So I am given $n$ points in the plane and I want to create a planar graph such that every face is bounded by $3$ edges (even the "outside" face). Hence if I have say $4$ vertices, then there is a unique way to triangulize it, make all edges adjacent to each other.
Let $L_n$ denote the number of triangulations on $n$ points. I was thinking a way to count $L_n$. To go from $L_4$ to $L_5$ we have an extra vertice that is placed in one of the faces, then in order to triangulize it we can match that point with the vertices of the triangule that points lives in. Since there are $4$ ways of picking which face to place the new vertex we have $L_5=4$. Note that the number of faces goes up by $2$ since we are destroying one face and adding $3$ new ones. Hence $L_6=6L_5$, and in general I would have that $L_n=2^{n-4}(n-3)!$.
I was wondering what is wrong with the above argument since my homework says "Prove $\lim \log L_n$ as $n\rightarrow \infty$ exists" (Since this is homework please only give me hints towards the correct solution, thanks)