Let the functions be defined as $g:R\to R$ and $f:R \to (1,\infty)$. Both f and g are continuous. Let $F_n$ denote the $n^{th}$ Fibonacci number, then for $n ∈ N$ let $\Gamma$ and $t_n$ be defined as follows;
$$t_n=\sqrt [n]{(2n-1)!! F_n}$$
$$\Gamma=\lim_{n\to\infty} \int_{t_n}^{t_{n+1}} \frac{(f(x-t_n))^{g(t_{n+1}-x)}}{(f(t_{n+1}-x))^{g(x-t_n)}+(f(x-t_n))^{g(t_{n+1}-x)}} \, dx$$
My initial attempt involved the recurrence relation of Fibonacci numbers for $n>1$;
$\boxed{F_n=F_{n-1}+F_{n-2}}$ and $\boxed{F_0=0, F_1=1}$. The closed form being; $\boxed{F_n=\frac{1}{\sqrt 5}\left[(\phi)^n-(\psi)^n\right]}$
$\phi$ being the golden ratio and $\psi$ being the conjugate of golden ratio.
Not being familiar with the double factorial notation, here's what I have understood/found;
$\boxed{n!!=\sum_{k=0}^{\lceil \frac{n}{2} \rceil-1}(n-2k)=n(n-2)(n-4)\cdots}$ and $\boxed{n!=n!!(n-1)!!}$
$$\boxed{t_n=\sqrt[n] {\frac{(2n-1)!!(\phi^n-\psi^n)}{\sqrt5}}}$$ $$\boxed{t_{n+1}=\sqrt[n+1] {\frac{(2n+1)!!(\phi^{n+1}-\psi^{n+1})}{\sqrt5}}}$$
Substituting,
$t_n=\sqrt [n]{(2n-1)!! F_n}$ and $t_{n+1}=\sqrt [n+1]{(2n+1)!! F_{n+1}}$.
Let $$\Gamma=\lim_{n\to\infty} \beta_n$$
where $\beta_n$ is;
$$\boxed{\int_{\sqrt [n]{(2n-1)!! F_n}}^{\sqrt [n+1]{(2n+1)!! F_{n+1}}} \frac{(f(x-\sqrt [n]{(2n-1)!! F_n}))^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)} \, dx}{(f(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x))^{g\left(x-\sqrt [n]{(2n-1)!! F_n}\right)}+(f(x-\sqrt [n]{(2n-1)!! F_n}))^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)}}}$$
On just a little simplification;
$$\boxed{\int_{\sqrt [n]{(2n-1)!! F_n}}^{\sqrt [n+1]{(2n+1)!! F_{n+1}}} \frac{1}{\left[\frac{\left(f\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)\right)^{g\left(x-\sqrt [n]{(2n-1)!! F_n}\right)}}{\left(f\left(x-\sqrt [n]{(2n-1)!! F_n}\right)\right)^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)}}\right]+1}\,dx}$$
I'm now left more confused than before, was this the right approach at all? If so, how does one continue from here?
Note - This question did not come with the solution, so I am not certain if the limit converges (if that is necessary)