Let $F$ be the functor $F:\mathcal{B}G \rightarrow Vect_{\mathbb{F}}$, where $Vect_{\mathbb{F}}$ is the category of vector spaces over a field $\mathbb{F}$. Prove that $\lim F \cong V^G$ is the fixed point subspace of the $G$-action.
I have proven in a different exercise that there is a one-to-one correspondence between functors from $\mathcal{B}G$ to $Vect_{\mathbb{F}}$ to vector spaces with a G-representation. Also, I know that for any functor $G: \mathcal{D} \rightarrow Set$ with $Obj(\mathcal{D})$ a set, then $\lim G$ exists and
$\lim G \cong \big\{(x_d \in G(d))_{d \in Obj(\mathcal{D}}):G(f)(x_d) = x_{d'}, \; \forall f \in \mathcal{D}(d,d')\big\}$
Although we are considering $Vect_{\mathbb{F}}$ instead of $Set$, it seems like this result can be applied somehow to prove the desired result. However, it may also be necessary to compute $\lim F$ by using the definition. I am stuck in that case as well.