Inverse limit of modules and tensor product Let $(M_n)_n$ be an inverse system of finitely generated modules over a commutative ring $A$ and $I\subset A$ an ideal.
When is the canonical homomorphism
$$\left(\varprojlim\nolimits_n M_n\right)\otimes_A A/I \rightarrow \varprojlim\nolimits_n \left(M_n \otimes_A A/I\right)$$
an isomorphism?
What does one need? E.g. all $M_n$ flat over $A$ or special conditions about $A$ and $I$?
 A: Let us show that the canonical morphism
$$
\left(\varprojlim\nolimits_n M_n\right)\otimes_A A/I \rightarrow \varprojlim\nolimits_n \left(M_n \otimes_A A/I\right)
$$
in the question is not always an isomorphism. (This fact was implicit in the question and in the accepted answer, but, as far as I can see, it was not proved.)
As in this answer, let $K$ be a field and $A$ the commutative $K$-algebra with one generated by the symbols $a$ and $b$ subject to the relations $ab^2=ab$ and $a^2=0$. 
Set $M_n:=(b^n)$ and $I:=(a)$. 
The above morphism becomes the obvious morphism 
$$
\left(\bigcap_n\ (b^n)\right)\Bigg/a\left(\bigcap_n\ (b^n)\right)\to\varprojlim\nolimits_n ((b^n)/a(b^n)).
$$ 
But we have 
$$
\bigcap_n\ (b^n)=(ab),\quad(ab)/a(ab)=(ab)/(0)\simeq(ab)\ne(0),
$$ 
and, for $n>0$, 
$$
(b^n)/a(b^n)=(b^n)/(b^n)=0,
$$ 
so that we get the morphism 
$$
0\ne(ab)\to0.
$$
A: It's not true in general that tensor product commute with projective limits.
E.g. consider $\mathbb Z_p := \projlim_n \mathbb Z/p^n.$
We have that $\mathbb Z_p \otimes_{\mathbb Z} \mathbb Q$ is non-zero;
it is the field $\mathbb Q_p$.  
On the other hand $\mathbb Z/p^n \otimes_{\mathbb Z} \mathbb Q = 0$ for each value of $n$.  
On the other hand, suppose that the modules $M_n$ are of finite length, and that $N$ is finitely presented.  Then $(\varprojlim_n M_n)\otimes_A N \to \varprojlim_{n} M_n\otimes N$ is an isomorphism.
To see this, choose a finite presentation $A^r \to A^s \to N \to 0$ of $N$.
Then we have to show that the cokernel of 
$\varprojlim_n M_n^r \to \varprojlim_n M_n^s$ is isomorphic to the projective limit of the cokernels of the maps $M_n^r \to M_n^s$.  This follows from the finite length assumption, which shows (applying Mittag--Leffler) that the projective limit of the cokernels is indeed the cokernel of the projective limits.  

Now suppose that $I$ is finitely generated (e.g. assume $A$ is Noetherian).  Then $A/I$ is finitely presented, and so if the $M_n$ are furthermore of finite length, the natural map you ask about is an isomorphism.
A: I would like to suggest another proof inspired by marlu's proof in direct product commutes with tensor product?. In the case of a surjective system $\{M_i\}$ and a flat finitely presented $A$-module $N$, we can relax the requirement that the members $M_i$ in the inverse system have to be of finite length in order for $(-)\otimes_AN$ to commute with the inverse limit. Alternatively, if we assume that $\{M_i\}$ is a surjective system of flat $A$-modules, then we can relax the flatness condition on $N$, as well. Then it suffices if $N$ is finitely presented over $A$. In the following, I consider both situations separately (Claim 1 and Claim 2).
Claim 1: Let A be a commutative ring. Let $\{M_i\}$ be a surjective system of $A$-modules. Let further $N$ be a finitely presented and flat $A$-module. Then, there is an isomorphism of $A$-modules
$$(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN).$$
Proof: Since $N$ is finitely presented, by definition, there are integers $m, n\geq1$  together with two surjective maps of $A$-modules
$$\varphi: A^n\twoheadrightarrow N$$
and
$$\psi:A^m\twoheadrightarrow\ker\varphi.$$
Both maps give rise to the short exact sequence of $A$-modules
$$0\to A^m/\ker\psi\hookrightarrow A^n\twoheadrightarrow N\to0.~~~(1)$$
Since $A^n$ is free, hence flat over $A$ and $N$ is by assumption flat over $A$, the $A$-module $\ker\varphi\cong A^m/\ker\psi$ is flat over $A$, too. This means that $$\mathrm{Tor}^k(N, M)=\mathrm{Tor}^k(A^n, M)=\mathrm{Tor}^k(\ker\varphi, M)=0$$ for every $k\geq1$ and every $A$-module $M$. Hence, from the $\mathrm{Tor}$-long exact sequence, we deduce that for each $i$, there is a short exact sequence of $A$-modules
$$0\to M_i\otimes_AA^m/\ker\psi\to M_i\otimes_AA^n\to M_i\otimes_AN\to0.$$
The right exactness of the functor $(-)\otimes_AA^m/\ker\psi$ implies that $\{M_i\otimes_AA^m/\ker\psi\}$ is a surjective system, too. Hence, by Proposition $10.2$ in Atiyah-MacDonald's book Introduction to Commutative Algebra, the sequence of $A$-modules
$$0\to\lim_{\leftarrow i}(M_i\otimes_A\frac{A^m}{\ker\psi})\to\lim_{\leftarrow i}(M_i\otimes_AA^n)\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0~~~(2)$$
is exact, too. For every $i$, there is a canonical isomorphism of quotient $A$-modules:
$$\frac{M_i}{\{0\}}\otimes_A\frac{A^m}{\ker\psi}\cong \frac{M_i^{\oplus m}}{M_i\otimes\ker\psi}.$$
As the inverse system $\{M_i\otimes_A\ker\psi\}$ is surjective, Proposition $10.2$ in Atiyah-MacDonald’s book implies the following isomorphism:
$$\lim_{\leftarrow i}(\frac{M_i^{\oplus m}}{M_i\otimes\ker\psi})\cong\frac{\lim_{\leftarrow  i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}.$$
Thus, we can write short exact sequence $(2)$ in the form
$$0\to\frac{\lim_{\leftarrow  i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}\to\lim_{\leftarrow i}(M_i^{\oplus n})\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0.~~~(3)$$
Applying the right exact functor $\lim_{\leftarrow i}(M_i)\otimes_A(-)$ on the short exact sequence $(1)$ yields the exact sequence
$$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{(\lim_{\leftarrow i}M_i)\otimes_A\ker\psi} \to(\lim_{\leftarrow i}M_i)^{\oplus n}\to (\lim_{\leftarrow i}M_i)\otimes_AN\to0~~~(4)$$
The canonical map
$$(\lim_{\leftarrow i}M_i)^{\oplus m}\to\lim_{\leftarrow i}(M_i^{\oplus m})~~~(5)$$
is an isomorphism by the fact that finite direct sums are limits and limits always commute with other limits. At the same time, that isomorphism maps the subspace $\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi$ to $\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)$. Hence, the canonical map $(5)$ descends to a surjective map
$$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi}\twoheadrightarrow\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}~~~(6)$$
Now, both exact sequences $(3)$ and $(4)$ fit in the following commutative diagram
$\require{AMScd}$
\begin{CD}
    \frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi} @>>> (\lim_{\leftarrow i}M_i)^{\oplus n} @>>> (\lim_{\leftarrow i}M_i)\otimes_AN@>>>0@>>>0 \\
    @VVV @VVV @VVV @VVV @VVV \\
    \frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)} @>>> \lim_{\leftarrow i}(M_i^{\oplus n}) @>>> \lim_{\leftarrow i}(M_i\otimes_AN)@>>>0@>>>0.
\end{CD}
The top and bottom rows are exact. The vertical arrows are the canonical maps. The first vertical arrow is the surjective map $(6)$. The second one is an isomorphisms, as already explained, by the fact that inverse limits commute with finite direct sums. The last two morphsims $0\to 0$ are trivially isomorphisms. Then, a direct application of the  Five Lemma implies that $(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN)$.  QED

Claim 2: Let A be a commutative ring. Let $\{M_i\}$ be a surjective system of flat $A$-modules. Let further $N$ be a finitely presented  $A$-module. Then, there is an isomorphism of $A$-modules
$$(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN).$$
Proof: Since $N$ is finitely presented, by definition, there are integers $m, n\geq1$  together with two surjective maps of $A$-modules
$$\varphi: A^n\twoheadrightarrow N$$
and
$$\psi:A^m\twoheadrightarrow\ker\varphi.$$
Both maps give rise to the short exact sequence of $A$-modules
$$0\to A^m/\ker\psi\hookrightarrow A^n\twoheadrightarrow N\to0.~~~(1)$$
Since by definition, every $A$-module $M_i$ is flat, for each $i$, there is a short exact sequence of $A$-modules
$$0\to M_i\otimes_AA^m/\ker\psi\to M_i\otimes_AA^n\to M_i\otimes_AN\to0.$$
The right exactness of the functor $(-)\otimes_AA^m/\ker\psi$ implies that $\{M_i\otimes_AA^m/\ker\psi\}$ is a surjective system, too. Hence, by Proposition $10.2$ in Atiyah-MacDonald's book Introduction to Commutative Algebra, the sequence of $A$-modules
$$0\to\lim_{\leftarrow i}(M_i\otimes_A\frac{A^m}{\ker\psi})\to\lim_{\leftarrow i}(M_i\otimes_AA^n)\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0~~~(2)$$
is exact, too. For every $i$, there is a canonical isomorphism of quotient $A$-modules:
$$\frac{M_i}{\{0\}}\otimes_A\frac{A^m}{\ker\psi}\cong \frac{M_i^{\oplus m}}{M_i\otimes\ker\psi}.$$
As the inverse system $\{M_i\otimes_A\ker\psi\}$ is surjective, Proposition $10.2$ in Atiyah-MacDonald’s book implies the following isomorphism:
$$\lim_{\leftarrow i}(\frac{M_i^{\oplus m}}{M_i\otimes\ker\psi})\cong\frac{\lim_{\leftarrow  i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}.$$
Thus, we can write short exact sequence $(2)$ in the form
$$0\to\frac{\lim_{\leftarrow  i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}\to\lim_{\leftarrow i}(M_i^{\oplus n})\to\lim_{\leftarrow i}(M_i\otimes_AN)\to0.~~~(3)$$
Applying the right exact functor $\lim_{\leftarrow i}(M_i)\otimes_A(-)$ on the short exact sequence $(1)$ yields the exact sequence
$$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{(\lim_{\leftarrow i}M_i)\otimes_A\ker\psi} \to(\lim_{\leftarrow i}M_i)^{\oplus n}\to (\lim_{\leftarrow i}M_i)\otimes_AN\to0~~~(4)$$
The canonical map
$$(\lim_{\leftarrow i}M_i)^{\oplus m}\to\lim_{\leftarrow i}(M_i^{\oplus m})~~~(5)$$
is an isomorphism by the fact that finite direct sums are limits and limits always commute with other limits. At the same time, that isomorphism maps the subspace $\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi$ to $\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)$. Hence, the canonical map $(5)$ descends to a surjective map
$$\frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi}\twoheadrightarrow\frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)}~~~(6)$$
Now, both exact sequences $(3)$ and $(4)$ fit in the following commutative diagram
$\require{AMScd}$
\begin{CD}
    \frac{(\lim_{\leftarrow i}M_i)^{\oplus m}}{\lim_{\leftarrow i}(M_i)\otimes_A\ker\psi} @>>> (\lim_{\leftarrow i}M_i)^{\oplus n} @>>> (\lim_{\leftarrow i}M_i)\otimes_AN@>>>0@>>>0 \\
    @VVV @VVV @VVV @VVV @VVV \\
    \frac{\lim_{\leftarrow i}(M_i^{\oplus m})}{\lim_{\leftarrow i}(M_i\otimes_A\ker\psi)} @>>> \lim_{\leftarrow i}(M_i^{\oplus n}) @>>> \lim_{\leftarrow i}(M_i\otimes_AN)@>>>0@>>>0.
\end{CD}
The top and bottom rows are exact. The vertical arrows are the canonical maps. The first vertical arrow is the surjective map $(6)$. The second one is an isomorphisms, as already explained, by the fact that inverse limits commute with finite direct sums. The last two morphsims $0\to 0$ are trivially isomorphisms. Then, a direct application of the  Five Lemma implies that $(\lim_{\leftarrow i}M_i)\otimes_A N\cong\lim_{\leftarrow i}(M_i\otimes_AN)$.  QED
