I'll use the example of an equalizer.
When seen as a limit, we can say that:
- $eq$, $f \circ eq$, $g \circ eq$ is the cone
- $E$ is the apex
- $D$ is the base
And saying $(E, eq)$ is the limit implies that all other cones factor through it. In this example it is $(O,m)$ with the unique factorization $u$, $m = eq \circ u$.
However in the case of diagrams that only have one entry point, isn't this entry point always the actual limit?
Taking the above example: Looking at the cone we realize, that every arrow in the cone is expressed as a composition of $eq$, now if we remove this $eq$, we get a new cone $(X, 1_X)$:
- $1_X$, $f$, $g$ is the cone
- X is the apex
- D is the base
To prove the universal property, we can say $eq = 1_X \circ eq$, and $m = 1_X \circ m$. How come this is not always the actual limit.
#/media/File:Equalizer-01.svg){kind=link}
The problem is that $(X,1_X)$ is (probably) not a cone: that would mean that $f\circ 1_X=g\circ 1_X$ or $f=g$, which is probably not true. In the special case that $f=g$, then you are correct that $(X,1_X)$ would be a limit of the diagram. More generally, if a diagram has an "entry point" (an object that maps to every other object) with the additional property that any two parallel maps out of the entry point in the diagram are equal, then the entry point is a limit.