(tax with) costs to first order 1) What does it mean that something (in this context the taxation) has (or doesn't) costs to first order?

So, below is an excerpt from Romer's Advanced Macro (p. 598) with such a statement that I've encountered and felt a bit lost. Altough the model developed later is pretty clear, I still don't get what the sentence with this 'costs to first order' really mean. 

2) Also: how the last sentence is implied by the sentence before the last one? That is: what exactly (this first-orderness? how?) implies that $C \approx T^2$?
1) "To first order" appears to be a shorthand for "to a first-order approximation", meaning in turn, "to a linear approximation".

2) It is not that "the one implies the other", it is that the one is consistent with the other.  

In attempting to arrive at a functional specification for the "tax distortion costs" function, say $C_D=h(T)$, first we reason that if tax revenue is zero (not due to inefficiency of tax collection and/or tax evasion, but because the tax rate is set to zero), then distortions due to tax will also be zero. This implies that $h(T)$ should not contain a constant term, so that $h(0) = 0$.  

Moreover if we reason (as D. Romer does) that "distortions by taxes are likely to increase more than proportionally with the amount of revenue raised", this implies that $h(T)$ should not be linear in $T$, but it should be convex (this is what it means to say "more than proportionally"). 

The natural step then is to consider a quadratic distortion cost function, without a constant term:

$$C_D =  aT^2,\;\; a&gt;0$$

which reflects both our theoretical arguments (and in particular, $h(0) =0$).

But this also implies that "distortion costs are zero to a first-order approximation", because we have

$$h'(T) = 2aT \implies h'(0)=0$$

so a first-order approximation (Taylor expansion) around zero would give

$$h(T) \approx h(0) + h'(0)\cdot T = 0+0\cdot T =0$$