The limit of an iterated sequence is a fixed point

1. Since $x_n\to L$ and $f$ is continuous at $L$, $f(x_n)\to f(L)$. But $f(x_n)=x_{n+1}$, which also tends to $L$ (it is the same sequence shifted). By uniqueness of limits, $f(L)=L$.

2. The function $f(x)=1+\frac1x$ is continuous on $(0,+\infty)$. All terms satisfy $x_n\ge1$ (indeed $x_0=1$ and $x_{n+1}=1+\frac1{x_n}>1$), so the limit satisfies $L\ge1$, in particular $L\ne0$ and $L$ lies in the domain. Then $L=1+\frac1L$, that is $L^2-L-1=0$, whose positive root is
$$L=\frac{1+\sqrt5}2.$$

Check. The first terms are $1,\,2,\,1.5,\,1.667,\,1.6,\,1.625,\dots$ and the golden ratio is $1.618\dots$

Warning. Part 1 does *not* prove that the sequence converges: it only says that *if* it converges, the limit must be a fixed point. Convergence has to be established separately. For this sequence, the terms alternate around $L$ and get closer.