Let $(X,d)$ be a metric space and let $f:X\to [-\infty, +\infty] $.
Now we fix a $x\in X$ and we want to find a sequence $(x_j)\subset X$ convergint to $x$ such that $$\limsup_{j\to +\infty} f(x_j) \le \bar f(x)$$ where $\bar f$ is defined as the greatest lower semicontinuous function not greater than $f$, i.e. for every $x\in X$ $$\bar{f}(x)= \sup \{ g(x) : g \mbox{ l.s.c., } g\le f \}.$$
What I have observed is that if $x$ is a point of of lower semicontinuity of $f$, then I can choose the constant sequence where every $x_j = x$: this way $$\limsup_{j\to +\infty} f(x_j) = \limsup_{j\to +\infty} f(x) = f(x)=\bar f(x)$$ (where $f(x) = \bar f(x)$ since $f$ is l.s.c. at $x$).
I cannot find a way for when $x$ is not a l.s.continuity point for $f$.