First, define $f^{(n)}{(x)}$ as the nth iterate of x. So, $f^{(0)}(x)=x$, $f^{(1)}(x)=f(x)$, $f^{(2)}(x)=f(f(x))$, and so on.
In this specific case, where $f(x)={a\over x}+1$, if you set $a=1$, then $\lim_{n \to \infty}f^{(n)}{(x)} = {1+\sqrt5 \over2}$ (the golden ratio) for all positive $x$.
I noticed that for all x, including non-integer values of x, $\lim_{n \to \infty}f^{(n)}{(x)}$ will always equal ${\sqrt{4a+1}+1 \over 2}$. I found it interesting how this always seems to occur, but I am not sure why such a nice formula would come out.
This is just out of interest, and any responses would be greatly appreciated!
Note: I tested this using Desmos, and after about 100 iterations the values seem to match up to 11 decimal points, the highest that Desmos shows.
For any such iterated sequence where $f$ is continuous, any possible limit point $\ell$ must satisfy $$ \ell = f(\ell) \tag{1} $$ since you have $x_{n+1} = f(x_n)$ and (as $f$ is continuous) $\lim_{n\to\infty} x_n = \ell$ implies $$ \lim_{n\to\infty} x_{n+1} = f(\lim_{n\to\infty} x_n)\,. $$ ((1) is thus a necessary condition)
Now, in your specific case solving $$ \ell = \frac{a}{\ell}+1 \tag{2} $$ or equivalently $\ell^2 - \ell - a = 0$ (for $\ell \neq 0$) yields $$ \ell = \frac{1\pm\sqrt{1+4a}}{2}\,. \tag{3} $$