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I'm having trouble with the limit approach to calculus ever since I heard about the infinitesimal definition. Maybe you can help me settle what's been bothering me this year.

Looking at the limit definition of the derivative equation makes sense. However, what trips me up is the fact that because the slope function is not defined when $\Delta x $ equals zero, how can we say the derivative is tangent instead of an infinitely accurate secant line? Because from my understanding, in order for it to be a tangent line, it intersects the curve at one point only, however $\Delta x$ approaches zero, it never reaches it, so $\Delta x$ must be greater than zero, however infinitesimally small, correct?

Mathematicians have generally abandoned this idea now from what I understand with the exception of non-standard analysis. Can somebody explain where my thinking is wrong?

rb612
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    The derivative isn't the tangent line itself, it's the slope of the tangent line, which is well-defined when the usual limit exists. – BrianO May 16 '16 at 04:53
  • Have only read the title. Answer - it is, with BrianO's qualification. –  May 16 '16 at 04:54
  • @selfawareuser you're saying it intersects at more than one point, just the distance between intersection is infinitely small? – rb612 May 16 '16 at 04:56
  • @BrianO yes you're right, fixed. But you didn't answer my question why it's the tangent and not secant line. – rb612 May 16 '16 at 04:57
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    @rb612 If the distance is infinitesimal then it's arbitrarily small (in colloquial usage it would be immeasureable). I think regardless of the exact 'philosophy' calculus has to start with ideas like these. –  May 16 '16 at 05:01
  • @selfawareuser definitely, and that's why my question has to do with why we've thrown those ideas out the window for general calculus. – rb612 May 16 '16 at 05:17
  • You've seen the usual drawings of $\Delta y, \Delta x$, yes? "tangent" means vertical distance $/$ horizontal distance. The derivative is the limit of slopes of tangent lines, I don't see what secants have to do with anything here. – BrianO May 16 '16 at 05:22
  • Just to point that zero is an infinitesimal quantity. Indeed is the unique infinitesimal quantity that is a real number. More important than infinitesimal is the concept of limit, it is one of the more important concepts in mathematics. – Masacroso May 16 '16 at 05:39
  • @BrianO the question is if the line with a slope of the derivative intersects at $f(x)$ and $f(\Delta x)$, which would make it a secant line? I'm saying if $|f(x)-f(\Delta x)| > 0$, it must intersect at $f(x)$ and $f(\Delta x)$. The tangent line is defined as being just touching at a specific point, so if it touched those two points, then it would be a secant line, no? – rb612 May 16 '16 at 05:57
  • It has to intersect $f$ at $f(x_0)$, because the tangent line itself is $$ y = f'(x_0)(x - x_0) + f(x_0). $$ – BrianO May 16 '16 at 06:01
  • @BrianO yes that's correct. I get that it intersects at $f(x_{0})$ I'm saying if it intersects at two points, $f(x_{0})$ and $f(x_{0}+\Delta x)$, because if $\Delta x$ is not zero, it must intersect there too from what I understand. Sorry my question is confusing. – rb612 May 16 '16 at 06:05
  • Ah ok. Sure, the tangent line at $x_0$ might intersect $f$ at another point $(x_1, f(x_1))$. But the $f'(x_1)$, if it exists, is in general different than $f'(x_0)$ — even if the tangent line at $x_0$ intersects $f$ at $(x_1, f(x_1))$. – BrianO May 16 '16 at 07:04
  • @BrianO "Tangent" does not mean $\Delta y/\Delta x$. In the standard definition of a derivative on a function such as $y=x^2$, all the lines that you get through a point on the curve by taking $\Delta x$ and $\Delta y$ to another point on the curve are called secant lines, and the tangent is the unique line the secant lines converge to as $\Delta x\to 0$. (You clearly know this, because you wrote the equation of that line correctly; I think all we are concerned with here is a difference in the meanings of a couple of words.) – David K May 17 '16 at 01:22
  • @DavidK I was thinking of "tangent" as the trig function, where it is that sort of ratio (as opposed to secant). And, yes, we're all concerned with fine points and trying to clarify the issue. – BrianO May 17 '16 at 02:29
  • @BrianO The multiple meanings of both "tangent" and "secant" (trig functions or geometric figures) really don't help make this discussion any easier, do they? But I see now what you meant. – David K May 17 '16 at 02:40
  • @DavidK I at first forgot the other (relevant) meaning of "secant". – BrianO May 17 '16 at 02:44

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Because from my understanding, in order for it to be a tangent line, it intersects the curve at one point only, however Δx approaches zero, it never reaches it, so Δx must be greater than zero, however infinitesimally small, correct?

You're right. We don't ever reach that point. We take a limit.

The colloquialism, "reaching the point" is a good anthropomorphic description. Limits allow us to stretch the constraints of the real numbers by pushing towards the infinite and infinitesimal. Technically, though, to venture into such territory, we need to properly define limits. This is often introduced with the epsilon-delta formalization.

Say there exists a limit $f'(x)=\lim_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$. Then for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $0<\Delta x<\delta$, we find $|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$.

We can heuristically think of the last paragraph as the following: our derivative exists if for every positive number $\epsilon$ and $\delta$, including the most ridiculously small numbers you can ever imagine, whenever $\Delta x$ is trapped between zero and any of these ridiculously small numbers, the difference between our derivative and the original expression is imperceptible.

But wait a minute, you say

...Δx must be greater than zero, however infinitesimally small, correct?

The epsilon-delta definition seems to hint that as well, but there's a catch: $$|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$$

This is not less than some real positive number $\epsilon$. This is less than ANY POSSIBLE real positive number $\epsilon$. Such a concept only exists within the formalism of a limit, and is by no means a measurable quantity. That's what is meant by infinitesimal.

Due to the limit, then, the derivative cannot represent any possible secant line. There are no two points corresponding to $x+\Delta x$ and $x$ that are indistinguishable! The value we reach has converged to that which represents the slope of the tangent.

Added note: $\Delta x\rightarrow 0$ doesn't just imply that $\Delta x$ is running through the positive numbers towards zero. For the limit to exist, we typically require it to be two-sided, meaning that $\Delta x\rightarrow0^+$ and $\Delta x\rightarrow0^-$ must produce the same result. In either case, the difference between $\Delta x$ and zero becomes vanishingly small.

zahbaz
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  • Thank you! So my next question is: you have something like an integral, where it's an infinite sum of derivative values (simplified a ton), why don't those infinitesimal values add up to be something significant? – rb612 May 16 '16 at 06:51
  • Do you mean, why do they add up to something significant? In the Riemann sum definition of an integral, loosely speaking, it's the infinite number of infinitesimally tiny pieces added together that gives rise to a substantial quantity. The infinite counters the infinitesimal, and you find something often in between. But, I realize that now it sounds like we're treating both infinity and the infinitesimals as real numbers. They are not. An integral is not merely a sum, but, you guessed it, a limit as well. The value we find is whatever value the limit converges to. – zahbaz May 16 '16 at 07:12
  • That'd make for another good question, though, and one which I haven't justly answered. Formulate it and post anew! – zahbaz May 16 '16 at 07:14
  • @rb612, you are absolutely right about the integral; it can also be treated using infinitesimals, and in terms of infinite sums of infinitely thin rectangles. This approach is explained in detail in Keisler's textbook Elementary Calculus. – Mikhail Katz May 16 '16 at 12:18
  • Thanks again zahbaz and thank you @MikhailKatz. Now my last followup question is this: Leibniz defined a tangent line as being the line between two infinitely close points on the curve. If I plot the graph $f(x)-f'x$, and it touches at exactly one point (meaning it's tangent), then by Leibniz's definition, is it also touching at another point infinitely close? – rb612 May 17 '16 at 05:08
  • The plot $f'(x)$ is tangent to $f$ at $x$. Leibniz's definition is an okay way to think about it, but his idea predates modern formalism. I suppose if you consider two points that are infinitely close to one another, then either a) you're really talking about one point (the limiting case) or b) you're talking about points whose coordinates cannot be distinguished by real numbers. – zahbaz May 17 '16 at 05:31
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    A number system that extends the reals to include infinitesimals would be the hyperreals. The link seems to suggest an immediate link to the definition of the derivative. You may also want to read about non-standard analysis.... But to really explore any of this, the prerequisite would be real analysis. – zahbaz May 17 '16 at 05:32
  • @rb612, the only prerequisite for learning calculus with infinitesimals is highschool algebra, reviewed in the first chapter of Keisler's textbook Elementary Calculus. As I mentioned in my answer below, Leibniz is working with a generalized notion of equality "up to". This means that the tangent line will not be exactly the same as the secant line for an infinitesimal increment, but rather it will be infinitely close to such a line. The secant line through two infinitely close points by definition passes through a second point. – Mikhail Katz May 17 '16 at 06:49
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You're treating the notion of "tangent" as pre-existing and asking how we can say that the derivative is (the slope of) the tangent. But the situation is more complex than that – by defining the derivative, we're defining what we mean by tangent in many cases in which there was no such pre-existing concept. So the task is to clarify the pre-existing, more restricted concept of "tangent" and check whether it coincides with the one induced by the derivative in the cases where it applies.

Your description of the tangent as intersecting the curve at one point doesn't capture the pre-existing concept of tangent. Let's consider four examples:

a) $f(x)=x^3$ at $x=1$. The tangent coincides with the curve at two points (at $x=1$ and at some negative $x$).

b) $f(x)=x$ at $x=0$. The tangent coincides with the curve everywhere. And what's worse, any other line through the origin intersects the curve only once – so by your definition the tangent is the only non-tangent.

c) $f(x)=x^3$ at $x=0$. Here the tangent indeed coincides with the curve in one point only, but the curve lies on different sides of it, and it's at least not obvious how to distinguish this from an ordinary line intersecting a curve without using calculus.

d) $f(x)=x^2\sin\frac1x$ at $x=0$. Here the tangent as defined by calculus (the $x$-axis) intersects the curve infinitely often in any neighbourhood of the origin. I don't know of any way of defining this as the tangent without calculus.

So in order to make your question more precise, you'd need to provide a clear definition of how "tangent" was used before the advent of calculus and what cases it applied to, and then we could check whether the calculus definition coincides with this (and expands it).

joriki
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  • Ah, thank you. I was referring to "just touching" as my definition of a tangent line, which would not be the case if it touched at 2 points, however small the distance between. – rb612 May 16 '16 at 06:51
  • @rb612: I'm not sure I understand what you mean by that. The two curve points that the upper horizontal tangent of $x^4-\lambda x^2$ touches are closer together the smaller $\lambda$ is. Is that not a tangent line according to your definition? Or if it is, what other scenarios did you intend to exclude by that? – joriki May 16 '16 at 09:09
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I like your starting line "I'm having trouble with the limit approach to calculus ever since I heard about the infinitesimal definition." +1 for that. The short answer to your question lies in that line because of the following non-mathematical theorem:

A non-mathematical theorem: There is no sound theory of infinitesimals available for a beginner in calculus and hence any approach based on infinitesimals is bound to cause confusion.

The fundamental problem with an infinitesimal is that it tries to describe something which is smaller than any positive number and not yet zero. No such thing exists in the real number system. There is a way to define "infinitesimals in sound manner" which is called "Non-standard Analysis" but it is not suitable for a beginner learning calculus.

It is somewhat ironical but also surprising that the rebirth of calculus happened with the idea of infinitesimals and many mathematicians faced similar confusion as OP and therefore there was a concerted effort by the mathematical community to discard the infinitesimals totally and present calculus in a manner which is sound and therefore far less confusing. There was another problem with calculus apart from the infinitesimals and it was that many important and deep theorems had no proofs and this last problem was fixed by a proper theory of real numbers.

In the previous paragraph I mentioned the "rebirth of calculus" because the birth of calculus happened long back in time of Archimedes and the Greeks were the ones who had developed calculus (mainly integral calculus in its basics) without infinitesimals and they also had a proper theory of real numbers. Unfortunately they did not have the differential calculus stuff which was championed so much by the likes of Newton and Leibniz that the very pure theory of calculus was forgotten for a long time.

It is our great luck that the infinitesimal approach has been discarded in our time. The definition of limit which you should focus on is the one involving $\epsilon, \delta$ because they are far less confusing if presented in proper manner. Also some calculus textbooks try to teach limits via jumping on to the notion of derivatives. A limit involved in the concept of a derivative is a "difficult type of limit" (namely the "indeterminate form" $0/0$) and it is better to avoid such an approach. A beginner is well off if he/she studies limit without knowing anything about derivatives and then later learn special kinds of limits which are called derivatives.

Anyway lets come back to tangents, secants, and derivatives. The geometrical concept of a secant is defined for any kind of curve without the need for calculus, but the notion of tangent to a point on a curve can not be defined in purely geometric manner without using calculus concepts. The concept of tangent does not exists apriori rather it is defined in terms of derivative. One exception is that a tangent to a circle can be defined without any appeal to calculus by simply defining it to be a line perpendicular to the radius at the point under consideration.

The idea of a secant line becoming a tangent as the points of secant come infinitely close to each other is a totally non-rigorous idea and it is used by instructors to somehow provide a geometric interpretation for the concept of derivatives. There is nothing like "infinitely" close. Two points are either same or at some specific distance to each other. A secant is line which essentially requires two points on curve. A tangent normally deals with one point on the curve. There is thus no way a secant can become a tangent.

What calculus has to offer here is the following. Let $f$ be a function defined on a certain interval $I$ and let $C$ be the curve which is the graph of the function and let $c$ be any specific point in interval $I$. Let the point $P$ on the curve $C$ be $(c, f(c))$. Consider another point $x \in I$ such that $x \neq c$ and let $Q = (x, f(x))$ be the corresponding point on the curve. The equation of the secant $PQ$ is given by $$Y - f(c) = \frac{f(x) - f(x)}{x - c}(X - c)$$ The slope of this secant $PQ$ is $$\frac{f(x) - f(x)}{x - c}$$ and sometimes for some functions $f$ it is possible that the limit $$\lim_{x \to c}\frac{f(x) - f(x)}{x - c} = f'(c)$$ exists. When this happens we say that the curve $C$ possesses a tangent at point $P$ and its equation is $$Y - f(c) = f'(c)(X - c)$$ There are some corner cases to consider when the above limit is $\pm\infty$ but apart from that the above constitutes the definition of a tangent to a curve at a point $P$.

Paramanand Singh
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    I wonder what your sources are for your sweeping and misleading claims. – Mikhail Katz May 16 '16 at 15:44
  • @MikhailKatz: It would be better if you can let me know some specific claim and then I can try to improve my answer. – Paramanand Singh May 16 '16 at 15:48
  • I was referring to your sweeping claim that there is no theory of infinitesimals suitable for a freshman calculus course. Having spent the past three years teaching calculus with infinitesimals to freshmen I find such an unsourced claim puzzling. – Mikhail Katz May 16 '16 at 15:56
  • @MikhailKatz: OK I got it. My point was not to disrespect any methodology of teaching. Also I was referring "beginner in calculus" and in our country this means students of age around 16 years. I think you have pointed out yourself about George Berkeley in your answer. I read most of such problems of calculus in early stages from the book "A Radical Approach to Real Analysis" by David M. Bressoud. – Paramanand Singh May 16 '16 at 16:13
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    @MikhailKatz: BTW I don't see how we can avoid the theory of limits based on $\epsilon, \delta$ definition (or something equivalent) and the theory of real numbers and instead justify the infinitesimals in some other manner. – Paramanand Singh May 16 '16 at 16:22
  • If this was not your intention then you should delete the surprising comment to the effect that A non-mathematical theorem: There is no sound theory of infinitesimals available for a beginner in calculus and hence any approach based on infinitesimals is bound to cause confusion. I think I showed above that your theorem is false. – Mikhail Katz May 16 '16 at 16:40
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    @MikhailKatz: After your comments I did have a reading of the introductory chapter on Hyperreals from Keisler's book and I find that it adds a lot of new stuff (finite, infinite, infinitesimals) and gives a lot of new rules to deal with these. I must say getting to reals from rationals itself is a very difficult step and add to that the complexity of construction of hyperreals and proving various principles (transfer, extension, standard part) its all very complex. I don't see how that much of stuff is simpler than traditional $\epsilon, \delta$. – Paramanand Singh May 16 '16 at 16:50
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    @MikhailKatz: I stand by what I have written in the post. I accept your downvote. I will try to study more from Keisler's book and if my views change I will change my post accordingly. – Paramanand Singh May 16 '16 at 16:59
  • Just because you "can't see" something doesn't mean you can make unsourced blanket claims. – Mikhail Katz May 16 '16 at 17:00
  • You can "stand by" whatever you wish of course but the question is whether you actually have any support for sweeping statements like "It is our great luck that the infinitesimal approach has been discarded". There is much evidence, including controlled studies like that by Sullivan, that contradict your dubious claims. – Mikhail Katz May 20 '16 at 07:23
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    @MikhailKatz: I found some nice discussion on the use of Non-standard analysis at http://math.stackexchange.com/q/51453/72031 – Paramanand Singh May 20 '16 at 09:57
  • Indeed this popular answer by user @Rachel seems to refute the unsupported claims you made in your answer here. – Mikhail Katz May 22 '16 at 09:08
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To answer the query contained in the title of your question, the derivative is the shadow of the ratio of infinitesimals $\frac{\Delta y}{\Delta x}$. Sometimes the shadow is referred to as the standard part. Thus even in the infinitesimal approach the derivative is not exactly the ratio but only the ratio rounded off to the nearest real number. This involves an infinitesimal change in value.

The issue of "how can you divide by zero?" is one that has bothered many authors historically, including George Berkeley. In a number of recent articles we have clarified the situation. The situation is that Leibniz already provided a satisfactory account in terms of his generalized relation of equality "up to" (a negligible term). Not everybody recognized this and in particular Berkeley did not recognize this. In retrospect it is clear that Leibniz's theoretical framework for justifying infinitesimals was more soundly based than Berkeley's empiricist (and somewhat confused) criticism thereof.

Mikhail Katz
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