I am hoping someone can help me determine the limit of two unique generalized Fibonacci sequences.
Most everyone is familiar with the much talked about $\lim_{x\to \infty}$ $\frac{F_{n+1}}{F_n}=\varphi$
where $\varphi$ is the golden ratio.
I would like to calculate $\lim_{x\to \infty}$ $\frac{G_n}{H_n}$
where $G_n$ and $H_n$ are two unique generalized Fibonacci sequences defined by $G_n=G_{n-2}+G_{n-1}=aF_{n-2}+bF_{n-1}$ where $G_1=a$ and $G_2=b$.
The same goes for $H_n$, albeit with different initial values $a$ and $b$.
The Binet-type formula for a generalized Fibonacci sequence is ${c\alpha^n-d\beta^n\over \sqrt5}$,
where $\alpha={1+\sqrt5\over 2}$ and $\beta={1-\sqrt5\over 2}$, in addition $c=a+(a-b)\beta$ and $d=a+(a-b)\alpha$.
I feel like I should be able to figure this out, but so far I haven't had much luck.
Any help with be very much appreciated!
Every two sequences $G_n,H_n$ satisfying the Fibonacci recurrences can be written as $G_n = g_\alpha \alpha^n + g_\beta \beta^n$ and $H_n = h_\alpha \alpha^n + h_\beta \beta^n$; you can find $g_\alpha,g_\beta$ ($h_\alpha,h_\beta$) by solving linear equations given your initial values $G_1,G_2$ ($H_1,H_2$). For a "generic" choice, you will get $g_\alpha,h_\alpha \neq 0$, and so, since $\beta^n \to 0$, you would have $$ \frac{G_n}{H_n} = \frac{g_\alpha \alpha^n + g_\beta \beta^n}{h_\alpha \alpha^n + h_\beta \beta^n} \approx \frac{g_\alpha \alpha^n}{h_\alpha \alpha^n} = \frac{g_\alpha}{h_\alpha}. $$ In fact, this formula holds even if $g_\alpha = 0$ (in which case you get $0$) or $h_\alpha = 0$ (in which case you get $\infty$).