Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$. Is $\widehat{A}[[x]]$ the completion of $A[x]$ at $(\mathfrak{m},x)$?
We certainly have surjections $$\widehat{A}[[x]]/(\mathfrak{m}^n\widehat{A} + x^n\widehat{A})\cong (A/\mathfrak{m}^n)[x]/(x^n)\longrightarrow A[x]/(\mathfrak{m},x)^n$$ which factors through surjections $\widehat{A}[[x]]/(\mathfrak{m}\widehat{A} + x\widehat{A})^n\longrightarrow A[x]/(\mathfrak{m},x)^n$, so this determines a map $\widehat{A}[[x]]\rightarrow \lim_n A[x]/(\mathfrak{m},x)^n$.
On the other hand, the kernel of this map is visibly $\bigcap_{n\ge 1} (\mathfrak{m},x)^n$, which is zero since $\widehat{A}[[x]]$ is local, so it seems like the question has a positive answer. Is this correct?
