Let $D \subset \mathbb{R}$ and $f: D \to \mathbb{R}$ and $t$ a limit point of $D$. Is the definition
$$L:=\liminf_{s \to t} f(s):= \lim_{\epsilon \to 0} (\inf \{ f(s)| s \in D \cap (t-\epsilon,t+\epsilon)\setminus \{t\} \})$$ equivalent to: For all sequences $(t_n) \subset D\setminus \{t\}$ with $\lim_{n \to \infty} t_n=t$ it holds that $$\liminf_{n \to \infty}f(t_n)=L?$$
No. Consider the following example. Let $D=\mathbb{R}$ and $f(x)$ be $1$ for all non zero $x$ and $f(0)=0$. Then, $$\lim_{\epsilon \to 0} (\inf \{ f(s)| s \in D \cap (0-\epsilon, 0+\epsilon) \}) = 0$$ but, for all sequences $(t_n) \subset D\setminus \{0\}$, with $\lim_{n \to \infty} t_n=0$, we have $$\liminf_{n \to \infty}f(t_n)=1$$
Edit after the correction in the definition
My counterexample doesn't work now. But still they are not equivalent. Consider $f(x)$ being $1$ for all $x<0$ and $f(x)=0$ for $x\ge 0$. Then $$\lim_{\epsilon \to 0} (\inf \{ f(s)| s \in D \cap (0-\epsilon, 0+\epsilon) \setminus \{0\} \}) = 0 $$ even though we can find $t_n$ with $t_n \to 0$ and $\liminf_{n \to \infty}f(t_n)=1$. For example, $t_n=-1/n$