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Most students meet limits as a procedure: substitute, factorise, cancel. Behind that procedure sits a question that troubled mathematicians for more than a century after Newton and Leibniz: what does it mean for a quantity to "approach" a value? Augustin-Louis Cauchy gave one of the first careful answers, in a textbook written for engineering students. This post covers what he wrote, where he went wrong, and how his way of working can sharpen your own calculus.
Who was Augustin-Louis Cauchy?
Cauchy was born in Paris on 21 August 1789, the year the French Revolution began, and died at Sceaux, near Paris, on 23 May 1857, according to the MacTutor biography from the University of St Andrews. He entered the École Polytechnique in 1805, trained as an engineer, and from 1810 worked on harbour works at Cherbourg, taking books by Laplace and Lagrange with him. In 1815 he became assistant professor of analysis at the École Polytechnique, responsible for its second-year course.
He was extraordinarily productive. MacTutor counts 789 mathematics papers, and his collected works fill 27 volumes. His name is attached to the Cauchy integral theorem, the Cauchy-Riemann equations and Cauchy sequences. After the 1830 revolution he lost his posts because he would not swear an oath of allegiance to the new regime.
What problem was Cauchy trying to solve?
Eighteenth-century calculus worked brilliantly but rested on shaky ground. Mathematicians handled infinite series as if they were long polynomials and often assumed that a formula true in one case was true everywhere. In his Cours d'analyse of 1821, Cauchy wrote that he wanted his methods to have "all the rigour which is demanded in geometry", and warned that most algebraic formulas "are valid only under certain conditions and for certain values of the quantities they contain" (translation from MacTutor's page on Cauchy's calculus). His aim was to state those conditions.
How did Cauchy define a limit?
In MacTutor's translation, his definition reads: "When the values successively attributed to the same variable approach indefinitely a fixed value, eventually differing from it by as little as one could wish, that fixed value is called the limit of all the others."
The key phrase is "as little as one could wish". It turns a vague picture into a challenge: name any tolerance, however small, and the variable must eventually stay within it. Cauchy built the rest on this. A function is continuous if an infinitely small change in x always produces an infinitely small change in f(x). A series converges if its partial sums approach a limit, and that limit is its sum. The derivative is the limit of the ratio [f(x + i) − f(x)] / i as i shrinks, which he wrote as f′(x). Today's epsilon-delta definitions are more formal, but the idea of a limit as a tolerance you can always meet is already in his text.
An English translation by Robert Bradley and Edward Sandifer, Cauchy's Cours d'analyse (Springer, 2009), makes the whole book available to modern readers.
A worked example: Cauchy's test on two series
Take the series 1/2 + 1/4 + 1/8 + ... . Its partial sum after n terms is S_n = 1 − (1/2)^n, so the gap between S_n and 1 is (1/2)^n.
For a tolerance of 0.001 we need (1/2)^n < 0.001, that is, 2^n > 1000. Since 2^9 = 512 and 2^10 = 1024, every n from 10 onwards works.
For a tolerance of 0.000001 we need 2^n > 1,000,000. Since 2^19 = 524,288 and 2^20 = 1,048,576, every n from 20 onwards works.
Whatever tolerance you name, there is a point after which every partial sum stays within it. That is Cauchy's definition, so the sum is 1.
Now take the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... . Its terms shrink to zero, and many students conclude that it must converge. Cauchy pointed out that terms tending to zero is necessary but not sufficient. He also required that sums of consecutive terms, "taken, from the first, in whatever number we wish", eventually stay below any assigned amount. Test that on the block of terms from 1/(n + 1) to 1/(2n). There are n terms and each is at least 1/(2n), so the block adds up to at least n × 1/(2n) = 1/2. However far along you go, such a block never drops below 1/2. Cauchy's condition fails, and the series diverges: its partial sums grow without limit.
Where did Cauchy go wrong?
Rigour did not make Cauchy infallible. MacTutor notes that he "proved", incorrectly, that a convergent series of continuous functions always has a continuous sum, and that Niels Henrik Abel produced a counterexample. Repairing that gap needed a stronger idea, now called uniform convergence. The episode is a useful warning: a careful definition does not protect you if you skip a step when applying it.
What can a student take from Cauchy's way of working?
State the condition, not just the formula. Cauchy's complaint about older calculus was that formulas were used outside the cases where they hold. Before using a result, ask when it is true. Does the series converge? Is the function continuous on the whole interval?
Turn "approaches" into a number. When a limit feels slippery, pick a tolerance such as 0.001 and work out how far you must go to stay within it, as in the example above.
Test intuitions on a counterexample. "The terms go to zero, so the series converges" sounds right and is wrong. Keep a short list of standard counterexamples, starting with the harmonic series.
Expect to be corrected. Even Cauchy's theorems needed repair. Treat a mistake found in your work as part of the method.
Our guides to limits and continuity for JEE and sequences and series apply these ideas to exam problems. If a definition such as continuity still feels abstract, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
A short practice routine
Write Cauchy's definition of a limit in your own words, without looking.
For the series 1 + 1/3 + 1/9 + ..., find the sum and the smallest n for which the partial sum is within 0.001 of it. (Answer: the sum is 3/2, and n = 7, since 3^7 = 2187 is the first power of 3 above 1500.)
Add 1/5 + 1/6 + 1/7 + 1/8 and check that the total is at least 1/2, as the argument above predicts for n = 4.
Tonight, take one limit from your last test and rewrite the answer as a tolerance that the function can always meet.
