More precisely, if $\mathcal F\_i$ is a system of sheaves, is it the case that 
$$
(\lim \mathcal F\_i)\_p = \lim ((\mathcal F\_i)\_p)
$$ 
and similarly for colimits? I can see how to get a map 
$$
(\lim \mathcal F\_i)\_p \rightarrow \lim ((\mathcal F\_i)\_p)
$$
by taking the stalks in the diagram for $\lim \mathcal F\_i$ and then using the universal property of $(\lim \mathcal F\_i)\_p$, but I can't see how I should use the sheaf condition to go the other way.


I learning algebraic geometry and taking a class with Hartshorne's book, and we showed that a map of sheaves is injective/surjective iff it is injective/surjective on stalks. Since one can check injectivity/surjectivity by computing the kernal/cokernal, I am thinking that what I wrote above would be a generalization of this fact. 


Looking at sheafication as a adjoint and how it interacts with limits and colimits really made me feel more comfortable with sheafication, so I'm hoping that looking the interaction between sheaves and their stalks in terms of limits and colimits will give me some intuition. 



Is the stalk of the (co)limit of sheaves equal to the (co)limit of the stalks?
Let $F$ be a sheaf on $X$ and $p \in X$. Then $F\_p$ is just the pullback $i^{-1} F$, where $i : \{p\} \to X$ is the inclusion of a point. Now $i^{-1}$ is left adjoint to $i\_\*$, thus cocontinuous, i.e. preserves all colimits. This shows that the canonical morphism $\mathrm{colim}\_i(F\_p) \to \mathrm{colim}\_i(F)\_p$ is an isomorphism.


Now for limits, we get a canonical morphism $(\mathrm{lim}\_i F)\_p \to \mathrm{lim}\_i(F\_p)$. This is almost never an isomorphism (neither injective nor surjective). Consider infinite products and see what happens. However, the (left) exactness of $F \mapsto F\_p$ means that this functor preserves *finite* limits.


