I have recently become comfortable with the definition of the inverse image sheaf in algebraic geometry. In particular, given ringed spaces $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ and a morhpism $f: (X, \mathcal{O}_{X}) \longrightarrow (Y, \mathcal{O}_{Y})$, then we can use the sheaf structure on $Y$ to define one on $X$. In particular, for $U \subseteq X$ open, we define the presheaf to be $$U \mapsto\varinjlim G(V)$$ where the colimit is taken over $V \subset Y$ such that $f(U) \subseteq V$. Intuitively, I interpret this as "approximating" $f(U)$ by smaller and smaller open sets, and the universal property of the colimit says that we want the smallest open set such that it's not smaller than $f(U)$. The drawback of this is that you need to sheafify the result, since sheafification is not preserved by colimits.
My question is, why not do the opposite? Why not take a limit (instead of a colimit) over $V \subset Y$ such that $V \subset f(U)$. That way you wouldn't need to sheafify since, by abstract nonsense, limits preserve sheafification. Is there a reason it is defined the first way rather than the second? Have I fundamentally misunderstood something important? I ask because I noticed the way you can use a limit to recover a sheaf structure on a space once you have it defined on a base of open sets.
I tagged this as a soft question too since I am not sure it has a definite answer.
Thanks
The section over $U$ of the inverse image sheaf can be tought of as pull backs of sections defined in some nbd of the image of $U$. Two such pullbacks are equal if they agree in a nbd of the image of $U$; thus we take the colimit.
How would you interpret the limit?
Moreover, usually there are no open subsets contained in $f(U)$ other then the empty set.