10 min read

Continuity: where the loop refuses to settle

Three ways a function can fail to have a limit at a point, and the one condition that makes substitution legal.

In the friendly examples so far, the limit at a point and the value at that point agreed, whenever the function had a value there at all. That agreement is not automatic. It is a property some functions have at some points, it has a name, and almost every convenient thing you will do with limits depends on it.

Here is a function that does not have it. A car park is free for the first 15 minutes. After that you pay 1.50 for every hour or part hour you have been there, counted from when you arrived.

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Add 15.0000001 to the list. However close you creep to 15 from above, the answer is still 1.50, and at 15 itself it is still 0.

Approach 15 minutes from below and the charges are all 0, so the left-hand limit is 0. Approach from above and they are all 1.50, so the right-hand limit is 1.50. The two sides disagree, so there is no limit at 15 at all, and no amount of creeping closer will change that.

The definition

A function $f$ is continuous at a point $a$ when three separate things are true. Each of them can fail on its own, which is why the definition is written as three.

$f(a)$ exists. The point is in the domain and the function has a value there.
$\lim_{x \to a} f(x)$ exists. Both one-sided limits exist and they agree.
The two are equal to each other.

\lim_{x \to a} f(x) = f(a)

All three conditions at once, since writing this down asserts that both sides exist.

Read that as a licence rather than as a description. Continuity at $a$ is precisely the permission to compute the limit by substituting. It is why so much limit arithmetic looks like no work: for continuous functions the answer is the value, and polynomials, sines, cosines and exponentials are continuous everywhere.

The three ways it fails

Kind	Example	What the numbers do
Removable	$\frac{x^2-9}{x-3}$ at $x = 3$	Both sides head for 6, but there is no value at the point
Jump	The car park charge at 15 minutes	The two sides head for different numbers, so there is no limit
Infinite	$\frac{1}{x}$ at $x = 0$	The values grow past any bound, so there is nothing to head for

Each failure breaks a different one of the three conditions.

Removable is the mildest, and the name says the cure: the limit exists, so you can define the missing value to be it and the function becomes continuous. Nothing forces you to choose that value, though, and choosing any other leaves a function that is perfectly well defined and still not continuous.

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Change the 6.0 to 5.0 and run it again. Nothing errors, the function is still defined everywhere, and it is now discontinuous at 3.

Probing a point from both sides

The pattern used throughout this section is worth extracting into something reusable: hand it a function and a point, and it prints what the function does just left and just right of that point, at shrinking distances.

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Change the last line to probe(lambda x: 3.0 * x + 1.0, 15.0). The two columns now close on the same number, which is what continuity looks like from a loop.

Be honest about what that tool can do. It probes finitely many points, and the previous lesson showed it cannot probe very close before the arithmetic betrays it. A function can look settled at every distance you tried and not be. The loop is evidence about where to look; the algebra is what decides.

A function that jumps at every whole number

f(x) = \lfloor x \rfloor \quad\text{(round down)},\qquad a = 2.5

The point

a

2.5How far the probe reaches0.1

Value at $a$: 2
From the left: 2
From the right: 2

Drag a across a whole number and watch the two probes disagree. Everywhere else the three numbers agree, and that agreement is continuity.

Let

f(x) = \frac{x^2 - 4}{x - 2}

for every

x \neq 2

. What single value must

f(2)

be given for

f

to be continuous at 2, and why is there only one choice?

It must be 4. Factoring gives $f(x) = x + 2$ for $x \neq 2$ , so $\lim_{x \to 2} f(x) = 4$ . Continuity requires the value to equal the limit, and the limit is a fixed number decided by the rest of the function, so there is exactly one value that works and every other choice leaves a discontinuity.

A function is defined as

f(x) = x + c

for

x < 1

and

f(x) = 4x

for

x \ge 1

. For which value of

c

f

continuous at 1?

The left-hand limit is $1 + c$ and the right-hand limit is 4, which is also the value $f(1)$ . Continuity needs all three equal, so $1 + c = 4$ and $c = 3$ . Problems of this shape are common on paper exams, and the method is always the same: write down the two one-sided limits and the value, then force them to agree.

Check yourself

$f(3)$ is defined and equals 5, and $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = 5$ . Which of these follows?

f

is continuous at 3, since the value exists, the limit exists, and they agree

f

is continuous everywhere, since it is continuous at 3Nothing follows without also knowing that

f

can be drawn without lifting the pen

f

is continuous at 3 only if

f

is also given by a single formula near 3

One thread runs through this whole section. You computed a difference quotient, watched it crowd around a number, found that the machine could not follow it all the way, and then learned the statement that says what the crowding means and the property that says when a value can be read off directly. That particular limit, the one where the step goes to zero, turns out to be worth a name and an algebra of its own, and that is where the next section starts.