Let $\mathcal{B},\mathcal{C}$ be finitely complete categories, $F:\mathcal{B}\rightarrow\mathcal{C}$ a functor that reflects finite limits and $\mathcal{D}$ a finitely generated category with $(f_i)_{1\leq i\leq n}$ generating its set of morphisms.
Let $G:\mathcal{D}\rightarrow\mathcal{B}$ be a functor. Suppose $(M,(q_D)_{D\in\mathcal{D}})$ be a cone over $G$ in $\mathcal{B}$ such that $(F(M),(F(q_D))_{D\in\mathcal{D}})$ is a limit of $F\circ G$ in $\mathcal{C}$. I have to show that $(M,(q_D)_{D\in\mathcal{D}})$ is a limit of $G$ in $\mathcal{B}$.
First of all, since $\mathcal{D}$ is finitely generated and $\mathcal{C}$ is finitely complete, there exists a limit $(L,(p'_D)_{D\in\mathcal{D}})$ of $G$ in $\mathcal{B}$. Thus it suffices to show that $M\cong L$. Note that $(F(L),(F(p'_D))_{D\in\mathcal{D}})$ is a cone over $F\circ G$. Hence there exists a unique morphism $m:F(L)\rightarrow F(M)$ in $\mathcal{C}$ such that $$F(p_D)=F(q_D)\circ m$$ for $D\in\mathcal{D}$.
I'm not sure how to go on from this point. I still haven't used the fact that $F$ reflects finite limits: what I seem to have here are finitely generated limits not finite limits, so I'm not sure how to utilize this assumption.
Any suggestions?
Edit:
There exists a morphism $u:F(M)\rightarrow\prod_{D\in\mathcal{D}} F(G(D))$ in $\mathcal{C}$ such that $(F(M),u)$ is the equalizer of the unique morphisms $\alpha',\beta':\prod_{D\in\mathcal{D}} F(G(D))\rightarrow\prod_{i=1}^nF(G(t(f_i)))$.
Similarly, there exists a morphism $l:L\rightarrow\prod_{D\in\mathcal{D}} G(D)$ in $\mathcal{B}$ such that $(L,l)$ is the equalizer of the unique morphisms $\alpha,\beta:\prod_{D\in\mathcal{D}} G(D)\rightarrow\prod_{i=1}^nG(t(f_i))$. It follows that $\alpha'=F(\alpha)$ and $\beta'=F(\beta)$.
Furthermore, there exists a unique morphism $r:M\rightarrow\prod_{D\in\mathcal{D}}G(D)$ such that $q_D=p'_D\circ r$ for $D\in\mathcal{D}$. How to show that $F(r)=u$?