Consider a complex-oriented multiplicative generalized cohomology theory $h^{*}(X)$. It is complex-oriented, if by the definition the following two conditions hold:

1) There exists an element $t\in h^{2}(\mathbb{C}P^{\infty})$, restricting (the abuse of notation) to the canonical generator $t$ of $h^{2}(\mathbb{C}P^{1})$. (Canonical means that we have $\widetilde{h}^{2}(S^{2}) \cong \widetilde{h}^{2}(\mathbb{C}P^{1}) \cong h^{0} $ from the suspension map $\Sigma^{2}:\ \lbrace x, y \rbrace \rightarrow S^{2}$, hence $h^{2}(\mathbb{C}P^{1})$ is free $h^{*}$ - module of rank $1$).

2) $t$ goes to $-t$ under the involution of $\mathbb{C}P^{1}$.

Using the Atiyah-Hirzebruch spectral sequence it isn't difficult to show that $$h^{*}(\mathbb{C}P^{n}) \cong h^{*}[t]/(\ t^{n+1}).$$ I want to deduce $$h^{*}(\mathbb{C}P^{\infty}) \cong h^{*}[[ t]].$$

The article below at p.3 tells me to use the Milnor exact sequence to make it: http://www.math.rochester.edu/people/faculty/doug/otherpapers/coctalos.pdf

$$0 → lim^{1}\ h^{∗−1}(\mathbb{C}P^{n}) → h^{∗}(\mathbb{C}P^{∞}) → lim\ h^{∗}(\mathbb{C}P^{n}) → 0.$$

$lim^{1} A_{i}$ is defined for an inverse system of groups $\lbrace A_{i},\ f_{i}\rbrace$ as follows. Let $\Delta:\ \prod\limits_{i=1}^{\infty} A_{i}\rightarrow \prod\limits_{i=1}^{\infty}A_{i}$ be defined by $\Delta(a_{i}) = a_{i} - f_{i+1}(a_{i+1}).$ Then $$lim^{1} A_{i} = \prod\limits_{i=1}^{\infty} A_{i}\Big/ \lbrace x_{i}\rbrace\sim *\\ \text{iff}\qquad \exists\lbrace a_{i}\rbrace:\ \lbrace x_{i}\rbrace = \Delta(\lbrace a_{i}\rbrace)$$

I haven't found any refences to the expalanation of the existence of that sequence. My questions are:

1) What is the Milnor exact sequence?

2) Why does the first part of the sequence vanish?

3) Are there any examples unveiling the meaning of $lim^{1}$?