Let $X$ be a metric space, $D$ be a nonempty subset of $X$ and $x_0$ be a limit point of $D.$ Assume $f: D\to \mathbb{R}.$ We define as usual the upper limit $\limsup_{x\to x_0} f(x)$ of $f$ as $x$ tends to $x_0$ as $$\limsup_{x\to x_0}:=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},$$ where $\hat{\mathbb{B}}(x_0,r)$ denote the punctured ball of $x_0$ with radian $r.$ Write \begin{gather*} L:=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},\\ \begin{aligned} L^*:=\sup\{ L'\mid \exists&\text{ sequence } (x_n) \text{ of points in } D\backslash\{x_0\}, \text{ such that } x_n\to x_0, f(x_n)\to L', \text{ as } n\to \infty\}. \end{aligned} \end{gather*} I want to prove that $L=L^*.$ It is easy for me to prove $L\geq L^*,$ and I have given my proof of this below. But I have trouble in proving the converse part, that is, $L\leq L^*.$
Proof of $``L\geq L^*"$ $\quad$ It suffices to prove for the case that $L\in\mathbb{R}.$ Since $L=\inf_{r>0}\sup\{f(x)\mid x\in D\cap\hat{\mathbb{B}}(x_0,r)\},$ by the definition of inf, we have:
$(\alpha)$ $\forall r>0, \sup\{f(x)\mid x\in D\cap \hat{\mathbb{B}}(x_0,r)\}\geq L,$
and
$(\beta)$ $\forall \epsilon>0~\exists r_{\epsilon}>0 : \sup\{f(x)\mid x\in D\cap \hat{\mathbb{B}}(x_0,r_{\epsilon})\}<L+\epsilon.$
By $(\beta),$ we have: $\forall \epsilon>0~\exists r_{\epsilon}>0 : \sup\{f(x)\mid x\in D\cap \hat{B}(x_0,r_{\epsilon})\}<L+\epsilon.$ Assume $(x_n)$ is a sequence of points in $D\backslash\{x_0\}$ with $x_n\to x_0$ such that $f(x_n)\to L'.$ Now that $x_n\to x_0,$ for the $r_{\epsilon},$ there exists $N_1\in\mathbb{N}$ such that for all $n\in\mathbb{N}:$ $n>N_1\implies x_n\in D\cap \hat{\mathbb{B}}(x_0,r_{\epsilon}).$ Thus
$f(x_n)\leq \sup \{f(x)\mid x\in D\cap \hat{B}(x_0,r_{\epsilon})\}<L+\epsilon,$ if $n>N_1.$ Since $f(x_n)\to L',$ letting $n\to \infty,$ we have $L'=\lim_{n\to\infty} f(x_n)\leq L+\epsilon.$ Since our choice of $(x_n)$ is arbitrary,
the last inequality implies that $L^*\leq L+\epsilon.$ Because $\epsilon$ is arbitrary, we conclude that $L^*\leq L.$
My question is: How to prove that $L^*\geq L?$