TITLE: Why are there no filtered limits?
QUESTION [15 upvotes]: I just read the Wikipedia-article about filtered category, and now I wonder why it mentions filtered colimits and cofiltered limits, but not filtered limits. On the nLab-article titled filtered limit a filtered limit is even defined as a limit over a cofiltered category.
Is there a reason for this? The first thing that comes to my mind when I try to think of a limit over a filtered (but not cofiltered) category is a pullback. So apart from this simple case, are limits over filtered categories not very useful, or well-behaved, or whatever?

REPLY [10 votes]: If a category $C$ has a terminal object, then, as far as limits in $C$ are concerned, "filtered" would be useless information.  To see this, consider any functor into $C$, say $F:I\to C$.  Define $I^+$ to be the category obtained by adjoining a terminal object to $I$.  (If $I$ already has a terminal object, ignore it and adjoin a new one.)  So the objects of $I^+$ are those of $I$ plus the new terminal object $1$, and the morphisms of $I^+$ are those of $I$ and, for each object of $I^+$, a single morphism from that object to $1$.  Extend $F$ to a functor $F^+:I^+\to C$ by sending the new $1$ of $I^+$ to the terminal object of $C$ and defining $F$ on the new morphisms of $I^+$ in the only possible way.  Then, unless I've stupidly overlooked something, $I^+$ is filtered, and the limit of $F^+$ agrees with that of $F$.  In other words, any limit at all (like that of $F$) can be turned into a filtered limit (like that of $F^+$) by a trivial modification.