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In the title, $\operatorname{id}$ is a functor. Or $\operatorname{id}(C)$, to explicitely write a category $C$. So, a limit of all objects and morphisms in $C$. I am in doubt, is it proper to take such a big limit? Below I explain that $C$ in the motivating example is the category of algebras which usually is infinite. The question resides above, but if you can tell me something about the motivating example, you are welcome.

I found a proof that if such a limit exists, it is an initial object in $C$. The apex of the limit is an initial object and the morphisms of the limit are catamorphisms (this is just a special name for that unique morphism that goes from any initial object). This occurs in programming. $C$ is the category of algebras, the limit $L$ of $\operatorname{id}(C)$ is a set (?) where every element is a function that picks for every algebra an element of the carrier of that algebra. Intuitively, the function folds hidden something by a given algebra. Sometimes it's desirable to keep this function rather then an element of some initial algebra $0$. We go from any element $x$ of $0$ to the element of $L$ by feeding $x$ to morphisms of $L$. We go from any element $y$ of $L$ to the element of $0$ by picking the component of $y$ that corresponds to $0$.

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It seems to me you've answered your own question. The limit of the identity functor on $C$ exists exactly when $C$ has an initial object, in which case the limit is an initial object. –  Charles Rezk Oct 15 '11 at 18:55
Although most of the space is taken up by discussion of a particular example that the poster has solved, it appears that this question is really about how to set up foundations so that limits over arbitrarily large diagrams make sense. If I were asking, I would put more emphasis on the point of doubt instead of the example, but it seems to be a reasonable question. –  S. Carnahan Oct 16 '11 at 0:57
@S. Carnahan: Yes, you are right. Usually I ignore foundations and call everything a “set.” $C$ is the category of algebras, so it is usually infinite. The second paragraph gives additional info (perhaps it will be relevant) and motivation. I do not like questions without motivation. –  beroal Oct 17 '11 at 14:04
@Charles Rezk: You say that if $C$ has an initial object $0$, then $0$ is a limit of $\operatorname{id}(C)$? I could not prove this, can you hint me? –  beroal Oct 17 '11 at 14:06
One can define limits over large diagrams and check that $\mathrm{id}_C$ has a limit iff there exist initial objects without any foundational issues at all, no? –  Mariano Suárez-Alvarez Oct 17 '11 at 18:51

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