$\displaystyle{\limsup_{x \to \infty} f(x) = L = \liminf_{x \to \infty} f(x)}$ proof In the following link at the end of the page there is presented a proof for the following theorem:

Let $f : (a, \infty) \to \mathbb{R}$ where $a \in \mathbb{R}$. Then
$\displaystyle{\lim_{x \to \infty} f(x) = L}$ if and only if
$\displaystyle{\limsup_{x \to \infty} f(x) = L = \liminf_{x \to
> \infty} f(x)}$.

Most of this proof seems to me like it is missing some pieces, for example in the "Hence..." paragraph how does the author conclude this $
\displaystyle{\liminf_{x \to \infty} f(x) = \lim_{x \to \infty} \inf_{t \geq x} \{ f(t) \} = L}$.
Can anyone explain this proof in a little more detail or maybe even show it on concrete example?
 A: Here's how their proof goes, let's analyse it line by line:

Proof: $\Rightarrow$ Suppose that $\lim_{x \to \infty} f(x) = L$. Then for all $\epsilon > 0$ there exists an $M \in \mathbb{R}$, $M \geq a$ such that $f(x) - L < \epsilon$.

This is correct, but not complete. What they should have said was
". . .such that $\color{red}{\lvert} f(x) - L \color{red}{\rvert} < \epsilon$."
This is just the definition of limit. In fact, only with this correction does their next sentence make sense:

So, if $x \geq M$, then $$L - \epsilon < f(x) < L + \epsilon \tag{3}$$

They could not have concluded that $L - \epsilon < f(x)$ just from $f(x) - L < \epsilon$. Moving on,

So for $x \geq M$ we have that $L + \epsilon$ is an upper bound for $f$ and $L - \epsilon$ is a lower bound for $f$. Thus for $x \geq M$ we have that:
  $$
L - \epsilon \leq \inf_{t \geq x}\{ f(t) \} \leq \sup_{t \geq x}\{ f(t) \} \leq L + \epsilon \quad \Leftrightarrow \quad \left\lvert \inf_{t \geq x}\{ f(t) \} - L \right\rvert < \epsilon, \left\lvert \sup_{x \geq t}\{f(t)\} - L \right\rvert < \epsilon \tag{4}
$$

First of all, they have a potentially confusing typo in the last inequality: it should be
$$
\left\lvert \sup_{\color{red}{t \geq x}}\{f(t)\} - L \right\rvert < \epsilon.
$$
Secondly, note that if you have that $L - \epsilon \color{green}{<} f(x)$ for all $x \geq M$, then $L - \epsilon \color{green}{\leq} \inf_{x \geq M}\{f(x)\}$. Think of how analogously you have $\inf\{ x : 2 < x < 3\} = 2$, for example. Similarly, 
if you have that $f(x) \color{green}{<} L + \epsilon$ for all $x \geq M$, then $\inf_{x \geq M}\{f(x)\} \color{green}{\leq} L + \epsilon$.
So, what have they really shown so far? What they have shown is that for every $\epsilon > 0$, there exists a real number $M \geq a$ such that for all $x \geq M$, we have
$$
\left\lvert \inf_{t \geq x}\{ f(t) \} - L \right\rvert < \epsilon
$$
and
$$
\left\lvert \sup_{t \geq x}\{f(t)\} - L \right\rvert < \epsilon.
$$
Hence, if we define the functions $g, h \colon (a,\infty) \to \mathbb{R}$ by
$$
g(x) = \inf_{t \geq x}\{ f(t) \} \quad \text{and} \quad h(x) = \sup_{t \geq x}\{f(t)\},
$$
then we have that $\lim_{x \to \infty} g(x) = L$ and $\lim_{x \to \infty} h(x) = L$, by the definition of limit. On the other hand, we also have the definitions for $\limsup$ and $\liminf$, so combining them we get:

Hence, $\liminf_{x \to \infty} f(x) = \lim_{x \to \infty} \inf_{t \geq x} \{ f(t) \} = L$ and $\limsup_{x \to \infty} f(x) = \lim_{x \to \infty} \sup_{t \geq x} \{ f(t) \} = L$. $\blacksquare$

