@user21820: can you please have a look at the validity / correctness of this answer?

@ParamanandSingh It's wrong for exactly the reason that Bram28 said. And it bodes terribly for any answer to recommend Phil SE.

@user21820 I have forgotten way too much about Hilbert style :P

@user21820: thanks for the feedback.

I will check if this answer can be posted in CURED so that it is dealt via reviews.

@ParamanandSingh I posted it already.

@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.

@user21820 x = y/(1+y). |y| < ε|v|. So now we have to find a upper bound on |x|?

@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.

@user21820 So far I was only looking at this graph.

It seems like I can always find a upper bound inside the interval (-1,1).

@Prithubiswas Upload an image. External links easily go bad.

It is the graph of f(x) = |x/(1+x)|.

Ok so yes, you can do so inside (−1,1), but it's ugly, and we are free to shrink the interval further.

In some sense, you cannot avoid this, because if ε = 1, you still cannot do anything else except cut the interval.

@user21820 Is o(1) - o(1) ⊆ o(1)?

@Prithubiswas Right.

@user21820 This guess might be horrible , but is |x| < min(1,ε|v|)?

This problem is hurting my brain. Not because it is hard , but I am unsure about if I know what I am doing.

@Prithubiswas You should work with the simple 1/(1+o(1)) ⊆ 1+o(1) first. No point making things complicated with the v.

@Prithubiswas It's confusing if you forget the goal.

You want to bound x to make 1/(1+x) = 1+y where y is small.

I didn't really think it was useful to draw the backwards graph.

In the first place, I didn't think this was the right approach.

But it's workable; x = y/(1+y) means that all the large y will be excluded if you exclude large x (that is incompatible with this equation).

But this really is quite confusing. I think it's better to get a proof by using the intuition from the stronger asymptotic expansion:

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 ...