Given a smooth manifold $M$ and a *relatively compact exhaustion* $M=\bigcup_{n\in\mathbb N} M_n$ with open and relatively compact $M_n\subseteq M_{n+1}$ (hence $M=\lim\limits_{\to}M_n$) do we always have $$H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)?$$

This looks so natural that the answer should be known in which case I would like to get a reference.

EDIT (inspired by Weibel's book and an article of Milnor from 1962). For fixed $k$ the projective spectrum $\Omega^k(M_n)$ of $k$-forms on $M_n$ satisfies the (strict) Mittag-Leffler condition so that the derived functor ${\lim\limits_{\leftarrow}}^1 \Omega^k(M_n)$ vanishes. We thus get an exact sequence $0\to \Omega^\ast(M)\to \prod_n \Omega^\ast(M_n)\to \prod_n \Omega^\ast(M_n)\to 0$ of cochain complexes (the last map is the difference map $(\omega_n)_n\mapsto (\omega_n-\rho_{n+1}(\omega_{n+1}))_n$ with the restriction operator $\rho_{n+1}$). The long cohomology sequence thus yields an exact sequence $$\cdots \to H^k(M)\to \prod_nH^k(M_n) \to \prod_nH^k(M_n) \to H^{k+1}(M)\to \cdots.$$ From this one gets short exact sequences $$0\to {\lim\limits_{\leftarrow}}^1 H^{k-1}(M_n) \to H^k(M)\to \lim\limits_{\leftarrow}H^k(M_n)\to 0.$$ It seems to me that the spectrum $H^0(M_n)$ of the spaces of locally constant functions again satisfies the strict Mittag-Leffler condition so that ${\lim\limits_{\leftarrow}}^1H^0(M_n)=0$. This means that $H^1(M)=\lim\limits_{\leftarrow} H^1(M_n)$ is always true. Another consequence is $H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)$ if all cohomologies $H^{k-1}(M_n)$ are finite dimensional (because this implies a Mittag-Leffler condition).

However, in general the question remains open (and I would rather expect a counterexample).