Let $\dotsc A_2 \to A_1 \to A_0$ be a sequence of surjective homomorphisms of commutative rings. Consider the projective limit $\varprojlim_i A_i$. If $S$ is an (infinite) set, then $\varprojlim_i (A_i^{\oplus S})$ is a module over this limit. It consists of families $(a_i^s \in A_i)_{s,i}$ with $a_{i+1}^s \mapsto a_i^s$ and for fixed $i$ almost all $a_i^s=0$. (Notice that this is larger than $(\varprojlim_i A_i)^{\oplus S}$!)
There is a canonical epimorphism of $A_0$-modules $$\alpha : \varprojlim_i (A_i^{\oplus S}) \otimes_{\lim_i A_i} A_0 \to A_0^{\oplus S},~ (a_i^s)_{s,i} \otimes u \mapsto (a_0^s u)_s.$$
Question: Is this an isomorphism? In other words, if $(a_i^s)_{i,s} \in \varprojlim_i (A_i^{\oplus S})$ is such that $a_0^s=0$ for all $s$, is this element then contained in $\ker(\lim_i A_i \to A_0) \cdot \varprojlim_i (A_i^{\oplus S})$?
Notice that $\alpha$ has a canonical section $\beta$, mapping $(a_s)_s \in A_0^{\oplus S}$ to $\sum_s (\delta_s)_i \otimes a_s$. Then $\alpha \beta = \mathrm{id}$ and the question is if $\beta \alpha = \mathrm{id}$.
I think the following gives a counterexample.
Let $k$ be a field, and let $A_0=k$, $A_i=k[x_1,\dots,x_i]$ modulo the ideal generated by all monomials of degree two, with the map $A_{i+1}\to A_i$ sending $x_{i+1}$ to zero. So $A=\varprojlim(A_i)$ consists of all infinite formal sums $\mu+\lambda_1x_1+\lambda_2x_2+\dots$.
Let $S=\mathbb{N}$, and for $i\in\mathbb{N}$ let $a_i=(a^0_i,a^1_i,\dots)$, where $$a_0=(0,0,0,0\dots),$$ $$a_1=(x_1,0,0,0,\dots),$$ $$a_2=(x_1,x_2,0,0,\dots),$$ $$a_3=(x_1,x_2,x_3,0,\dots),$$ and so on.
Then $(a^s_i)_{i,s}\in\varprojlim_i\left(A_i^{\oplus S}\right)$ with all $a^s_0=0$.
$K=\ker(A \to A_0)$ consists of all formal sums $\lambda_1x_1+\lambda_2x_2+\dots$.
If $(a^s_i)_{i,s}\in K \cdot \varprojlim_i (A_i^{\oplus S})$, or even $(a^s_i)_{i,s}\in K \cdot A^S$, then there is a finitely generated $A$-submodule of $K$ that contains all the $x_i$.
But every finitely generated submodule of $K$ is finite-dimensional.