I am not sure what is your question. But you can construct a linear functional in this way.
The limit does not exist for some $f$ in $L^\infty$. But you can use the Banach limit.
Consider the bounded linear functional on $C(R)$ defined as the limit as $x\to 0$.
By Hahn-Banach this has an extension to a bounded linear functioal on $L^\infty(-\infty,0)$.
and also has an extension to $L^\infty(0,\infty)$.
Now define your functional as you propose, using these Banach limits.

But all this is a kind of trivial nonsense. Hahn Banach says you from the beginning that
you have (very many) linear functionals on $L^\infty$ which are zero on $C$.
This is a pure existence theorem. But your "construction" really adds nothing to it,
because the existence of a Banach limit is again a pure existence theorem.