I understood the definition of $\liminf$ and $\limsup$ for sequences well. But there is a little bit of confusion when it comes to functions.
If I did learn it correctly, the definition of $\limsup_{x \to l} f(x)$ is given as the following: Let $A(\delta)$ a set with $A(\delta) = \left\{f(x): 0<| x-l | < \delta \right\}$ for each $\delta > 0$. Then it is $\limsup_{x \to l} f(x) = \lim_{\delta \to 0^{-}} \sup A(\delta)$.
For a regular limit $\lim_{x \to l}f(x)$ it is not important whether $f$ is defined at $l$ or not. For example let $f(x)$ be a well behaved, continuous function except only at $l$ where it makes a pointwise "jump". Then it is $\lim_{x \to l}f(x) = a$ where $a$ is the value which makes $f$ completely continuous if it were $f(l)=a$.
My question is whether the same is valid for $\limsup$ and $\liminf$. Let's say that we calculate $\limsup_{x \to l} f(x)$ for the same $f(x)$ which makes a jump at $l$. Since the set $A(\delta)$ at the definition does not cover $0$, $f(l)$ is not included in $A(\delta)$. But since we take the limit of $\sup A(\delta)$s as $\delta$ approaches to $0$, $\limsup$ again becomes $a$ which is equal to $\lim_{x \to l}f(x)$. Is this logic correct here?
