Limits

The last chapter ended with two questions that turn out on inspection to be one. We found the tangents to the ellipse and the parabola by a trick, through mirrors, but the graph of $y = x^3$ or a sine wave has no mirrors. And a planet, as the law of areas requires, flies faster near the Sun than far from it, but how fast is it flying now, at this very instant? Average speed is distance divided by time. In one instant there’s neither distance nor time, and speed “at an instant” looks like zero divided by zero.

This question is where calculus begins, the part of the course about how quantities change. Everything in it rests on one notion, the limit. For a century and a half mathematicians used limits without being able to say what they were, and got correct answers from arguments that didn’t stand up to criticism. A precise definition was finally found, and it’s built as a game against a stubborn opponent. In this chapter we’ll learn its rules and play a few rounds.

Ghosts of departed quantities

Drop a stone from a tall tower. Ignoring air resistance and rounding the acceleration due to gravity to $10$ m/s², in $t$ seconds the stone falls $s(t) = 5t^2$ metres: the familiar parabola from the chapter on quadratic equations. How fast is it falling exactly one second after it’s released?

There’s nothing to measure instantaneous speed with. But average speed is easy to compute. Take a short stretch of time, from $1$ to $1 + h$ seconds. In it the stone falls $s(1 + h) - s(1) = 5(1 + h)^2 - 5 = 10h + 5h^2$ metres, and the average speed over this stretch is

$$v(h) = \frac{10h + 5h^2}{h} = 10 + 5h.$$

The shorter the stretch, the closer the average speed is to ten:

The answer suggests itself: the speed after one second is $10$ m/s; just substitute $h = 0$ into $10 + 5h$. But you can’t substitute. The expression $10 + 5h$ came from dividing by $h$, and you can divide only while $h \ne 0$. At $h = 0$ the original fraction becomes $\frac00$ and means nothing: in zero time the stone falls zero metres.

That’s exactly how Isaac Newton and Gottfried Wilhelm Leibniz calculated when, in the second half of the seventeenth century, they independently created what we now call calculus. Leibniz published his rules first, in 1684, in the paper “A New Method for Maxima and Minima”. He called increments like our $h$ infinitely small. Newton, in the Mathematical Principles of Natural Philosophy (1687), spoke of “the ultimate ratios of evanescent quantities”: the ratio that quantities have not before they vanish and not after, but at the very moment they vanish. The methods worked: orbits, tides and the flight of projectiles were computed with them. But what the moment of vanishing was, nobody could properly explain.

In 1734 the Irish philosopher and bishop George Berkeley published a pamphlet, The Analyst, addressed “to an infidel mathematician” (usually taken to mean Edmond Halley). Berkeley struck exactly at our spot. First the increment is taken to be not zero, or else you couldn’t divide by it. Then it’s taken to be zero, and the term $5h$ is thrown away. One or the other: either $h$ is zero or it isn’t, and here it’s been both in a single argument. Then comes the famous passage:

And what are these Fluxions? The Velocities of evanescent Increments. And what are these same evanescent Increments? They are neither finite Quantities nor Quantities infinitely small, nor yet nothing. May we not call them the Ghosts of departed Quantities?

Berkeley didn’t dispute Newton’s answers; he disputed that they had been proved, and he was right. The answer to his objection was found only in the nineteenth century, and it’s unexpected: $h$ is never set equal to zero. Instead one proves that $10 + 5h$ differs from $10$ by as little as you like, provided $h$ is small enough. And the words “as little as you like” and “small enough” are turned into moves in a game.

There are two players. The first is us: we claim a quantity tends to some number. The second is an opponent who takes nothing on trust. Let’s call them the Skeptic, in Berkeley’s honour. Each round goes the same way: the Skeptic says what accuracy they demand, and we show where that accuracy is reached. Beating the Skeptic whatever their move is exactly what it means to prove.

The rules of the game

Let’s play the first rounds on sequences from chapter 13: there time is measured by the indices $n = 1, 2, 3, \dots$ Take the sequence $a_n = \frac{n + 1}{n}$, whose terms are $2$, $1.5$, $1.333\ldots$, $1.25$, $1.2$, …

We claim it tends to one. The Skeptic: “How close? Here’s a tolerance, $\varepsilon = 0.01$. Show me from which point on the terms differ from one by less than that.” We compute: $a_n - 1 = \frac1n$, and $\frac1n < 0.01$ for $n > 100$. We answer: “Every term after the hundredth.” The first hundred terms don’t get into the band, but they don’t have to: the game is about what happens further on. The Skeptic narrows the tolerance to $0.0001$, and we answer “after the ten-thousandth”. For any tolerance $\varepsilon$ we have an answer ready: an index $N$ no smaller than $\frac1\varepsilon$. The Skeptic has lost, and in all rounds at once.

In 1821 Augustin-Louis Cauchy, in the Course of Analysis written for the École Polytechnique, defined the limit in words: if the values of a variable approach a fixed number without bound, so that they end up differing from it by as little as you like, that number is called the limit. The game turns each word of this sentence into a move.

The Skeptic’s move: “for every $\varepsilon > 0$”. They name a tolerance, as small as they like but positive. The symbol $\forall$ reads “for every”. Our answer: “there is an $N$” (the symbol $\exists$). We choose the index after hearing $\varepsilon$, so it may depend on $\varepsilon$. “For all $n > N$”: we answer for the whole tail of the sequence, not for one lucky term. The first $N$ terms can behave however they like. The term lies in the band of width $2\varepsilon$ around $a$: it differs from $a$ by less than $\varepsilon$. Example: for $a_n = \frac{n+1}{n}$ and $\varepsilon = 0.01$, $N = 100$ will do: for $n > 100$ we have $|a_n - 1| = \frac1n < \frac1{100}$. Any index greater than a hundred will do too.

Behind the answer “$N$ no smaller than $\frac1\varepsilon$” hides a property of numbers that seems obvious: whatever positive number is named, there’s a natural number larger than it. Take the whole-number part of its decimal and add one. This property is called the Archimedean property: for Archimedes it was an axiom about lengths, that any segment, however short, laid off enough times, exceeds any given one. From it follows the simplest and most needed limit: $\frac1n \to 0$. Indeed, if $N \ge \frac1\varepsilon$, then for $n > N$ we get $\frac1n < \frac1N \le \varepsilon$.

The sequence $a_n = 3 - \frac2n$ tends to $3$. The Skeptic names $\varepsilon = 0.01$. What is the smallest $N$ you can answer with?

$|a_n - 3| = \frac2n$, and $\frac2n < 0.01$ exactly when $n > 200$. So $N = 200$: after the two-hundredth term all the rest are in the band, and the two-hundredth itself, $a_{200} = 2.99$, lies exactly on its edge and isn’t in the band, but it isn’t required to be. Any $N$ greater than two hundred will do too.

Not every sequence can be brought to a number. Take $(-1)^n$: $-1, 1, -1, 1, \dots$ Whoever declares its limit to be some $a$, the Skeptic answers with the tolerance $\varepsilon = 1$. The band $(a - 1, a + 1)$ has width $2$, and two numbers a distance $2$ apart could fit in it only at its very edges, and the edges aren’t in the band. The numbers $-1$ and $1$ occur in the tail after any $N$, and at least one of them will be outside the band. Here the Skeptic wins.

And there’s the answer to Berkeley’s ghosts. There’s no number among the reals that’s smaller than every positive number and yet not zero: if $x > 0$, then $\frac x2$ is a smaller positive number. An infinitesimal is a process: a quantity that in time becomes smaller than any tolerance. The sequence $\frac1n$ is infinitesimal, and each of its terms is a perfectly ordinary positive number.

Before playing on, the Skeptic is entitled to ask: couldn’t one sequence tend to two different numbers at once? Then the notation $\lim a_n = a$ would mean nothing.

The trick is to choose a tolerance at which the bands around the two “limits” don’t overlap.

Let the sequence tend both to $a$ and to $b$, with $a \ne b$. The terms of the sequence are plotted as points above their indices. Play the Skeptic and name the tolerance $\varepsilon = \frac{|a - b|}{2}$, half the distance between $a$ and $b$. Since $a_n \to a$, there’s an index $N_1$ after which all the terms lie in the band $(a - \varepsilon, a + \varepsilon)$. Since $a_n \to b$, for the same tolerance there’s an index $N_2$ after which all the terms lie in the band $(b - \varepsilon, b + \varepsilon)$. The bands don’t overlap. Each reaches from its centre half the distance between $a$ and $b$, so they meet only at the midpoint between $a$ and $b$, and that point belongs to neither: the edges aren’t in the bands. Take an index $n$ larger than both $N_1$ and $N_2$. The term $a_n$ must lie in both bands at once, and they have no common points. A contradiction: there can be only one limit.

Let’s test intuition on a sequence where it often lets you down.

What does $\left(1 + \frac1n\right)^n$ tend to?

To $1$: after all $1 + \frac1n \to 1$, and one to any power is one To infinity: the exponent grows without limit To a number around $2.718$ To $2$

The terms are: $2$, $2.25$, $2.370$; at $n = 10$ already $2.5937$; at $n = 100$, $2.7048$; at $n = 1000$, $2.7169$. They grow, but ever more slowly, and close in on the number $e = 2.71828\ldots$, one of the main numbers of mathematics. Why the limit exists, and where else this number hides (in a bank, in radioactive decay, in the hat-check problem), is told in the chapter on the number $e$.

An old debt: Fibonacci’s rabbits

In chapter 13 we saw that the ratios of neighbouring Fibonacci numbers, $1, 2, 1.5, 1.667, 1.6, 1.625, \dots$, oscillate around the golden ratio $\varphi = \frac{1 + \sqrt5}{2} \approx 1.618$. There we managed to prove half of it: if the ratios have a limit, it’s $\varphi$. That the limit exists was left owing. Now we have the language to say it precisely, and the game to win it.

The bound is generous: in fact $r_{10} = \frac{89}{55} = 1.61818\ldots$ differs from $\varphi$ by only $0.00015$, while the bound promises less than $0.0082$. But generosity doesn’t prevent a win: the game doesn’t require the best index, just some index.

Along the way let’s pay back the debt of chapter 12. There it was said that the powers $2^{1.4} \approx 2.639$, $2^{1.41} \approx 2.657$, $2^{1.414} \approx 2.6647$ “close in” on the number $2^{\sqrt2} = 2.66514\ldots$ Now this can be said precisely: for any $\varepsilon$ there’s an index after which all the powers with exponents from the decimal approximations of $\sqrt2$ differ from $2^{\sqrt2}$ by less than $\varepsilon$. Why such a sequence has a limit at all (it increases and doesn’t exceed $2^{1.5}$) is proved in chapter 53.

A game on a graph

Back to the stone. There $h$ approaches zero smoothly, without any indices, and the rules have to change a little. The Skeptic still names a tolerance $\varepsilon$: a band around the number $10$ on the speed axis. But we have to answer with a window around zero on the time axis: a number $\delta > 0$ such that for any $h$ in the window $(-\delta, \delta)$, except zero itself, the average speed lies in the band.

The deviation of the average speed from ten is $|10 + 5h - 10| = 5|h|$. It’s less than $\varepsilon$ when $|h| < \frac\varepsilon5$, and we answer $\delta = \frac\varepsilon5$. The Skeptic demands an accuracy of $0.01$ m/s: we take stretches of time shorter than $0.002$ s. They demand $0.000001$: shorter than $0.0000002$ s. For every move there’s an answer, and the stone’s speed one second after release is $10$ m/s. Nowhere did we set $h$ equal to zero: zero is punched out of the window. No ghosts were needed.

The Skeptic names a tolerance: a band of width $2\varepsilon$ around $L$, vertically. We answer with a window of width $2\delta$ around $a$, horizontally; $\delta$ may depend on $\varepsilon$. All the $x$ in the window except the point $a$ itself. The inequality $0 < |x - a|$ punches it out: what happens at the point $a$ is none of the limit’s business. The value of the function lies in the band. In the picture: the piece of the graph above the window fits entirely in the band. Example: the stone, $v(h) = \frac{5(1 + h)^2 - 5}{h}$, $a = 0$, $L = 10$. To $\varepsilon = 0.1$ we answer $\delta = 0.02$: for $0 < |h| < 0.02$ we have $|v(h) - 10| = 5|h| < 0.1$.

This notation, with the symbols $\varepsilon$ and $\delta$ and the strict order “for every… there is…”, was brought into use by Karl Weierstrass in his Berlin lectures of the 1860s; it reached print largely through his students. Since then every university course in analysis has started with it.

Sometimes the neighbours on the left and on the right behave differently. The function $\operatorname{sgn} x$ (“the sign of $x$”) equals $-1$ for negative $x$, $1$ for positive $x$ and $0$ at zero. Whatever $L$ we name, the Skeptic says “$\varepsilon = 1$”: every window around zero contains both values $-1$ and values $1$, and they won’t fit together in one band of width $2$, exactly as with $(-1)^n$. There’s no limit at zero. But if you look only at the neighbours on the right, they all equal one.

For the sign function $\lim\limits_{x \to 0^+} \operatorname{sgn} x = 1$, and $\lim\limits_{x \to 0^-} \operatorname{sgn} x = -1$: the one-sided limits exist but differ, so there’s no two-sided one.

The game can be played around things other than numbers. The notation $\lim\limits_{x \to 0^+} \frac1x = +\infty$ means a round in which the Skeptic names a height $M$, and we must find a window to the right of zero where the function is above $M$: $\delta = \frac1M$ will do. And the notation $\lim\limits_{x \to +\infty} \frac1x = 0$ means a round where instead of a window around a point we name a ray $x > R$ on which $\frac1x$ lands in the band $(-\varepsilon, \varepsilon)$: $R = \frac1\varepsilon$ will do. The form of the moves changes; the meaning of the game stays the same: to every challenge from the Skeptic we have an answer.

Find $\lim\limits_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.

The fraction isn’t defined at $3$, but a limit is about the neighbours. For $x \ne 3$ the numerator factors: $x^2 - 9 = (x - 3)(x + 3)$, and the fraction equals $x + 3$. Values near three are close to $6$: for a tolerance $\varepsilon$, $\delta = \varepsilon$ will do, since $|(x + 3) - 6| = |x - 3|$. Answer: $6$.

Winning strategies

Playing every round from scratch is tiring. Chess players remember openings, and we’ll have strategies that win whole classes of rounds at once. The main one assembles complicated limits from simple ones.

The idea: split the Skeptic’s tolerance in half and give one half to $f$, the other to $g$.

The Skeptic names a tolerance $\varepsilon$ for the sum. The deviation of the sum $f(x) + g(x)$ from $A + B$ is made up of two deviations: $\bigl(f(x) - A\bigr) + \bigl(g(x) - B\bigr)$. Play a round with $f$, setting ourselves the tolerance $\frac\varepsilon2$. Since $f(x) \to A$, there’s a $\delta_1$: for $0 < |x - a| < \delta_1$ the value $f(x)$ differs from $A$ by less than $\frac\varepsilon2$. Likewise with $g$: there’s a $\delta_2$ at which $g(x)$ differs from $B$ by less than $\frac\varepsilon2$. Take $\delta = \min(\delta_1,\ \delta_2)$: in this window both conditions hold. The deviation of the sum is two arrows placed end to end, each shorter than $\frac\varepsilon2$. Together they’re no longer than the sum of their lengths, and so shorter than $\varepsilon$: $|(f - A) + (g - B)| \le |f - A| + |g - B| < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$. To every tolerance $\varepsilon$ we’ve answered with a window $\delta$, so $f(x) + g(x) \to A + B$. For sequences, instead of the smaller window you take the larger index, $N = \max(N_1,\ N_2)$.

From this theorem and two of the simplest limits, $\lim\limits_{x \to a} c = c$ for a constant and $\lim\limits_{x \to a} x = a$ (for a tolerance $\varepsilon$, $\delta = \varepsilon$ will do), a great deal can be assembled. A polynomial is built from $x$’s and numbers by additions and multiplications, so its limit at any point equals its value: $\lim\limits_{x \to 2} (x^3 - 4x + 1) = 8 - 8 + 1 = 1$. The same goes for a fraction of polynomials, if the denominator isn’t zero at the point.

When it is zero, the quotient rule is silent: the expression $\frac00$ says nothing about the limit; it only signals that another strategy is needed. The first is to cancel a common factor, as with $\frac{x^2 - 9}{x - 3}$. Cancelling is legitimate because the limit doesn’t look at the point itself, and near it the factor isn’t zero. The second strategy, at infinity, is to divide the numerator and denominator by the highest power:

$$\frac{2x^2 + 1}{5x^2 - x} = \frac{2 + \frac{1}{x^2}}{5 - \frac1x} \to \frac{2 + 0}{5 - 0} = \frac25 \qquad (x \to \infty).$$

Find $\lim\limits_{x \to \infty} \dfrac{(x + 1)(2x - 3)}{4x^2 + x}$.

Expand the brackets: the numerator is $2x^2 - x - 3$. Divide the numerator and denominator by $x^2$: $\frac{2 - \frac1x - \frac3{x^2}}{4 + \frac1x} \to \frac24 = \frac12$. The leading terms decide everything: at infinity the fraction behaves like $\frac{2x^2}{4x^2}$.

If you come across a limit that won’t give in, the limit solver will take it apart step by step and explain which strategy worked.

Under the microscope: the sine and the arc

The most famous limit of school calculus doesn’t give in to our strategies:

$$\lim_{x \to 0} \frac{\sin x}{x}.$$

Both the numerator and the denominator tend to zero, and there’s nothing to cancel: the sine isn’t a polynomial, and no factor $x$ is in sight. Let’s compute (angles in radians):

It looks like one. To understand why, let’s look at the graph under a microscope.

The circle shows where the one comes from: for a small angle the arc hardly differs from the half-chord. To turn this “hardly” into a proof we need one more strategy. If a quantity is hemmed in on both sides by two others and both tend to the same number, it has nowhere else to go.

In English it’s called the squeeze or sandwich theorem; elsewhere students know it jokingly as the theorem of the two policemen: two policemen lead a detainee by the arms, and wherever they go, the detainee goes too.

If the Skeptic has lost to both policemen, the detainee takes the Skeptic down with them.

The graph of $g$ is below, $h$ above, $f$ between them. The Skeptic names a tolerance $\varepsilon$: a band around $L$. Since $g(x) \to L$, there’s a window $\delta_1$ in which $|g(x) - L| < \varepsilon$, in particular $g(x) > L - \varepsilon$: the lower policeman doesn’t drop below the band. Since $h(x) \to L$, there’s a window $\delta_2$ in which $h(x) < L + \varepsilon$: the upper one doesn’t rise above the band. In the smaller window, $\delta = \min(\delta_1,\ \delta_2)$, both conditions hold, and $L - \varepsilon < g(x) \le f(x) \le h(x) < L + \varepsilon$. The graph of $f$ above the window lies entirely in the band. For every tolerance there’s a window, so $f(x) \to L$.

The idea: nest three figures inside one another whose areas are expressed through $\sin x$, $x$ and $\tan x$, and squeeze the fraction between two policemen.

Take $0 < x < \frac\pi2$ and lay off the angle $x$ from the radius $OA$ of the unit circle; its other arm is the radius $OB$. The tangent to the circle at the point $A$ meets the ray $OB$ at the point $T$. Triangle $OAB$: base $OA = 1$, and the height from $B$ is $\sin x$. Its area is $\frac12 \sin x$. The sector $OAB$. The area of a sector is proportional to its angle: the whole disc of radius $1$, with angle $2\pi$, has area $\pi$ (chapter 19), so the sector with angle $x$ has area $\frac x2$. Triangle $OAT$: the tangent is perpendicular to the radius, so it’s right-angled, with legs $OA = 1$ and $AT = \tan x$. Its area is $\frac12 \tan x$. Triangle $OAB$ lies inside the sector, and the sector inside triangle $OAT$, so $\frac12\sin x < \frac x2 < \frac12\tan x$, that is, $\sin x < x < \tan x$. Divide by the positive $\sin x$: $1 < \frac{x}{\sin x} < \frac{1}{\cos x}$, and turn the fractions over: $\cos x < \frac{\sin x}{x} < 1$. For negative $x$ it’s the same, because $\frac{\sin(-x)}{-x} = \frac{\sin x}{x}$ and $\cos(-x) = \cos x$. The upper policeman is the constant $1$. The lower one, $\cos x$, also tends to one. By the double-angle formula for the cosine, which follows from the addition formula (chapter 21), $\cos x = 1 - 2\sin^2\frac x2$, and the inequality $|\sin t| < |t|$ we’ve just obtained gives $0 \le 1 - \cos x < 2\left(\frac x2\right)^2 = \frac{x^2}2$. So $1 - \frac{x^2}{2} < \cos x \le 1$, and by the same theorem $\cos x \to 1$. The fraction $\frac{\sin x}{x}$ is squeezed between $\cos x$ and $1$, and both policemen tend to one. By the squeeze theorem $\frac{\sin x}{x} \to 1$. The half-chord: the height of the point $B$ of the unit circle above the radius $OA$. The arc $AB$ of the same circle: the angle in radians. For small angles the half-chord and the arc are almost the same: $\sin x \approx x$. Example: $\sin 0.1 = 0.099833\ldots$, a ratio of $0.99833$. A pendulum swung out by $10^\circ \approx 0.1745$ radians: $\sin 0.1745 \approx 0.1736$, and replacing the sine by the angle is off by only half a per cent. That’s why the formulas for a pendulum’s small swings are so simple.

Isn’t this circular?

It’s sometimes said that this proof goes round in a circle: the area of a disc is supposedly derived from the very limit we’re proving. That’s not so. Archimedes found the area of a disc by the method of exhaustion, inscribing and circumscribing polygons, with no sine at all (chapter 19). The radian measure is also defined without a limit: an angle is measured by the length of an arc of the unit circle. The weak spot is elsewhere: to speak of the length of an arc and the area of a sector with full rigour, these notions themselves need precise definitions. They’re given through limits and integrals (chapter 28), and the sine can be defined by a series (chapter 30). None of these definitions relies on the fundamental trigonometric limit.

And if the angle is measured in degrees? What is $\lim\limits_{x \to 0} \dfrac{\sin x^\circ}{x}$?

$1$ $\frac{\pi}{180} \approx 0.0175$ $\frac{180}{\pi} \approx 57.3$ There’s no limit

$x$ degrees is $t = \frac{\pi x}{180}$ radians, so $\frac{\sin x^\circ}{x} = \frac{\pi}{180} \cdot \frac{\sin t}{t}$, and $t \to 0$ along with $x$. The limit is $\frac{\pi}{180} \approx 0.0175$. That’s why calculus measures angles in radians: only there is the limit one, and the derivative of the sine, as we’ll see in the next chapter, the cosine with no extra factor.

Find $\lim\limits_{x \to 0} \dfrac{\sin 3x}{2x}$.

Make the denominator match the sine’s argument: $\frac{\sin 3x}{2x} = \frac32 \cdot \frac{\sin 3x}{3x}$. As $x \to 0$, $t = 3x \to 0$ too, and $\frac{\sin t}{t} \to 1$. Answer: $\frac32$. Numerical check: at $x = 0.001$ the fraction is $1.49999775\ldots$

A gallery of discontinuities

For most of the functions we’ve dealt with, the limit at a point is simply the value: plug in and you have it. This property is called continuity.

The limit exists: the neighbours of the point $a$ have agreed where to go. The value at the point itself exists and coincides with where the neighbours are going. Example: $f(x) = x^2$ at the point $3$. The limit is $9$ by the arithmetic of limits, and so is the value: the function is continuous. But $\frac{x^2 - 9}{x - 3}$ isn’t defined at $3$, and it can’t be called continuous there, although it has a limit.

By the arithmetic of limits, polynomials are continuous everywhere, and fractions of them wherever the denominator isn’t zero. The sine is continuous too. The inequality $|\sin t| \le |t|$ holds for all $t$: for $|t| < \frac\pi2$ we’ve just proved it, and for $|t| \ge \frac\pi2$ it follows from $|\sin t| \le 1 < \frac\pi2$. Hence also $1 - \cos t = 2\sin^2\frac t2 \le \frac{t^2}{2}$ for all $t$. By the addition formula $\sin(a + h) - \sin a = \sin a\,(\cos h - 1) + \cos a \sin h$, so $|\sin(a + h) - \sin a| \le \frac{h^2}{2} + |h|$. For $|h| \le 1$ this is at most $\frac32|h|$, and for a tolerance $\varepsilon$, $\delta = \min\left(1,\ \frac\varepsilon2\right)$ will do. The other functions we’ve met in the course, roots, exponential functions, logarithms, are also continuous on their domains; the proofs are in chapters 29 and 53.

On a graph, continuity is visible at once: you can draw it without lifting the pencil from the paper. Where the pencil has to be lifted, discontinuities live.

Discontinuities aren’t only found in textbooks. The postage on a letter rises with its weight in steps: a letter just over a price boundary costs noticeably more. The function $\frac1x$ runs off to infinity near zero. And $\sin\frac1x$ oscillates faster and faster near zero: on any interval to the right of zero, however tiny, it runs through all values from $-1$ to $1$ infinitely many times, and no candidate for its limit survives a round with the Skeptic.

The most instructive exhibit is $x \sin\frac1x$. It too oscillates infinitely often, but the amplitude shrinks: $-|x| \le x\sin\frac1x \le |x|$. The policemen $-|x|$ and $|x|$ tend to zero, and by the squeeze theorem the limit at zero is $0$. At zero itself the function isn’t defined, but the discontinuity is removable: set $f(0) = 0$ and you get a function continuous everywhere. No hand can draw it “without lifting the pencil”: near zero the graph trembles infinitely many times. The pencil rule is a good hint, but the definition remains the limit.

A root that can’t be hidden

Continuous functions can do what you intuitively expect of them: if in one place a function is negative and in another positive, somewhere in between it’s zero. A temperature that dropped overnight from plus five to minus three was exactly zero at some moment. It sounds obvious, and for a long time it was taken as such. In 1817 Bernard Bolzano, a priest and mathematician in Prague, published a “purely analytic proof” of this fact: he realised that a picture proves nothing here, and that one has to rely on the properties of the numbers themselves. In 1821 Cauchy gave his own proof in the Course of Analysis.

Why it’s a matter of numbers can be seen from an example. The function $f(x) = x^2 - 2$ on the interval $[1, 2]$ is continuous, $f(1) = -1 < 0$, $f(2) = 2 > 0$. If the line contained only fractions, there’d be no root: $\sqrt2$ isn’t a fraction (chapter 6). The graph would slip across the axis through a hole. The theorem rests on the real numbers having no holes, and this property has to be written down explicitly.

In chapter 6 this property already flashed by: each new digit of a decimal narrows the window tenfold, and a real number is what’s left when the windows shrink without end. For fractions the principle fails: the intervals $[1, 2]$, $[1.4, 1.5]$, $[1.41, 1.42]$, … around $\sqrt2$ are nested, but they have no common fraction. We accept this principle as an axiom: it’s one form of the property called the completeness of the real numbers. In chapter 53 it will be derived from a rigorous construction of $\mathbb R$.

Keep halving the interval, each time keeping the half whose ends still have opposite signs.

For definiteness let $f(a) < 0 < f(b)$. Write $[a_0, b_0] = [a, b]$. Take the midpoint $m$. If $f(m) = 0$, the root is found. If $f(m) > 0$, keep the left half $[a, m]$; if $f(m) < 0$, the right half $[m, b]$. In both cases $f$ is negative at the left end of the remaining interval and positive at the right. Repeat the same again and again (unless a root turns up earlier). We get nested intervals $[a_n, b_n]$ of length $\frac{b - a}{2^n}$, and on each $f(a_n) < 0 < f(b_n)$. By the nested interval principle they have a common point $c$. Let’s show that $f(c) = 0$. Suppose $f(c) < 0$. Play against continuity with the tolerance $\varepsilon = |f(c)|$: there’s a window $(c - \delta, c + \delta)$ in which $f(x)$ differs from $f(c)$ by less than $|f(c)|$ and so is negative. But for large $n$ the length $\frac{b - a}{2^n}$ is less than $\delta$, and the interval $[a_n, b_n]$, which contains $c$, lies entirely in the window. Then $f(b_n) < 0$, yet we know $f(b_n) > 0$. A contradiction. In the same way $f(c) > 0$ is impossible: then $f$ is positive in a window around $c$, and for large $n$ we’d get $f(a_n) > 0$. That leaves $f(c) = 0$: the root exists. The proof also suggests how to find it.

Hence the theorem’s name. Apply it to the function $f(x) - C$: if a continuous function takes the values $f(a)$ and $f(b)$, it also takes every intermediate value $C$ between them.

The method in the proof is called bisection, or interval halving. It’s slow but reliable: each step adds one correct binary digit, and ten steps narrow the interval by a factor of $2^{10} = 1024$. Many root-finding programs use it as a safety net under clever fast methods: where a clever method might miss, bisection won’t.

Against the cunning Skeptic you can’t do better than bisection. They don’t even have to choose a function in advance: it’s enough that their answers don’t contradict one another. After each question they leave the root in the longer part of the interval, and it’s always longer than half unless you asked exactly at the midpoint. So in $k$ questions the interval can’t be narrowed by more than a factor of $2^k$, and for an accuracy of $0.001$ on an interval of length $1$ you need ten questions: $2^9 = 512 < 1000 < 1024 = 2^{10}$.

The intermediate value theorem pays back one more debt, from chapter 14.

That’s what the chapter on polynomials promised: the graph of a cubic can cross the axis at one, two or three points, but it can’t slip past it.

Practice

Rounds with the Skeptic are needed to prove the rules. Computing limits by those rules is a separate craft, and it’s learned by hand. The trainer has three levels: substitution and cancelling fractions, limits at infinity, and limits that reduce to the fundamental trigonometric one.

A step-by-step breakdown of any limit of your own is in the solver, and all the course’s trainers are on the practice page.

Where next

We started with the stone: the average speed over the stretch of time $[1, 1 + h]$ is $10 + 5h$, and its limit as $h \to 0$ is the speed one second after release, $10$ m/s. The same move works for any motion and any function. The average rate of change of $f$ over the stretch from $x$ to $x + h$ is $\frac{f(x + h) - f(x)}{h}$, and its limit as $h \to 0$ answers the question the chapter began with: how fast is a quantity changing at a given instant. On a graph the same limit is the slope of the tangent, for the parabola, for $y = x^3$ and for a sine wave alike, with no mirrors. This limit has a name: the derivative. A speedometer needle shows it, and a few rules help to compute it, with no need for rounds against the Skeptic. Where those rules come from, and what a speedometer really shows: chapter 26.