Why $\lim_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n-1)} =\frac{9}{8}$? $$F(m,n)= \begin{cases}
1, & \text{if $m n=0$ }; \\
 \frac{1}{2} F(m ,n-1) + \frac{1}{3} F(m-1,n )+ \frac{1}{4} F(m-1,n-1), & \text{ if $m n>0$. }%
\end{cases}$$
Please a proof of:
$$\lim_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n-1)} =\lim_{n\rightarrow \infty}\frac{F(n,n-1)}{F(n-1,n-1)}=\frac{9}{8}$$
$$\lim_{n\rightarrow \infty}\frac{F(n-1,n)}{F(n-1,n-1)}=1$$

We will compute the generating function, and use the method described in section 2 of this paper.
Let $F_{m,n}=F(m,n)$. Consider the generating function
$$G(x,y)=\sum_{m=0}^\infty\sum_{n=0}^\infty F_{m,n}x^my^n.$$
Then the recurrence gives
\begin{align*}
&G(x,y)=\sum_{m=0}^\infty F_{m,0}x^m+\sum_{n=1}^\infty F_{0,n}y^n+\sum_{m=1}^\infty\sum_{n=1}^\infty F_{m,n}x^my^n\\
&=\frac{1}{1-x}+\frac{y}{1-y}+\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\frac{1}{2}F_{m,n-1}+\frac{1}{3}F_{m-1,n}+\frac{1}{4}F_{m-1,n-1}\right)x^my^n\\
&=\frac{1-xy}{(1-x)(1-y)}+\frac{y}{2}\sum_{m=1}^\infty\sum_{n=0}^\infty F_{m,n}x^my^n+\frac{x}{3}\sum_{m=0}^\infty\sum_{n=1}^\infty F_{m,n}x^my^n+\frac{xy}{4}G(x,y)\\
&=\frac{1-xy}{(1-x)(1-y)}+\frac{y}{2}\left(G(x,y)-\frac{1}{1-y}\right)+\frac{x}{3}\left(G(x,y)-\frac{1}{1-x}\right)+\frac{xy}{4}G(x,y)\\
&=\frac{1-xy-\frac{y}{2}(1-x)-\frac{x}{3}(1-y)}{(1-x)(1-y)}+\left(\frac{x}{3}+\frac{y}{2}+\frac{xy}{4}\right)G(x,y)\\
&=\frac{1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{6}}{(1-x)(1-y)}+\left(\frac{x}{3}+\frac{y}{2}+\frac{xy}{4}\right)G(x,y).
\end{align*}
Solving for $G(x,y)$ gives
$$G(x,y)=\frac{1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{6}}{(1-x)(1-y)\left(1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{4}\right)}.$$
Let $R(x,y)$ denote this rational function.
We have shown that $G(x,y)$ converges to $R(x,y)$ in some neighborhood of the origin.
Then for fixed small $x$, the Laurent series $G(x/y,y)$ will converge to $R(x/y,y)$ in some annulus around $y=0$.
Furthermore, $H(x)=\sum_{m=0}^\infty F_{m,m}x^m$ is the constant term of $G(x/y,y)$, and can be found via residue calculus as
\begin{align*}
H(x)&=\frac{1}{2\pi i}\int_\gamma\frac{1}{y}G(x/y,y)\,dy\\
&=\frac{1}{2\pi i}\int_\gamma\frac{1}{y}R(x/y,y)\,dy\\
&=\sum_k\mathrm{Res}\left[\frac{1}{y}R(x/y,y),y=z_k\right]
\end{align*}
where $\gamma$ is a counterclockwise contour in the annulus, and where $z_k$ are the singularities of $R(x/y,y)$ lying inside of $\gamma$.
We can compute
\begin{align*}
\frac{1}{y}R(x/y,y)&=\frac{y-\frac{x}{3}-\frac{y^2}{2}-\frac{xy}{6}}{(y-x)(1-y)\left(y-\frac{x}{3}-\frac{y^2}{2}-\frac{xy}{4}\right)}.
\end{align*}
This rational function has poles at the following points:

*

*$y=1$. This pole does not lie inside of $\gamma$.


*$y=x$. This pole lies inside of $\gamma$, and has residue $\frac{8}{8-9x}$.


*$y=(1-\frac{x}{4})+\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}$. This pole does not lie inside of $\gamma$.


*$y=(1-\frac{x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}$. This pole lies inside of $\gamma$, and has residue
$$\frac{\frac{x}{12}\left((1-\frac{x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)}{\left((1-\frac{5x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)\left(\frac{x}{4}+\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)\left(\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)}$$
which Wolfram Alpha can simplify to
$$\frac{x\left(13\sqrt3\,x-12\sqrt3+\sqrt{3x^2-56x+48}\right)}{(7x-6)(9x-8)\sqrt{3x^2-56x+48}}.$$
Putting this all together gives
$$H(x)=\frac{8}{8-9x}+\frac{x\left(13\sqrt3\,x-12\sqrt3+\sqrt{3x^2-56x+48}\right)}{(7x-6)(9x-8)\sqrt{3x^2-56x+48}}.$$
The second summand is actually holomorphic at $x=6/7$ and $x=8/9$.
Then the singularity of $H(x)$ closest to the origin is $x=8/9$, and we obtain the asymptotic
$$F_{m,m}\sim\left(\frac{9}{8}\right)^m$$
which proves the first limit.
The remaining limits can be solved in a similar way, by first determining the asymptotics of $F(m,m-1)$ and $F(m-1,m)$.
