Find $\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}! $ where $x_n=x_{n-1}+x_{n-2},x_1=1,x_2=2$ Find $\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}! $ where $x_n=x_{n-1}+x_{n-2} ,(n>2),x_1=1,x_2=2$
$x_n=x_{n-1}+x_{n-2}$
$x_{n+1}=x_{n}+x_{n-1}$
From the first recurrence relation, $$x_n=\frac{3+\sqrt{5}}{5+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n+\frac{3-\sqrt{5}}{5-\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$$
From the second recurrence relation,
$$x_{n+1}=c_2({x_1}^{'})^{n+1}+c_3({x_2}^{'})^{n+1}$$
where 
$${x_2}^{'}=\frac{1+\sqrt{5}}{2},{x_3}^{'}=\frac{1-\sqrt{5}}{2}$$ 
$$c_2=\frac{x_3-{x_3}^{'}x_2}{{x_2}^{'}({x_2}^{'}-{x_3}^{'})}=\frac{2(2+\sqrt{5})}{5+\sqrt{5}}$$
$$c_3=-\frac{x_3-{x_2}^{'}x_2}{{x_3}^{'}({x_2}^{'}-{x_3}^{'})}=\frac{2(2-\sqrt{5})}{5-\sqrt{5}}$$
gives 
$$x_{n+1}=\frac{2(2+\sqrt{5})}{5+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}+\frac{2(2-\sqrt{5})}{5-\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}$$
This gives $$\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}! =\frac{2(2+\sqrt{5})}{3+\sqrt{5}}!$$
Is this correct?
 A: This is perfectly valid and correct. An alternative technique can be used, however, to conclude this more quickly and easily...

Important Prerequisites:
First, bear in mind that $x_n$ is the Fibonacci recurrence, and thus $x_{n+1} / x_n \to \varphi = (1 + \sqrt 5)/2$ as $n \to \infty$. Next, we also reference the gamma function, a generalization of the factorial to all complex numbers (except nonpositive integers). It satisfies the relation
$$\Gamma(n) = (n-1)!$$
for positive integers $n$, and is continuous for all real $n>0$.
Finally, if $f$ is a continuous function, we note that
$$\lim_{x \to a} f(g(x)) = f \left( \lim_{x \to a} g(x) \right)$$
i.e. we can bring the limit inside continuous functions.

The Calculation
We see that, applying the definition of the gamma function,
$$L := \lim_{n \to \infty} \left( \frac{x_{n+1}}{x_n} \right)! = \lim_{n \to \infty} \Gamma \left( \frac{x_{n+1}}{x_n} + 1 \right)$$
Applying continuity,
$$L = \Gamma \left( \lim_{n \to \infty} \frac{x_{n+1}}{x_n} + 1 \right)$$
Obviously the limit for the $1$ summand is $1$, so with the fact that $x_n$ is the Fibonacci recurrence,
$$L = \Gamma( \varphi + 1 ) = \varphi!$$
This is equivalent to your solution since
$$\varphi := \frac{1 + \sqrt 5}{2} \cdot \frac{(3 + \sqrt 5)/2}{(3 + \sqrt 5)/2} = \frac{4 + \sqrt 5}{3 + \sqrt 5}$$
The value of $L$ can be approximated by your means of preference. Wolfram gives the approximation
$$L \approx 1.4492296$$

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
