@Prithubiswas I do Hilbert-style by translating it to Fitch-style. It is actually quite clear that Hilbert was thinking in Fitch-style when he came up with his axiom schemas. But he was concerned only about producing the right theorems, and not at all about practical usefulness.
@Prithubiswas What I wrote was for 1/(1+o(1)) ⊆ 1+o(1). You want a bound on |x| such that it makes |y| < ε.
The x = y/(1+y) will guide you to the right kind of bound.
It tells you that for y to be bounded by ε, you would need x to be bounded by ... Of course this doesn't produce any proof, since it is logically backwards, but then you try bounds of that kind and see what you can get.
It's just like if you believe that you can prove a theorem using induction in a certain manner, you would then tell yourself that for it to work you would need to prove the inductive step. This is logically backwards; even if you fail it doesn't mean that the desired theorem was false. But if you succeed then you succeed.
Alternatively, and more generally, you can take as an axiom 1/(1+x) ∈ 1−x+o(x) for x ≈ 0. Actually you have higher expansions as well, such as 1/(1+x) ∈ 1−x+O(x^2) for x ≈ 0.
It's easy to do this by simply replacing the asymptotic remainder with the actual remainder: 1/(1+x) = 1−x+x^2/(1+x).
So you know that if abs(x) < d, then 1/(1+x) = 1+y where abs(y) ≤ abs(x)+abs(x^2/(1+x)) < d+...
The ... is clearly small if d is small enough so that 1+x is close enough to 1.
Such as if d < 1/2.
In particular, if you bound the ... under assumption d < 1/2 by something simple in terms of d, then you can see how to prove 1/(1+o(1)) ⊆ 1+o(1) in ε-δ form.
Because to make abs(y)<ε you just need to make abs(x)<δ where δ = min(1/2,...) for some simple ...