Does the output tell me that the software cannot evaluate the sign because it does not know the sign of A?
In a way, yes. But it does not even assume
A to be real. What if
And does it tell me that the solution is for sure plus or minus infinity?
In a way, yes. But this is not generally correct. If
A==4, then the answer is zero.
BTW newer versions give the answer as
(-2 - A/2) ∞
Also, in newer versions of
Limit can do this:
Limit[-A/2 x - 2 x, x -> +Infinity, GenerateConditions -> True]
(* ConditionalExpression[(-2 - A/2) ∞, A != -4] *)
Finally, note that
Infinity is also just
DirectedInfinity in disguise.
(* DirectedInfinity *)
(* DirectedInfinity[-1] *)