Let $f:X\rightarrow\mathbb{R}$ be a function, $x_0\in\mathbb{R}$ be an adherent point of $X$ and $L$ be a real number.
We know that $\lim_{x\rightarrow x_0}f(x)=L$ iff for every sequence $(a_n)_{n=0}^{\infty}$ in $X$ which converges to $x_0$ the sequence $(f(a_n))_{n=0}^{\infty}$ converges to $L$.
I was wondering whether there are similar sequential statements for the $\limsup$ and the $\liminf$ of the function $f$ which are equivalent to the definitions: $$\limsup_{x\rightarrow x_0}f(x):=\inf_{r>0}\sup_{|x-x_0|<r}f(x)$$ respectively $$\liminf_{x\rightarrow x_0}f(x):=\sup_{r>0}\inf_{|x-x_0|<r}f(x).$$
My first guess was something like this:
$\limsup_{x\rightarrow x_0}f(x)=L$ iff for every sequence $(a_n)_{n=0}^{\infty}$ in $X$ which converges to $x_0$ we have $\limsup_{n\rightarrow\infty}f(a_n)=L$.
But I quickly found a counterexample to this statement, namely $$f:(0,\infty)\rightarrow\mathbb{R}, x\mapsto\sin(1/x).$$ Here we have $\limsup_{x\rightarrow 0}f(x)=1$ but taking countably many zeros of $f$ in the interval $(0,1]$ as we go from the right to the left on the x-axis, we get a sequence $(a_n)_{n=0}^{\infty}$ in $(0,\infty)$ which converges to $0$ but we have $\limsup_{n\rightarrow\infty}f(a_n)=0$.
Are there any better ideas?