Continuity

A limit at a point is deliberately blind to the value of the function at that point, and the functions worth working with are precisely the ones where that blindness costs nothing.

The previous lesson built the limit as a promise about tolerances and showed that for polynomials and similar functions the limit is just the value, so the whole calculation is a substitution. That coincidence deserves a name and an investigation, because it is not automatic, and because the theorems that follow from it are what later lessons quietly rely on when they assert that a maximum exists or that an area is a number.

The definition, in three parts

A function $f$ is continuous at $a$ when

and that single equation is really three demands stacked together. The right side must exist, so $a$ must be in the domain of $f$. The left side must exist, so the function must approach some single value from both sides. And the two must be equal.

Any one of the three can fail on its own. The function $(x^2-9)/(x-3)$ has a perfectly good limit of $6$ at $x=3$ and no value there at all. A function defined to be $x^2$ everywhere except at $x=1$, where it is declared to be $7$, has both a limit ($1$) and a value ($7$), and they disagree. A step function has a value at the step and no limit.

The picture everyone is taught, that a continuous function can be drawn without lifting the pen, is a decent guide and a poor definition. It fails to say anything useful about a function on the rationals, or about $\sin (1/x)$, which cannot be drawn at all near zero yet is continuous at every point where it is defined. The tolerance definition survives all such cases, which is why it is the one that is used.

A function is continuous on an interval when it is continuous at every point of it, with the endpoints requiring only the one-sided limit that stays inside.

Example. For which value of $c$ is the function equal to $x^2$ when $x\le 2$ and to $cx+1$ when $x>2$ continuous everywhere?

Away from $x=2$ each piece is a polynomial and so is continuous. At $x=2$ the value and the limit from the left are both $4$. The limit from the right is $2c+1$. Continuity requires $2c+1=4$, so $c=1.5$. With that choice the right-hand formula at $x=2$ gives $1.5\times 2+1=4$, matching, and the graph has no step in it.

Now you. The function equals $(x^2-25)/(x-5)$ for $x\ne 5$ and equals $k$ at $x=5$. What $k$ makes it continuous?

The three ways it fails

Discontinuities come in a small number of shapes, and naming them is useful because the repair, if any, differs.

A removable discontinuity is a hole: the limit exists but the value is missing or wrong. It is removable in the literal sense that redefining the function at that one point fixes it, which is what the previous exercise did. Every cancelled factor in an indeterminate form leaves one of these behind.

A jump discontinuity is two one-sided limits that exist and disagree. These are not repairable by any single value, and they are common in real data rather than pathological: a postage rate as a function of weight, an income tax rate as a function of income, the density of water as a function of temperature at 0 degrees Celsius. Physical laws are usually written for continuous quantities, and a jump is where such a law needs separate treatment on each side.

An infinite discontinuity is a vertical asymptote, as in $1/x$ at zero, where no value assigned at the point can help because the function is unbounded nearby. Oscillation, as in $\sin (1/x)$ at zero, is the remaining case and is equally unrepairable.

Example. Classify the discontinuities of $f(x)=\frac{x^2-x-6}{x^2-9}$.

The denominator vanishes at $x=3$ and $x=-3$, so those are the only candidates. Factoring gives $(x-3)(x+2)$ over $(x-3)(x+3)$, so for $x\ne 3$ the function equals $(x+2)/(x+3)$. At $x=3$ that expression is $5/6=0.8333$, a perfectly good limit with no value attached, so the discontinuity is removable. At $x=-3$ the reduced form still has a zero denominator with a non-zero numerator, so the function blows up: an infinite discontinuity, and no redefinition will fix it.

Now you. Classify the discontinuities of $f(x)=\frac{x^2-1}{x^2-3x+2}$.

Almost everything is continuous

The limit laws convert directly into a catalogue. Sums, differences, products and quotients of continuous functions are continuous wherever they are defined, and so is a composition of continuous functions. Polynomials are continuous everywhere. Rational functions are continuous except where the denominator vanishes. Roots, sines, cosines, exponentials and logarithms are continuous throughout their domains.

The practical consequence is that continuity is the default. Any function written down by combining the standard ones is continuous except at points you can find by inspection: division by zero, an even root of a negative number, a logarithm of something non-positive, or a definition given in pieces. Everything else is safe.

That fact is what makes the next two theorems worth having. They are hypotheses-and-conclusion statements whose hypothesis is almost always satisfied, so in practice they hand out conclusions for free.

A sign change guarantees a root

The Intermediate Value Theorem says that if $f$ is continuous on the closed interval from $a$ to $b$, then $f$ takes every value between $f(a)$ and $f(b)$ somewhere in that interval. In particular, if $f(a)$ and $f(b)$ have opposite signs, there is a point where $f$ is zero.

It sounds like a restatement of the pen-drawing picture, and it is much more than that: it is an existence theorem that converts a comparison of two numbers into a guarantee about the existence of a solution. It says nothing about where the root is or how many there are, and it needs no formula for solving anything.

Turned into a procedure, it gives bisection. Take $f(x)=x^3-x-1$, which no factoring will crack since it has no rational roots. Then $f(1)=-1$ and $f(2)=5$, so a root exists between $1$ and $2$. Evaluate at the midpoint: $f(1.5)=0.875$, positive, so the sign change is now between $1$ and $1.5$. Then $f(1.25)=-0.296875$, so the root is between $1.25$ and $1.5$. Then $f(1.375)=0.224609$, then $f(1.3125)=-0.051514$, then $f(1.34375)=0.082611$, then $f(1.328125)=0.014576$. Six steps have trapped the root between $1.3125$ and $1.328125$, and each further step halves the interval. The root, to seven figures, is $1.324718$.

Bisection is slow and utterly reliable: the error is guaranteed to fall by half per step, so twenty steps take an interval of width one down to below a millionth. Later in the course a much faster method appears, one that uses derivatives and can fail, and the comparison between them is the usual trade in numerical work.

Example. Show that $\cos x=x$ has a solution between $0$ and $1$, and bracket it in three bisection steps.

Apply the theorem to $g(x)=\cos x-x$, which is continuous everywhere. Here $g(0)=1$ and $g(1)=0.5403-1=-0.4597$, opposite signs, so a solution exists. The midpoint gives $g(0.5)=0.377583$, so the root lies in $[0.5,1]$. Then $g(0.75)=-0.018311$, so it lies in $[0.5,0.75]$. Then $g(0.625)=0.185963$, so it lies in $[0.625,0.75]$. The true value is $0.739085$.

Now you. Show that $x^3=10$ has a solution between $2$ and $3$, and take two bisection steps.

Why the theorem needs the real numbers

The Intermediate Value Theorem is not a triviality, and the cleanest evidence is that it is false if the number system has gaps.

Work entirely within the rational numbers and consider $f(x)=x^2-2$. It is continuous as a function of a rational variable, $f(1)=-1$, and $f(2)=2$. The signs are opposite, yet there is no rational number where $f$ is zero, since the previous course proved by contradiction that $\sqrt{2}$ is irrational. Continuity and a sign change are not enough. What is also needed is that the line has no holes in it.

That property, called completeness, is the defining difference between the rationals and the reals, and every existence theorem in this course descends from it. Continuity is the hypothesis you check; completeness is the hypothesis that comes with the number system and is easy to forget. Cauchy, Weierstrass, Dedekind and Cantor spent the middle of the nineteenth century making it precise, and calculus was placed on solid ground only when they had.

A closed interval guarantees a maximum

The second existence theorem is the Extreme Value Theorem: a continuous function on a closed bounded interval attains a maximum and a minimum somewhere on it. Not approaches, attains, at actual points of the interval.

Both hypotheses are load-bearing, and dropping either kills the conclusion. The function $1/x$ on the half-open interval $(0,1]$ is continuous there and has no maximum, since it exceeds any bound near zero. The function $f(x)=x$ on the open interval $(0,1)$ is continuous and bounded, and still has no maximum: it gets arbitrarily close to $1$ and never attains it, and no point of the interval is the winner. Take the same function on the closed interval $[0,1]$ and the maximum exists, at the endpoint. And a function with a jump can miss its own supremum even on a closed interval.

This theorem is the licence for the whole optimisation lesson later. When a problem asks for the shape that minimises cost, the answer is presumed to exist, and this is where that presumption comes from: continuous function, closed interval, therefore a minimum exists, and the derivative can then be used to find where it is. Without the theorem, the derivative would only be able to say that if a minimum exists then it is at one of these points.

Continuity is not enough for a tangent

Continuity guarantees no holes, no jumps and no blowups. It does not guarantee that a curve has a direction at each point, and that gap is what the next lesson has to close.

The function $f(x)=|x|$ is continuous everywhere, including at zero, where the limit and the value are both zero. But it has a corner there. Approach the origin along the left branch and the slope is $-1$; approach along the right branch and it is $+1$. Any line through the origin is a bad fit to at least one side, and there is no sensible tangent. The secant slopes from the previous lesson do exist here, but the one-sided limits disagree, which is exactly the jump failure applied to the difference quotient rather than to the function.

How much worse this can get was settled by Weierstrass in 1872, when he exhibited a function continuous at every point of the real line and differentiable at none. It is built as an infinite sum of ever faster, ever smaller cosine waves, so that at every scale, no matter how far you zoom in, the graph is still wrinkled and never settles down to look like a line. Until then most mathematicians assumed a continuous function must have a tangent almost everywhere, and the example ended that assumption permanently. It also demonstrates that intuition about curves is not a reliable guide, which is a large part of why the tolerance definitions were adopted.

So a curve can be unbroken and still have no slope. Having a slope at a point is a strictly stronger condition, and defining it, computing it and using it is what the next several lessons are about.