>>1
I've likely got it. Δv/Δx and Δu/Δx have standard-parts dv/dx and du/dx respectively. This implies Δu and Δv/Δx are in the domain of the standard-part function and so we can use its identities. st(ΔuΔv/Δx) = st(Δu) * st(Δv/Δx) = 0. I have no idea why the book mentions the increment theorem in this proof as it is neither necessary nor useful so far as I can tell, but I would love a correction on this.
>>2 aye.
>>3
The standard-part function is undefined for infinite numbers like 1/Δx. An infinitesimal divided by another infinitesimal is indeterminate e.g. 𝛿=𝜀2 such that 𝜀/𝛿 = 1/𝜀. As you don't know if this fraction is in the domain of the standard-part function and you can't use its identities (which were presumably proven for its domain). In this case we know the real part of Δu/Δx so we just evaluate it to that, but for non-zero infinitesimals in general if you find a hole in the graph given the assumption that the infinitesimal is zero you should factor etc. so as to remove this hole or to prove that the standard-part is undefined independent of this.