A functor $G:\mathcal{C}\rightarrow\mathcal{D}$ is final when the following conditions are satisfied for every category $\mathcal{A}$ and every functor $F:\mathcal{D}\rightarrow\mathcal{A}$:
- if the limit $(L,(p_D)_{D\in\mathcal{D}})$ of $F$ exists, then $(L,(p_{G(C)})_{C\in\mathcal{C}})$ is the limit of $F\circ G$;
- if the limit $(L,(q_C)_{C\in\mathcal{C}})$ of $F\circ G$ exists, then the limit $F$ exists as well.
Observe that in condition (2), applying condition (1) implies immediately that the limit of $F$ has the form $(L,(p_D)_{D\in\mathcal{D}})$ with $p_{G(C)}=q_C$ for $C\in\mathcal{C}$.
I don't see why applying (1) to (2) implies that $p_{G(C)}=q_C$ for $C\in\mathcal{C}$.
Let $(L,(q_C)_{C\in\mathcal{C}})$ be the limit of $F\circ G$. By (2), there exists a limit $(L,(p_D)_{D\in\mathcal{D}})$ of $F$. By (1), $(L,(p_{G(C)})_{C\in\mathcal{C}})$ is the limit of $F\circ G$. Thus there exists a unique isomorphism $r:K\rightarrow L$ such that $p_{G(C)}=q_C\circ r$ for $C\in\mathcal{C}$. But this doesn't show that $p_{G(C)}=q_C$. Abuse of notation, maybe?