Let $A$ be a ring (commutative and noetherian if it helps).
Suppose we are given an inverse system $M\_i$ of complexes of $A$-modules (where $i$ is a natural number),
and integers $a<b$
such that for each $i$, the complex $M\_i$ has non-zero cohomologies only in degrees $a<j<b$.


Consider the complex $\varprojlim M\_i$. Does this complex have bounded cohomology?


Note that I am not assuming that this system satisfy a Mittag-Leffler condition.



Does the inverse limit of complexes with bounded cohomology have a bounded cohomology?
No, even for $A=\mathbf{Z}$.


Take your favourite example of a complex of short exact sequences of abelian groups whose projective limit is not exact (see for example [this math.stackexchange question](https://math.stackexchange.com/questions/1153414/example-that-inverse-limit-is-not-exact)).


Now just make a complex of long exact sequences by splicing together the short exact sequences a la


$$\to0\to A\to B\to C\to 0\to 0\to A\to B\to C\to 0\to\cdots$$


and you get failure of exactness of the projective limit all over the place, even though all the sequences in the complex are exact.


