I just a have problem regarding a definition i found here on limit superior of a function at a point.
Definition of limsup of function ( i found it here on math exchange) : \begin{equation} \limsup _{x\rightarrow x_{0} ;x\in E} f(x)=\inf\Bigl\{\sup \bigl\{f(x):x\in E\land |x-x_{0} |< \delta \bigr\} :\delta \in \mathbf{R}^{+}\Bigr\} \tag{A} \end{equation}
the following are equivalent (for $L\in \mathbb R$):
$\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =L$
For every $\displaystyle \epsilon >0$ and every sequence $\displaystyle ( x_{n}) \subset E-\{x_{0} \}$ converging to $\displaystyle x_{0}$, there exists $\displaystyle N\in \mathbb{N}$ such that $\displaystyle n\geq N\Longrightarrow $$\displaystyle f( x_{n}) < L+\epsilon $ and $\displaystyle f( x_{m}) >L -\epsilon $ for infinite $\displaystyle m$.
But i have an example for which I can't seems to figure it out
here is the function i defined in desmos
By above definition, limsup at 2 is 3 right? So, we can find a sequence for example (1.9, 1.99, 1.999...) which converges to 2 from left but there exist epsilon let's take if 0.1 such that $\displaystyle f(x_{m})$ doesn't exceed 3 - 0.1 for any m or doesn't exceeds infinitely, so $\displaystyle f( x_{m}) >L -\epsilon $ for infinite $\displaystyle m$ doesn't hold, can someone help me with this, what i am missing etc.
Ps: my first time posting here, so please excuse me for any silly mistakes.