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Yes, lim inf is equal to $2$. I can't say what the lim sup is going to be. The lower bounds for double blocking sets in the case of non-square $q$ are not sharp. In particular, the working conjecture (unless there has been some very recent progress on this) seems to be that for prime $p$ a double blocking set in $PG(2,p)$ has size at least $3p$. Thus, for prime $p$ we probably can't improve $s(p) \leq 3p^2 - 3p$ using double blocking sets.
I made a mistake in my calculation. This approach actually gives $s(q) \geq (2q-2)(q-1) + 3(q-1) + 1 = 2q^2 - q$, which is the original bound I gave in my question. As Douglas Zare has shown, we can improve the lower bound to $s(q) \geq 2q^2 - 1$. So, this approach might not be very helpful here.
Valid point. I'll think about it. Just to point out the obvious: even if we can always find polynomials of that form meeting the bound, there is no guarantee that the $2q-1$ points on each of the hyperplanes will form a blocking set there. And so we need to find "nice" polynomials meeting the bound.
No, not always. The bounds obtained by this result, for example on multiple blocking sets, can be improved for many cases. Also, the value of $s(3)$ violates this bound.
$(2q-2)(q-1) + 3(q-1) = 2q^2 - 1$ by Ball and Serra's punctured combinatorial nullstellensatz with multiplicity: link.springer.com/article/10.1007%2Fs00493-009-2509-z. Thus giving another proof of Douglas Zare's lower bound given below. (Nice catch. I should have thought about this before)
I would like to add that a friend of mine is running a computation which has shown that $s(4) = 34 < 36$ (in case you are wondering about the sharpness of this bound $s(q) \leq 3q^2 - 3q$). He'll hopefully post more values he has computed by tomorrow. By interpolating from $s(2), s(3), s(4)$ we could make a bold (and probably wrong) conjecture that $s(q) = (5q^2 - 3q)/2$ ... But may be one can try constructing examples of size $(5q^2 - 3q)/2$ to show $s(q) \leq (5q^2 - 3q)/2$.
Thanks. This is how I understand the second part: all lines contained in the hyperplanes $H_1, \dots, H_q$ are blocked once you have chosen a pair of intersecting lines in each of them. And every transverse which doesn't contain $P$ is blocked by $H_1 \setminus \{P\}$. Any transverse through $P$ other than the line $\ell$ intersects every $H_i$ ($i > 2$) in a point $R_i$ other than $Q_i = \ell \cap H_i$ and the line $Q_iR_i$ is parallel to one of the $q+1$ lines through $P$ in $H_1$. So, we ensure that for each line through $P$ in $H_1$, we take a parallel line through $Q_i$ in $H_i$.
