Does a long exact sequence of flat modules remain exact after tensoring with an arbitrary module? In Liu's Algebraic Geometry and Arithmetic Curves, Proposition 1.2.6 states that given any short exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ with $M''$ flat, taking the tensor product of this with any module $N$ gives an exact sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$.
Later in the book, in Exercise 5.2.8, page 193, he suggests using this Proposition to show that given a long exact sequence which is eventually $0$:
$$ 0 \rightarrow M_0 \rightarrow M_1 \rightarrow \cdots $$
of flat modules, taking the tensor product with an arbitrary module still gives an exact sequence 
$$ 0 \rightarrow M_0 \otimes N \rightarrow M_1 \otimes N \rightarrow \cdots$$
How does this generalization follow? It appears that in the proof of the Proposition, he doesn't really use the surjectivity of $M \rightarrow M''$, but then again, all he proves is that $M' \otimes N \rightarrow M'' \otimes N$ is injective; the full exactness being something that always happens for tensor products of short exact sequences.
 A: In fact, Liu says (on page 193, Exercise 2.8(b)) that the sequence is "zero from a finite rank on". Then you start the breaking from the rightmost side of the sequence: $0\to X_{n-1}\to M_{n-1}\to M_n\to 0$, and so on. You know that $M_{n-1}$ and $M_n$ are flat, so $X_{n-1}$ is also flat, and so on.  
A: If the base ring is Noetherian, then the answer is yes: 


*

*It is sufficient to show that all syzygies $Z^k := \text{ker}(M^k\to M^{k+1})$ of the given long exact sequence $$(\ddagger)\quad 0\to M^0\to M^1\to\ldots$$ are again flat, since then, as user1 pointed out, one can break the sequence into short exact sequences of flat modules and apply the exercise you referred to.

*It suffices to treat the case of a Noetherian local ring $R$, since flatness is a local property.

*Over a Noetherian local ring, any module of finite projective dimension has projective dimension at most $\text{depth}_RR$ by the Auslander-Bridger formula. In particular, the big finitistic projective dimension of $R$ is finite, and by a result of Jensen, any module of finite flat dimension has finite projective dimension. Hence $\text{fl.dim}_RM<\infty$ implies $\text{fl.dim}_RM\leq\text{depth}_RR$ for any $R$-module $M$.

*Viewing $0\to M^0\to M^1\to\ldots\to M^{k-1}\to Z^k\to 0$ as a flat resolution of $Z^k$ shows that $Z^k$ has finite flat dimension $\text{fl.dim}_R Z^k\leq k$, and 3. shows that even $\text{fl.dim}_R Z^k\leq\text{depth}_RR$. 

*Any interval $0\to Z^k\to M^k\to\ldots\to M^{k+l-1}\to Z^{k+l}\to 0$ can be viewed as a partial flat resolution of $Z^{k+l}$ of length $l$. Combining this with 4. allows to conclude that $Z^k$ is flat by choosing $l>\text{depth}_RR$.
Of course, if we know that $(\ddagger)$ is bounded, user26857's argument is the one to apply and does not need any Noetherianness hypothesis.
Note also the result is wrong if $M^{\ast}$ is unbounded to both sides: Take $R = k[x]/(x^2)$ and consider $$\ldots\to R\xrightarrow{\cdot x} R\xrightarrow{\cdot x} R\to \ldots$$ It is exact, but tensoring with $k := R/(x)$ yields the non-exact sequence $$\ldots\to k\xrightarrow{0} k\xrightarrow{0}\ldots$$
A: let $$0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow \cdots$$ be exact. 
then there is exact sequences (break):
$$0 \rightarrow M_0 \rightarrow M_1 \rightarrow C_1 \to 0$$ and $$0\to C_1 \rightarrow M_2 \rightarrow C_2 \to 0$$  and so on...  
now tensor with $N$ and join sequences again.
