Upper Limit Definition The upper and lower limits can be defined as follows:
$$U:=\limsup_{x\to x_0} f(x)\\
L:= \liminf_{x\to x_0} f(x)$$
$L$ is characterized by the properties:
i) There is a sequence $\{y_k\} \subset E\setminus\{x_0\}$ such that $\lim y_k=x_0$ and $\lim f(y_k)=L$;
ii) For any sequence $\{y_k\} \subset E\setminus\{x_0\}$ converging to $x_0$ we have $\liminf f(y_k)\geq L$.
I was reading some upper and lower limit and came up on this site.
I am wondering how would I prove the above two properties since the site didn't say and I am a little confused about these defintions.
 A: Definitions
Provided that $x_0$ is an accumulation point of the set $E$, for $f \colon E \to \mathbb R$,
$$
U=\limsup_{x \to x_0} := \inf_{r >0} \sup_{x \in \check B_E (x_0;r)} f(x), L=\liminf _{x \to x_0} := \sup_{r >0} \inf_{x \in \check B_E (x_0;r)} f(x), 
$$
where
$$
\check B_E (x_0;r) :=\{y \in E\colon 0<|y - x_0|<r\}. 
$$
Propositions
For $\liminf f$, we have the following claims:

*

*There is a sequence $(x_n) \subseteq E \setminus \{x_0\}$ that $x_n \to x_0$ and $f(x_n) \to L$ as $n \to \infty$;

*For each sequence $(x_n) \subseteq E\setminus \{x_0\}$ that $x_n \to x_0$ as $n \to \infty$, $\liminf_{n\to \infty} f(x_n) \geqslant L$;

*Equivalently, $$ \liminf_{x\to x_0} f(x)= \inf \{\ell \in \mathbb R\cup \{-\infty, +\infty\}\colon \exists (x_n) \to x_0 \text{  s.t. } f(x_n) \to \ell \}.  $$
Proofs
We first investigate a function
$$
g\colon (0,1]\to \mathbb R\cup\{-\infty, +\infty\}, r \mapsto \inf_{x \in \check B(x_0;r)} f(x).  
$$
Note that $g$ is monotonically decreasing, since for $0<r_1 <r_2<1$, the sets
$$
\check B_E (x_0;r_1) < \check B_E (x_0; r_2), 
$$
then
$$
\inf_{\check B_E (x_0;r_1)} f(x) \geqslant \inf_{\check B_E (x_0; r_2)} f(x). 
$$
Thus $g(0^+)$ exists, and $g(0^+) = L$ by the properties of monotonic functions.

*

*For each $n$, consider $g(1/n)$, then $g(1/n) \nearrow g(0^+) = L$. For each $n$, by definition of $\inf$, there is an $x_n \in \check B_E(x_0; 1/n)$ that $g(1/n)\leqslant  f(x_n) \leqslant g(1/n)  + 1/n$. Thus by the squeezing theorem, $$ \lim_n g(1/n) \leqslant \lim_n f(x_n)  \leqslant \lim_n g(1/n) + 1/n,$$ which means $\lim_n f(x_n) = L$.

*By the definition of $g(r)$, for each $(x_n) \subseteq E\setminus \{x_0\}$ that $x_n \to x_0$, if $r_n := |x_n - x_0|+1/n$, then $f(x_n) \geqslant g(r_n)$. By taking $\liminf$, we get $$\liminf f(x_n) \geqslant \liminf g(r_n) = \lim g(r_n) = g(0^+) = L.$$

*Let the RHS be $L'$. According to No. 1, $L \geqslant L'$. According to No. 2, if we take a sequence $(x_n) \to x_0$ that $f(x_n) \to \ell$, then $\ell \geqslant L$, so $L \leqslant L'$. Therefore $L = L'$.

