Recursive Reciprocal Function Property

Consider the case when $x$ is a non-negative integer $n$ . The $n$ th term of the sequence $\{f(n)\}$ is given by $f(n) = \dfrac {F_n}{F_{n+1}}$ . where4 $F_n$ denotes the $n$ th Fibonacci number . Let us prove this claim by induction . As given $f(0) = 0 = \dfrac {0!}{1!}$ , the claim is true for $n=0$ . Assuming that the claim is true for $n$ , then

$f(n+1) = \frac 1{f(n)+1} = \frac 1{\frac {F_n}{F_{n+1}}+1} = \frac {F_{n+1}}{F_n + F_{n+1}} = \frac {F_{n+1}}{F_{n+2}}$

The claim is also true for $n+1$ , therefore true for all $n \ge 0$ . Then $\displaystyle \lim_{n \to \infty} f(n) = \frac {F_n}{F_{n+1}}$ . Since $\displaystyle \lim_{n \to \infty} \frac {F_{n+1}}{F_n} = \varphi$ , where $\varphi = \dfrac {1+\sqrt 5}2$ denotes the golden ratio , $\displaystyle \lim_{n \to \infty} f(n) = \frac 1\varphi = \varphi - 1 = \frac {\sqrt 5 - 1}2$ and $a+b+c = -1+5+2 = \boxed 6$ .

As $x \to \infty$ , $f(x) \sim f(x-1)$ and thus we may write $f(x) \sim \dfrac{1}{f(x)+1}$ , from here , we get the asymptotic value of $f(x)$ , which is $\dfrac{-1+\sqrt{5}}{2}$ , and thus $a=-1,b=5,c=2$ making the answer $\boxed{6}$ .