We are given - $$x_n = \begin{cases} (n+1)/n & \text{if $n$ ids odd}, \cr 0 & \text{ if $n$ is even.}\end{cases}$$
We have to compute lim inf and limsup of the sequence.
In the step, $$\lim_{n \to \infty } x_n = \lim_{n \to \infty}(sup\{x_k: k\ge n\}). $$
Then, the next step after this is -
$$\sup\{x_k : k \ge n\} =\begin{cases} (n+1)/n & \text{if $n$ is odd}, \cr ((n+2)/(n+1)) & \text{if $n$ is even.} \end{cases}$$
I don't understand how they got the equation for if n is even.
consider $x_n$ where $n$ is odd, we note that this is a decreasing positive subsequence.
Also, $x_n=0$ if $n$ is even, hence the supremum will not take value $0$ due to the presence of odd subsequence which is positive and greater than $1$. The biggest term in $\{ x_k : k \geq n\} $ is $x_j$ where $j$ is the smallest odd number that is at least $n$.
$$\sup \{x_k: k \geq n \}=\begin{cases} x_n & \text{if $n$ is odd}\\ x_{n+1} & \text{if $n$ is even}\end{cases}$$
if $n$ is even, $n+1$ is odd.
$$x_{n+1}=\frac{(n+1)+1}{n+1}=\frac{n+2}{n+1}$$ is the largest terms for the set $\{x_k: k \geq n \}$