# 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.

• – Hans Lundmark Jan 5 '18 at 15:38

I'm going to deal with the specific point you highlighted, and please comment if you'd like to move toward the broader issues you've raised.

Specifically, you're pointing to a step in the argument where we move from $$(\forall \varepsilon>0)(A - \varepsilon < \gamma < B + \varepsilon)$$ to $$A \leq \gamma \leq B$$

The argument is precisely the MacLaurin reductio ad absurdum argument you pointed to in the paper. It runs as follows (in modern terminology and notation).

First, suppose $A \not\leq\gamma$, i.e., $\gamma < A$. Then $A-\gamma > 0$. Now, by the density property of real numbers, we can find some real number $\varepsilon$ between $A-\gamma$ and $0$, i.e., $(\exists\varepsilon)(A-\gamma>\varepsilon>0)$. Since $\varepsilon>0$, we know from the first equation above that $A - \varepsilon <\gamma$, i.e., $A - \gamma < \varepsilon$, which contradicts $A - \gamma > \varepsilon$. So we must have $A \leq \gamma$.

Now suppose $B \not\geq \gamma$; then $\gamma-B>0$, and the rest of the argument runs similarly to the last paragraph (with $\gamma-B$ replacing $A-\gamma$).

• What you mean to say in the sentence that begins, 'Since $\varepsilon>0$...' is not clear. Seems some editing is needed. – Allawonder Jan 5 '18 at 16:24
• @Allawonder I'm not the best writer. What's not clear about it? – BallBoy Jan 5 '18 at 16:25
• For one, how does $A-\varepsilon<\gamma$ imply that $A-\varepsilon>\gamma$? – Allawonder Jan 5 '18 at 16:36
• @Allawonder Ah, that would definitely be unclear :) Edited – BallBoy Jan 5 '18 at 16:38
• Thank you for above explanation. I understand where $A \leq \gamma \leq B$ now comes from. This tells me that $\gamma$ is bound by $A$ and $B$? When I was reading Colin MacLaurin's example I was under the impression that $y =nx^{n-1}$ because there is no other value possible. However, I don't think Cauchy's formula is quite the same as it gives us a region so to speak which contains the "true value" but there is an error because no matter how close we get we never reach the definitive value? When I say "true value" I am thinking that a tangent line to a curve is a unique point on the curve. – MathQuest Jan 5 '18 at 16:41

See Y. Forman's answer for a formal argument as you wish to clarify, but as for the general intuitive notion of finding the limit of a function at a point of its domain (or as the independent variable becomes arbitrarily large or small), here:

As you said above, the idea is that of better and better approximations of the function as one moves towards the 'point' of interest. In this process, some functions tend to a particular goal, so to speak. There's a specific pattern to the approximations, namely they get nearer and nearer to a certain value. It is this value which they never quite strike, but which they always approach, that we call the limit of the function as the independent variable nears a certain value, or becomes very large or very small, as the case may be.

This is the same thing with differentiation. Whenever the limit of the ratio of interest (which approximates the gradient of the tangent line to the graph of the function at a point) seems to always point in a particular direction, so to speak, we say that we can find the derivative of this function. All this can be made rigorous, but that's the basic idea.

This is why the notion of limits is the fundamental one in all of analysis. It's worth thinking of for long before moving on to other stuff.

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 $m$.