Convergence to the golden ratio

1. $x_1=2$. If $x\in\left[\frac32,2\right]$, then by monotonicity $f(x)\in\left[f(2),f\!\left(\frac32\right)\right]=\left[\frac32,\frac53\right]\subset\left[\frac32,2\right]$. By induction $x_n\in\left[\frac32,2\right]$ for $n\ge1$.

2. The fixed point satisfies $\varphi=1+\frac1\varphi$, i.e. $\varphi^2=\varphi+1$, and $\varphi\approx1.618\in\left[\frac32,2\right]$. For $x\ge\frac32$, $|f'(x)|=\frac1{x^2}\le\frac49$. The mean value theorem gives
$$|x_{n+1}-\varphi|=|f(x_n)-f(\varphi)|\le\frac49|x_n-\varphi|.$$
Hence $|x_n-\varphi|\le\left(\frac49\right)^{n-1}|x_1-\varphi|\to0$.

Sharper rate. Near $\varphi$ the ratio is $|f'(\varphi)|=\frac1{\varphi^2}\approx0.382$.

Remark. The terms are $x_n=\frac{F_{n+1}}{F_n}$, ratios of consecutive Fibonacci numbers: $1,2,\frac32,\frac53,\frac85,\dots$ (see Binet's formula).