On inverse limits of $\pi$-adically complete algebras

Question

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, 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, Compos. Math. 156, No. 2, 299-324 (2020). ZBL1441.14085.

Answer 1

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!