How is the concept of the limit the foundation of calculus?

Solution 1:

Note: The concept of a limit had an enormous unifying impact in the beginning of the $19$th century at a time when there was a dissatisfaction with the logical status of analysis.

Some reasons were according to Morris Klines Mathematical Thought from Ancient to Modern Times; chap 40: The Installation of Rigor in Analysis :

• The very concept of a function was not clear, the use of series with regard to convergence and divergence had produced paradoxes and disagreements; the controversy about the representations of functions by trigonometric series had introduced further confusion and the fundamental notions of derivative and integral had never been properly defined.

• He cites Abel who complained in a letter of $1826$ about:

  the tremendous obscurity which one unquestionably finds in analysis. It lacks so completely all plan and system that it is peculiar that so many men could have studied it. The worst of it is, it has never been treated stringently. There are very few theorems in advanced analysis which have been demonstrated in a logically tenable manner. Everywhere one finds this miserable way of concluding from the special to the general and it is extremely peculiar that such a procedure has led to so few to the so-called paradoxes.

Rigorous analysis begins with the work of Bolzano, Cauchy, Abel and Dirichlet and was furthered by Weierstrass. We concentrate on Cauchy who introduced the concept of a limit systematically into analysis.

Cauchy's program:

Central to Cauchy's successful rigorization of the calculus was his simultaneous realization of two facts. First, that the eighteenth-century limit concept could be understood in terms of inequalities (given an epsilon, to find an $n$ or a delta).

Second, and more important, that once this had been done, all of the calculus could be based on limits thereby transforming previous results on continuous functions, infinite series, derivatives, and integrals into theorems in his new rigorous analysis. Though there were occasional gaps in his reasoning, he nevertheless far outdistanced his predecessors. And his work provided the necessary groundwork for the eventual complete rigorization of analysis by the school of Weierstrass.

Cauchy's sources:

The techniques of the algebra of inequalities came in large part from the works on approximations, Lagrange's systematic Equations numériques. Most of the other information Cauchy needed, and used, came from four other works from Lagrange and S.F. Lacroix. Lacroix's procedure in treating a topic was to summarize all the major work on it by the leading mathematicians. He, like most mathematicians of the time, wanted to show how to solve problems, therefore his Traité included whatever techniques were applicable to this end. Precisely because Lacroix's book is a mathematical museum of diverse methods and results, presented in full complexity, it could be of service to Cauchy.

Cauchy's definition of limit:

Cauchy defined the limit concept in these words:

When the successivley attributed values of the same variable indefinitely approach a fixed value, so that finally they differ from it by as little as desired, the last is called the limit of all the others.

This concept, translated into the algebra of inequalities, was exactly what Cauchy needed for his calculus. The very language of this verbal definition is sometimes taken to show the superiority of Cauchy's limit concept over all previous work.

Cauchy's definition is

• free from the idea of motion

• it does not depend on geometry

• it does not retain the unnecessary restriction, often included in the earlier definitions, that a variable could never surpass its limit.

This and much more information can be found in The Origin of Cauchy's Rigorous Calculus by Judith V. Grabiner.

Epilogue: One of the many benefits of the new limit concept was, that formerly difficult tasks could then be solved in a nearly automated process. I'd like to show at least one example, which nicely demonstrates the power and elegance of limits.

Let's have a look at the interesting improper integral

\begin{align*} \int_{0}^{\infty}\frac{1}{x^3-1}=-\frac{\pi \sqrt{3}}{9} \end{align*} We observe, that the integrand has a singularity at $x=1$ and it is blowing-up to minus infinity if $x$ approaches $1$ from values less than $1$ and it is blowing-up to plus infinity if $x$ approaches $1$ from values greater than $1$. But, what about the nice finite value on the RHS? It seems, that since the integral is signed, the infinities cancel away somehow. But how to show this rigorously? That's where the limits come into play!

Note that partial fraction decomposition gives

$$\frac{1}{x^3-1}=\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2\left[\left(x+\frac{1}{2})^2+\frac{3}{4}\right)\right]}$$

And integrating the individual terms on the RHS gives

\begin{align*} \int\frac{dx}{x^3-1}&=\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+x+1)-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\\ &=\frac{1}{6}\ln\left[\frac{(x-1)^2}{x^2+x+1}\right]-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\\ \end{align*}

The argument of the $\log$ function in the integration interval is well-behaved except at $x=1$ where we get $\ln(0)$. Therefore we consider

\begin{align*} \int_{0}^{\infty}\frac{dx}{x^3-1}&=\lim_{\varepsilon\rightarrow 0}\int_{0}^{1-\varepsilon}\frac{dx}{x^3-1} +\lim_{\varepsilon\rightarrow 0}\int_{1+\varepsilon}^{\infty}\frac{dx}{x^3-1}\\ &=\lim_{\varepsilon\rightarrow 0}\left(\left.\left[\frac{1}{6}\ln\left[\frac{(x-1)^2}{x^2+x+1}\right]-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]\right|_0^{1-\varepsilon}\right)\\ &+\lim_{\varepsilon\rightarrow 0}\left(\left.\left[\frac{1}{6}\ln\left[\frac{(x-1)^2}{x^2+x+1}\right]-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]\right|_{1+\varepsilon}^{\infty}\right)\\ &=\lim_{\varepsilon\rightarrow 0}\left(\frac{1}{6}\ln\left[\frac{\varepsilon^2}{(1-\varepsilon)^2+(1-\varepsilon)+1}\right]\right.\\ &\qquad-\left.\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2(1-\varepsilon)+1}{\sqrt{3}}\right) +\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)\\ &+\lim_{\varepsilon\rightarrow 0}\left(-\frac{1}{\sqrt{3}}\tan^{-1}(\infty)-\frac{1}{6}\ln\left[\frac{\varepsilon^2}{(1+\varepsilon)^2+(1+\varepsilon)+1}\right]\right.\\ &\qquad\left.+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2(1+\varepsilon)+1}{\sqrt{3}}\right)\right)\\ &=\lim_{\varepsilon\rightarrow 0}\left(\frac{1}{6}\ln\left[\frac{\varepsilon^2+3\varepsilon+3}{\varepsilon^2-3\varepsilon+3}\right]\right)-\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}\right)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ &-\frac{1}{\sqrt{3}}\tan^{-1}(\infty)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}\right)\\ &=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\tan^{-1}(\infty)\\ &=\frac{1}{\sqrt{3}}\left(\frac{\pi}{6}-\frac{\pi}{2}\right)\\ &=-\frac{\pi\sqrt{3}}{9} \end{align*}

I found this example in Paul J. Nahin's Inside Interesting Integrals

Solution 2:

The OP asked (1) how the limits save the calculus, and more specifically (2) how the epsilon-delta definitions save the calculus. The distinction is well-taken and is helpful in clarifying the issues.

It is well known that there are two main tracks for the development of the calculus: (A) the approach exploiting an Archimedean continuum, where for example limits and continuity are defined via epsilon-delta; and (B) the approach exploiting what could be called a Bernoullian continuum, i.e., a continuum featuring infinitesimals. Johann Bernoulli was the first one to use systematically infinitesimals and differentials to found the calculus, whereas his teacher Leibniz exploited both (A-track) exhaustion techniques and (B-track) infinitesimal techniques.

The distinction between the OP's questions (1) and (2) is significant because limits are actually present in both the A-track and the B-track. Thus, one can define the limit of a function $f(x)$ as $x\to 0$ via epsilon, delta, but one can also define it as the standard part of $f(\alpha)$ where $\alpha\not=0$ is an infinitesimal.

This is a point that is often overlooked in historical commentary as well as in some of the answers above. Thus, to Cauchy the primitive concept was that of a variable quantity, and he defined both limits and infinitesimals in terms of this primitive concept. Cauchy's definition of continuity of $f$ is not at all what one would expect it to be based on traditional historiography. Cauchy says that $f$ is continuous if every infinitesimal increment $\alpha$ always produces an infinitesimal change $f(x+\alpha)-f(x)$ in the function.

To answer the OP's question, the reason epsilon-delta definitions so to speak "save" the calculus is because they are formulated in an A-track fashion. This "saves" the calculus because mathematicians around 1870 have not yet developed a theory of infinitesimals that was thought satisfactory. Therefore some of them thought that the A-track was the only one available. As mentioned in some of the other answers, this foundational difficulty was overcome about 60 years ago by Abraham Robinson, so today it is the infinitesimal B-track that "saves" the calculus as far as both the students and the professors are concerned.