I have one doubt and I don´t know how to solve it.
Let $R$ a notherian, commutative, integral ring, and let $U=\{\mathfrak{p}_{1},...,\}$ a set formed by proper prime ideals of $R$, (it can be infinite or finite).
Let $S_{U}=\{r\in R : r\notin \mathfrak{p}_{i} \forall i\}=A-U$.
Is it true that
$R_{S_{U}}\simeq \varprojlim_{i}R_{\mathcal{p}_{i}}?$
Where $R_{\mathfrak{p}}$ is the localization of $R$ by the multiplicative system $R-\mathfrak{p}$.
This is not true. Notice that we can consider a more general question: let $\{S_i\}$ be a collection of multiplicative systems in $R$. From here we can form the index category where we consider the localizations $S_i^{-1} R$ and for each inclusion $S_i \subseteq S_j$ we have the natural map $S_i^{-1} R \to S_j^{-1} R$. Then we can form $S = \cap S_i$ and ask whether $S^{-1} R = \lim_{\leftarrow i} S_i^{-1} R$. (We pass to the case in your question by setting $S_i = P_i^c$.
In this case we see that this should not work, because if it did every map from $S^{-1} R$ would extend to a map from $S_i^{-1} R$, but this could not happen if for example the map from $S^{-1} R$ is nonzero but has something from $S_i \setminus S$ in the kernel - any map with a unit in the kernel is zero. Observe, however, that $S^{-1} R$ has a map to every $S_i^{-1}R$ and thus should map to the inverse limit...
Now for the counterexample (I'll construct one that applies to your case of prime ideals). Set $R = k[x,y]$ $P = (x)$ and $Q = (y)$. Then $S = ((x) \cup (y))^c$. Consider the map $S^{-1} R \to S^{-1}R/(x)$. Then this map is nonzero ($y$ maps to something nonzero) and this cannot extend to a map from $R_{(y)}$.
As a little side-note: observe that in this example there are no arrows in the index category, as $(x) \not \subseteq (y)$ and $(y) \not \subseteq (x)$. This shows that the inverse limit (in the category of $R$-modules) is in fact $R_{(x)} \oplus R_{(y)}$
Your problem is either that you have the wrong formula or you have the wrong hypothesis.
The relevant fact is that for any domain $A$ with fraction field $K$, we have a canonical way to view $A_\mathfrak{p}$ as a subgroup of $K$, and
$$ A = \bigcap_{\mathfrak{p}} A_\mathfrak{p}$$
where $\mathfrak{p}$ ranges over all prime ideals. In terms of limits, $A$ is the limit of the diagram
$$ \begin{matrix} A_{\mathfrak{p}_1} & & A_{\mathfrak{p}_2} & & A_{\mathfrak{p}_3} & & \cdots \\ & \searrow & \downarrow & \swarrow & & \cdots \\ & & F \end{matrix} $$
which, in general, doesn't look anything like the limit you consider.
You would want to apply this in the case that $A = R_S$, since the prime ideals of $R_S$ are precisely the prime ideals that are subsets of $\bigcup U$. You would hope to show
$$ \bigcap_{\mathfrak{p}} A_\mathfrak{p} = \bigcap_{\mathfrak{p}\in U} A_\mathfrak{p}$$
I don't know off hand if this is always true or if you need additional conditions.
Your language and notation, however, suggests you might only mean to consider the case of a limit over an inverse system — what we would today call a "cofiltered limit".
In this special case, I believe you should have
$$ \bigcap_{\mathfrak{p}} A_\mathfrak{p} = \lim_{\mathfrak{p}} A_\mathfrak{p} $$
where the diagram on the right ranges over all prime ideals (and the arrows being the opposites of inclusions)
Your difficulty working through the problem, then, is that you haven't realized that you're putting this condition on $U$, and are mistakenly considering an arbitrary set of prime ideals.