Why isn't $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[[x]]$? We just computed in class a few days ago that $$H^*(\mathbb{R}P^n,\mathbb{F}_2)\cong\mathbb{F}_2[x]/(x^{n+1}),$$ and it was mentioned that $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[x]$, but I  (optimistically?) assumed that the cohomology ring functor would turn limits into colimits, and so $$\mathbb{R}P^\infty=\lim\limits_{\longrightarrow}\;\mathbb{R}P^n$$ would mean that we'd get the formal power series ring $$\lim\limits_{\longleftarrow}\;\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]].$$ 
My professor said that the reason we get $\mathbb{F}_2[x]$ is just that the cohomology ring, being the direct sum of the cohomology groups, can't have non-zero elements in every degree, which certainly makes sense. 
Even though that should settle the matter, for some reason, I'm still having a bit of trouble making this make click for me. Is the explanation just that the cohomology ring functor doesn't act as nicely as I'd hoped? Is there an intuitive explanation of what's going on?
 A: This is probably something that should be a comment, mainly because I'm not sure it is correct. Feel free to tell me it is rubbish! 
1) I have definitely seen it written $H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$. For example see Jacob Lurie's notes here. 
2) Cohomlogy doesn't play well with inverse limits in general. There is the Milnor exact sequence, which is this case should give:
$$0 \to \text{lim}^1 H^{\ast-1}(\mathbb{R} P^\infty;\mathbb{F}_2) \to H^*(\mathbb{R} P^\infty;\mathbb{F}_2) \to \underset{n}{\text{lim}} H^*(\mathbb{R}P^\infty;\mathbb{F}_2) \to 0$$
where $\text{lim}^1$ is the first dervied functor of the inverse limit. 
In this case the $\text{lim}^1$ terms should all be zero by the Mittag-Leffler criteria (which follows since the maps $H^*(\mathbb{R} P^{n+1};\mathbb{F}_2)\to H^*(\mathbb{R} P^{n};\mathbb{F}_2)$ are surjective). Thus we can conclude that $$H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$$
As an aside this is the calculation usually given to show $E^*(\mathbb{C} P^\infty) \simeq E_*[\![ x ]\!]$ for any complex oriented cohomology theory $E$, which basically sets up the whole relationship between complex oriented cohomology theories and formal group laws. 
Edit: Please see the comments by Mariano below. 
Possibly also this Math Overflow link is relevant
