Differentiation - Understanding the basic idea of a limit I have been working on Differentiation, and though I can proceed mechanically with the process I can't reconcile how taking the limit works. 
I understand that if I take the $ \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$ then as $h$ tends to 0 the answer comes closer to the derivative at the point $x$. The issue I have is that if $h = 0$ then it all falls apart but if $h$ doesn't equal $0$ then there will always be an error in my answer.  
I have spent a lot of time reading different books and the closest I have come to finding an explanation I understand is by reading Judith Grabiner - The Origins of Cauchy's Rigorous Calculus and also her paper at MAA Grabiner Newton Calculus
In the paper (pg. 398-399), Grabiner explains how Colin MacLaurin used reductio ad absurdum to prove that the derivative of $x^n$ was $nx^{n-1}$.
In the book Grabiner goes on to show the following (which I have interpreted to be based along a similar idea to what Colin MacLaurin did - it is a translation from the work of Cauchy here from page 44 onwards.
Extract 1 from book
Extract 2 from book
Extract 3 from book
Extract 4 from book
I follow the broad strokes of what is happening up to the last part - I've highlighted in yellow - where the $\epsilon$ is dropped from A and B and the $\gt$ and $\lt$ become $\ge$ and $\le$.
The way I am reading it (and I accept that I could be completely misreading it all), is that the current system of limits we use is built on Cauchy's work and that by removing the $\lt$ $\gt$ and putting in $\le$ $\ge$ it gives us that A must equal B and so the derivative at the point in question is exact and there is no error.
However, I can't understand how this happened.

The trick is that in a neighborhood of the point considered, if the function is differentiable you can write
$$f(x+h)=f(x)+mh+o(h).$$
In other words, the function is quasi-linear.
Then
$$\frac{f(x+h)-f(x)}h=m+\frac{o(h)}h$$
(you get the angular coefficient of the line with a perturbation), but when you take the limit, the perturbation vanishes and you get exactly
$$\lim_{h\to0}\frac{f(x+h)-f(x)}h=m.$$
Then you define $f'(x)$ to be that $m$.
