On some iterated inequalities and $x \geq 5$ Let $x_i \in \mathbb{N}$, $i = 1, 2, \ldots, n$.
Suppose that I have a function $f:\mathbb{N}\rightarrow\mathbb{Q}$, with initial bounds 
$$2 - \frac{2}{x_0} < f(x_0) = \frac{2{x_0}}{x_0 + 1} \leq 2 - \frac{5}{3x_0}.$$
Assume further that $f$ is iteratively defined as
$$f(x_{i+1}):=2-\frac{f(x_i)}{x_i}.$$
Questions

(1) What is the lower bound for $f(x_n)$ at the $n^{th}$ iteration?
(2) What is the upper bound for $f(x_n)$ at the $n^{th}$ iteration?
(3) What is
  $$\lim_{x_n \rightarrow \infty}{f(x_n)}?$$
(4) Will it be possible to improve on the lower bound $x_i \geq 5$ at some point $j$ in the iterative process?  (That is, must $x_{j+1} > x_j$ hold for some $j \geq \overline{n}$, where $\overline{n} > 1$?) 

 A: We start from a simple 
Lemma.  For each $n\ge 0$, $0<f(x_n)<2$. 
Proof.  We shall prove the claim by induction. $$f(x_0)=\frac {2x_0}{x_0+1}=2-\frac{2}{x_0+1}$$ and $$1\le 2-\frac{2}{x_0+1}<2.$$ 
Now assume that we have already proved that $0<f(x_n)<2$. Then $$f(x_{n+1}):=2-\frac{f(x_n)}{x_n}$$ and $$0=2-\frac{2}{1}<2-\frac{f(x_n)}{x_n}<2.\square$$
But Lemma allows us answer your Questions. 
(1) The upper bound $2$ for $f(x_n)$ is tight, because when we tend $x_{n-1}$ to infinity, $f(x_n)$ will tend to $2$. 
(2) The lower bound $0$ for $f(x_n)$ (for $n\ge 1$) is also tight, because when put $x_{n-1}=1$ and tend $x_{n-2}$ to infinity, 
$f(x_{n-1})$ will tend to $2$ and  $f(x_n)$ will tend to $0$.
(3) The answer depends on the sequence $\{x_n\}$. For instance, if $x_n$ tends to infinity, then Lemma implies that $f(x_n)$ tends to $2$.
If $x_n$ eventually stabilize at some number $k>1$ (that is, if there exists a number $\bar{n}$ such that $x_n=k$ for each $n\ge\bar{n}$), then we can easily show that $x_n$ tends to 
$$2\sum_{i=0}^\infty\frac{(-1)^i}{k^i}=\frac{2k}{k+1}.$$
If $x_n$ eventually stabilize at $1$, then the sequence $\{f(x_n)\}$ becomes periodic $$\dots, a, 2-a, a, 2-a,\dots.$$ 
Moreover, the existence of a limit $a$ of the sequence $\{f(x_n)\}$ imposes strict conditions on the sequence $\{x_n\}$. Indeed, in this 
case for each $\varepsilon>0$ exists natural $\bar n$ such that 
$\left|a-2+\frac{a}{x_n}\right|<\varepsilon$ for each $n\ge\bar n.$ 
Thus the sequence $\{\frac{a}{x_n}\}$ converges to $a-2$. Since $x_n$ are natural, they either tend to infinity (iff $a=2$) or stabilize (iff $a\ne 2$). Both cases were considered above. 
(4) I don’t understand the origin and the purpose of “lower bound $x_i \geq 5$”. But if $x_{n+1} > x_n$ hold for all $n \geq \overline{n}$, where $\overline{n} > 1$ then we can improve the lower bound, because if 
$$x_ \overline{n}< x_ {\overline{n}+1}<\dots< x_ {\overline{n}+n},$$
then $x_ {\overline{n}+n}\ge n+1$ and $f(x_ {\overline{n}+n+1})=2-\frac{f(x_ {\overline{n}+n})}{x_ {\overline{n}+n}}>
2-\frac{2}{n+1} $.
