Prove that the sequence $\{x_n\}$ diverges to $+\infty$ if and only if $\limsup\limits_{n\to\infty} x_n=\liminf\limits_{n\to\infty} x_n=+\infty$ 
Prove that the sequence $\{x_n\}$ diverges to $+\infty$ if and only if $$\limsup_{n\rightarrow\infty} x_n=\liminf_{n\rightarrow\infty} x_n=+\infty$$

My attempt:
$(\rightarrow)$ By hypothesis we know $\{x_n\}$ diverge.
Let $$E(x_n)=\{x\in \bar{R}: \text{Exists at least a subsequence } \{{x_n}_k\}_k\subset\{x_n\}_n \text{   such that} \{{x_n}_k\}\rightarrow+\infty \text{ when } k\rightarrow \infty\}$$
Then $\sup(x_n)=+\infty$

For $\inf(x_n)$ i'm stuck.
Can someone help me?

For the other implication I don't have a clear idea on how to start. Can someone help?
 A: Suppose that $\lim_{n\to\infty}x_n=\infty$. Take $M\in\mathbb R$. Then there is some $N\in\mathbb N$ such that $n\geqslant N\implies x_n>M$. Therefore, for each $n\geqslant N$, $\inf_{k\geqslant n}x_k\geqslant M$ and so$$\lim_{n\to\infty}\inf_{k\geqslant n}x_k\geqslant M.$$In other words, $\liminf_nx_n\geqslant M$. Since this occurs for each $M\in\mathbb R$, $\liminf_nx_n=\infty$. And, since $\limsup_nx_n\geqslant\liminf_nx_n$, $\limsup_nx_n=\infty$ too.
Now, suppose that $\liminf_nx_n=\infty$. Take $M\in\mathbb R$. Then, since$$\liminf_nx_n=\lim_{n\to\infty}\inf_{k\geqslant n}x_k=\infty>M,$$you know that there is some $N\in\mathbb N$ such that $\inf_{k\geqslant n}x_k>M$ for each $n\geqslant N$. But then $n\geqslant N\implies x_n>M$ and therefore, since this takes place for every $M\in\mathbb R$, $\lim_{n\to\infty}x_n=\infty$.
A: If you can use the following theorem:

A sequence $(x_n)$ has limit $l \in \mathbb{R} \cup \{ \pm \infty \}$ iff $\liminf x_n = \limsup x_n =l$

you're done.
If not, you have to prove directly the equivalence.
Your proof starts well, for if $\lim x_n = +\infty$, then $(x_n)$ is unbounded from above and therefore $\limsup x_n = +\infty$ by definition.
Let us show that $\liminf x_n= + \infty$ as well.
By contradiction, assume $\liminf x_n < +\infty$, so that either $\liminf x_n = -\infty$ or $\liminf x_n = l \in \mathbb{R}$. By definition, $x_n \to +\infty$ implies that $x_n > |l|+1$ for every $n>\nu$; this in turn implies that each possible subsequence $(x_{n_k})$ of $(x_n)$ contains infinitely many elements $x_{n_k} > |l|+1$. It is possible to find at least a subsequence $(x_{n_k})$ from $(x_n)$ s.t. $x_{n_k} \to \liminf x_n$ hence either:


*

*(if $\liminf x_n = -\infty$) there should exists $\kappa$ s.t. $x_{n_k}< -|l| < |l|+1 $ for all $k>\kappa$,

*or (if $\liminf x_n = l \in \mathbb{R}$) there should exists $\kappa$ s.t. $l-1/2 < x_{n_k}< l+1/2 < |l|+1 $ for all $k >\kappa$,
a contradiction.
Regarding to the converse, you should observe that $\liminf x_n = + \infty$ suffices to conclude $x_n\to +\infty$.
