Is an equalizer really a special case of a limit? Let $I$ be a category with $2$ objects $A,B$ and four morphisms $1_A, 1_B, u,v$ where $u,v \in Hom_I(u,v)$. Define a functor $F: I \to \mathcal{C}$ by 
$$F(A) = X, F(B) = Y, F(u) = f, F(v) = g$$
My notes claim that a limit of the functor $F$ is an equalizer of the pair of morphisms $f,g: X \to Y$. I have trouble figuring out why this is the case.

A cone on the functor $F$ is formally a pair $(C, \{p_1: C \to X, p_2: C \to Y\})$ such that $f \circ p_1 = p_2$ and $g \circ p_1 = p_2$ commute.
A limit is a special case of a cone.
An equalizer is formally a pair $(C,p: C \to Y)$
Both "datatypes" are not compatible, so how can an equalizer be a special case of a limit if they are not the same datatype?
It seems to me rather that if $(C, \{p_1: C \to X, p_2: C \to Y\})$ is a limit of $F$, then $(C,p_1)$ is an equalizer of the morphisms $f,g$ (and conversely). Is that correct? Any insight will be appreciated!

To your cone $(C,p_1,p_2)$ you can associate the pair $(C,p_1)$, which is the limit. The morphism $p_2$ is just extra data which is useless since it is determined by $p_1$.
Conversely, starting from a pair $(C,p)$ you can associate a cone $(C,p,p_2)$ with $p_2=f\circ p_1=g\circ p_1$.
The pair is an equalizer iff the cone is a limit.
