{}$? Here the answer in general is “no”. I.e. there are some things you could put in place of $\text{“}\cdots\cdots\text{''}$ for which the first statement would be true and the second false. However, in the definition of "limit" the thing in place of $\text{“}\cdots\cdots\text{''}$ is “if a certain thing is $<\delta$, then a certain thing follows.” Given a certain $\varepsilon>0$ suppose we know there exists $\delta>0$ such that if a certain thing is less than $\delta$ then a certain conclusion follows. If the given value of $\delta$ is small enough, then the minimum of that value of $\delta$ and $2.9$ is small enough, because everything less than $\min\{\delta,2.9\}$ is less than $\delta$. So when the statement in place of $\text{“}\cdots\cdots\text{''}$ has the form “if a certain thing is $<\delta$, then a certain thing follows.”, then it is true that if $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$ then $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$.