Non-vanishing $\mathrm{lim}^1$-term for the cohomology of a CW-complex Let $h$ be an additive cohomology theory. If we want to compute $h^*(X)$ for an infinite CW-complex $X$, a standard method is to use the Milnor sequence
$$ 0 \to \mathrm{lim}^1_k h^{n-1}(X^{(k)}) \to h^n(X) \to \mathrm{lim}_k h^n(X^{(k)}) \to 0, $$
where $X^{(k)}$ is the $k$-skeleton of $X$. If $h$ is singular cohomology, then the $\mathrm{lim}^1$-term always vanishes. 
What are examples of CW-complexes $X$ and cohomology theories $h$, where this $\mathrm{lim}^1$-term does not vanish?
 A: Take $X$ to be the Moore space for the group $\mathbb Z[\frac1p]$ in dimension $n$ (realized by a telescope of $n$-spheres mapping to each other vie the times-$p$ map), and take $h^*$ to be ordinary cohomology with coefficients in $\mathbb Z$.
The universal coefficient theorem gives you a sort exact sequence
$$
0\longrightarrow \mathrm{Ext}(H_n(X,\mathbb Z),\mathbb Z) \longrightarrow H^{n+1}(X,\mathbb Z)
\longrightarrow \mathrm{Hom}(H_{n+1}(X,\mathbb Z),\mathbb Z)\longrightarrow 0.
$$
The last term is zero, and so we get
$$H^{n+1}(X,\mathbb Z)=\mathrm{Ext}(H_n(X,\mathbb Z),\mathbb Z)=\mathrm{Ext}(\mathbb Z[\tfrac1p],\mathbb Z)=\mathbb Z_p/\mathbb Z,
$$
the quotient of the $p$-adics by the integers.
That same computation can be done using the Milnor sequence.
The limit of
$$
\mathbb Z\stackrel p \leftarrow\mathbb Z\stackrel p \leftarrow\mathbb Z\stackrel p \leftarrow\ldots
$$
is zero, but the $lim^1$ of that inverse system is not zero. It's again $\mathbb Z_p/\mathbb Z$.
A: For our CW-complex I'm going to take $X = \Bbb{CP}^\infty$ (as a based space), whose skeleta are $\Bbb{CP}^n$.  The cohomology theory will be more difficult to construct.
For any $k \geq 1$, let $E_k$ be the spectrum which is the homotopy fiber of the map
$$
Sq^{2^k} \cdots Sq^8 Sq^4 Sq^2: \Sigma^2 H\Bbb{Z}/2 \to \Sigma^{2^{k+1}} H\Bbb{Z}/2
$$
so that, for any space $Y$, we have a long exact sequence in cohomology with at least the terms
$$
\cdots \to H^1(Y) \to H^{2^{k+1}-1}(Y) \to (E_k)^0(Y) \to H^2(Y) \stackrel{Sq^{2^k} \cdots Sq^2}{\longrightarrow} H^{2^{k+1}}(Y) \to \cdots
$$
(The cohomology operation is actually taking the $(2^k)$'th power of the elements in degree 2.) In particular, we find that:
$$
(E_k)^0(\Bbb{CP}^n) = \begin{cases}
\Bbb{Z}/2 &\text{if }n < 2^{k}\\
0 &\text{if }n \geq 2^{k}\\
\end{cases}
$$
Now I'll put all these cohomology theories together by taking $E = \bigvee E_k$, so that $E^0(Y) = \bigoplus (E_k)^0(Y)$ for $Y$ a finite CW-complex.
We then find that
$$
E^0(\Bbb{CP}^n) = \bigoplus_{n < 2^k} \Bbb{Z}/2.
$$
This is a decreasing family of abelian groups with a nonzero $lim^1$-term.  To prove that, you can use the short exact sequence of inverse systems
$$
0 \to \bigoplus_{n < 2^k} \Bbb{Z}/2 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \bigoplus_{n \geq 2^k} \Bbb{Z}/2 \to 0,
$$
which gives a nontrivial $lim^1$-sequence
$$
0 \to 0 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \prod_{k \geq 1} \Bbb{Z}/2 \to lim^1 \to 0.
$$
This illustrates the basic issue: we can explicitly build our cohomology theory so that no cohomology classes survive to the whole of $\Bbb{CP}^\infty$, but it can take arbitrarily long to figure it out.  You can imagine how to build more general examples than this, but this one is convenient to calculate.
