Consider the following situation, let $\mathcal{R}$ be a discrete valuation ring with uniformizer $\pi$ (say the valuation ring of a finite extension $K$ of $\mathbb{Q}\_{p}$. Let $\{ A\_{n}\}\_{n\in\mathbb{N}}$ be a projective family of $\mathcal{R}$ algebras equipped with maps $A\_{n+1}\rightarrow A\_{n}$, considering their $\pi$-adic completions, we get an associated projective family $\{ \widehat{A\_{n}}\}\_{n\in\mathbb{N}}$, along with short exact sequences:
\begin{equation}
0\rightarrow A\_{n}\rightarrow \widehat{A\_{n}}\rightarrow \widehat{A\_{n}}/A\_{n}\rightarrow 0
\end{equation}
together with transition maps between the sequences. We will denote $B\_{n}=\widehat{A\_{n}}/A\_{n}$ for simplicity. Assume that the morphism $\varprojlim\_{n}A\_{n}\rightarrow \varprojlim\_{n}\widehat{A\_{n}}$ is an isomorphism and that all $\widehat{A\_{n}}$ are compact with respect to the $\pi$-adic topology. I want to be able to conclude that if this is the case, then for sufficiently $n$ the map $A\_{n}\rightarrow \widehat{A\_{n}}$ is already an isomorphism. For this, it would be enough to show two things:


1. The first right derived functor $R^{1}\varprojlim\_{n}A\_{n}= R^{1}\varprojlim\_{n}A\_{n}$.
2. If $\varprojlim\_{n} B\_{n}=0$ then $B\_{n}=0$ for sufficiently high $n$.


Notice that all morphisms considered here are continuous when all modules are endowed with the $\pi$-adic topology. Furthermore, the $\mathcal{R}$-modules $B\_{n}$ are compact, multiplication by $\pi$ is an isomorphism and the closure of $0$ is the whole $B\_{n}$. I do not expect complete answers to any of these questions, but I seem to lack the tools to tackle these kinds of problems, so suggestions of references where something like this might come up are very much appreciated.


Context: If $X=\mathrm{Sp}(A)$ is a smooth connected affinoid space over $K$ and $f\in A$, I would like to show that for $X(\frac{\pi^{n}}{f})=\mathrm{Sp}(C\_{n})$ the rings $C^{\circ}\_{n,A}$ defined in [MO Question: Noetherianess of some subalgebras of an affinoid algebra](https://mathoverflow.net/questions/432242/on-the-noetherianess-of-some-subalgebras-of-an-affinoid-algebra), agree with $A^{\circ}$ for sufficiently high $n$. In this setting the inverse limit
$\varprojlim\_{n}C^{\circ}\_{n,A}\rightarrow \varprojlim\_{n}\widehat{C^{\circ}\_{n,A}}=A^{\circ}$ by
theorem 2.6 in the reference.


*Hansen, David*, [**Vanishing and comparison theorems in rigid analytic geometry**](http://dx.doi.org/10.1112/S0010437X19007371), Compos. Math. 156, No. 2, 299-324 (2020). [ZBL1441.14085](https://zbmath.org/?q=an:1441.14085).



On inverse limits of $\pi$-adically complete algebras
Not sure about the main question, but regarding the "context" question, one can show the following:


If $A$ is any affinoid, and $f \in A$ any element with nowhere-dense zero locus, then $A^\circ = A\times\_{A\langle \pi^n/f \rangle} {A\langle \pi^n/f \rangle}^\circ$ for all sufficienly large $n$.


The point is that $A \to {A\langle \pi^n/f \rangle}$ is strict for the supermum seminorms as soon as the image of $\mathrm{Spa} {A\langle \pi^n/f \rangle} \to \mathrm{Spa} A$ contains the Shilov boundary of $\mathrm{Spa} A$, which it will for all sufficiently large $n$ because the Shilov boundary is finite and (by our assumption on $f$) disjoint from $V(f)$.


Hope this helps!


