Show that the sequence $X=(x_n)$ in $\mathbb{R}$ diverges to $+\infty$ if and only if $\lim \text{inf}(x_n)=+\infty$. Similarly, show that $\lim\text{sup}(x_n)=+\infty$ if and only if there is a subsequence $x_n'$ of $x_n$ such that $\lim(x_n')=+\infty$
My attempt:
Suppose $\lim(x_n)=+\infty$. Then for a given $\epsilon$ there exists an $N$ such that $n>N$, then by defintion $x_n>\epsilon$. Also, by the defintion of a limit inferior, there is a supremum of $x_n$, which we denote as $p_n=\text{sup}\{x_n: n\in \mathbb{N}\}$, such that there are at most a finite number of $m\in \mathbb{N}$ such that $x_m< w$ and an infinite amount of $m$ such that $w-\epsilon <x_m$ ...
Am I on the right track? If so, how can I continue in completing the proof?
Suppose $x_n$ diverges to $\infty$. This means for all $L$, there is some $N$ such that if $n \ge N$, then $x_n \geq L$. In particular, $\inf_{k \ge N} x_k \ge L$, and since $k \mapsto \inf_{k \ge N} x_k$ is increasing, we have $\liminf_n x_n \ge L$. Since $L$ was arbitrary, it follows that $\liminf_n x_n = \infty$.
Now suppose $\liminf_n x_n = \infty$. Choose $L$. Then for some $n$, we have $\inf_{k \ge n} x_k \ge L$, from which we get $x_k \ge L$ for all $k \ge n$. It follows that $x_n$ diverges to $\infty$.
Now suppose $\limsup_n x_n = \infty$. This means $\lim_n \sup_{k \ge n} x_k = \infty$. For each $n$, choose $k_n \ge n$ such that $x_{k_n} \ge \sup_{k \ge n} x_k-1$, and $k_{n+1} > k_n$. It follows that $x_{k_n} \to \infty$.
Now suppose some subsequence $x_{k_n} \to \infty$. Choose $L$. Then there exists $N$ such that for $n \ge N$, we have $x_{k_n} \ge L$. Hence we have $\sup_{k \ge n} x_k \ge L$. It follows that $\limsup_n x_n = \infty$.