Math Core

Informally, a function is continuous if you can draw its graph without lifting your pencil. Limits let you make that idea precise, and the precise version pays off: continuous functions are exactly the ones where limits can be found by plugging in, and they come with a powerful guarantee about hitting every in-between value.

Continuity at a point

Where could a pencil be forced to lift at $x = a$ ? There could be a hole in the graph, the graph could jump, or it could shoot off toward infinity. In each case something goes wrong with $f(a)$ , with the limit, or with how the two match up. The definition rules out all three.

Definition

Continuous at a point

A function $f$ is continuous at $x = a$ if all three of these conditions hold:

$f(a)$ is defined.
$\displaystyle \lim_{x \to a} f(x)$ exists.
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

If any condition fails, $f$ is discontinuous at $a$ .

Condition 3 is the heart of it: the value the function is heading toward is the value it actually has. A function is continuous on an interval if it is continuous at every point of that interval.

Three kinds of discontinuity

The graph below shows one of each kind.

A removable discontinuity at x = −3, a jump at x = −1 and an infinite discontinuity at x = 3.Open in grapher →

Removable discontinuity (at $x = -3$ ). The limit exists, but the function is undefined there or has the wrong value. The graph has a hole. It is called removable because you could fix it by defining (or redefining) a single value: set $f(a)$ equal to the limit.
Jump discontinuity (at $x = -1$ ). The one-sided limits exist but are different, so the graph jumps from one height to another. No single value can repair it.
Infinite discontinuity (at $x = 3$ ). The function grows without bound near $a$ , and the graph has a vertical asymptote.

Notice that the graph is continuous at $x = 2$ , even though the formula changes there: the flat piece reaches height $-1$ and the curve $y = \dfrac{1}{x - 3}$ starts at height $\dfrac{1}{2 - 3} = -1$ . The pieces meet, so there is no break.

Which functions are continuous?

Here is the good news. Almost every function you have studied is continuous wherever it is defined.

Continuous families

Polynomials are continuous everywhere.
Rational functions are continuous at every point of their domain (everywhere except where the denominator is $0$ ).
Root, exponential, logarithmic and trigonometric functions are continuous on their domains.
Sums, differences, products, quotients (where the denominator isn't $0$ ) and compositions of continuous functions are continuous.

This is why direct substitution works: for a continuous function, $\displaystyle \lim_{x \to a} f(x) = f(a)$ is exactly the definition. When you look for discontinuities, focus on the places where a formula breaks down (zeros of denominators) or where a piecewise function switches formulas.

Worked example: Classifying discontinuities

Find and classify the discontinuities of $f(x) = \dfrac{x^2 - 1}{x^2 - x - 2}$ .

Solution. A rational function is continuous except where its denominator is $0$ . Factor both parts:

f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 1)}.

The denominator is $0$ at $x = -1$ and $x = 2$ .

At $x = -1$ the factor $x + 1$ cancels, so for $x \ne -1$ , $f(x) = \dfrac{x - 1}{x - 2}$ . The limit is $\dfrac{-2}{-3} = \dfrac{2}{3}$ . The limit exists but $f(-1)$ does not, so this is a removable discontinuity (a hole at $\left(-1, \tfrac{2}{3}\right)$ ).
At $x = 2$ the numerator of the simplified form is $1$ , not $0$ , so the values blow up. This is an infinite discontinuity (a vertical asymptote).

Worked example: A piecewise function

Is $g(x) = \begin{cases} x^2, & x < 1 \\ 3 - x, & x \ge 1 \end{cases}$ continuous at $x = 1$ ?

Solution. Check the three conditions. First, $g(1) = 3 - 1 = 2$ , so it is defined. Next, the left-hand limit is $1^2 = 1$ and the right-hand limit is $3 - 1 = 2$ . They differ, so the limit does not exist and condition 2 fails. The function has a jump discontinuity at $x = 1$ .

Worked example: Choosing a constant for continuity

Find the value of $k$ that makes $f(x) = \begin{cases} kx + 1, & x < 2 \\ x^2 - k, & x \ge 2 \end{cases}$ continuous everywhere.

Solution. Each piece is a polynomial, so the only possible trouble spot is $x = 2$ . For continuity, the left-hand limit, the right-hand limit and $f(2)$ must all match:

\begin{aligned} \lim_{x \to 2^-} f(x) &= 2k + 1, \\ \lim_{x \to 2^+} f(x) &= f(2) = 4 - k. \end{aligned}

Set them equal: $2k + 1 = 4 - k$ , so $3k = 3$ and $k = 1$ . Check: both pieces give $3$ at $x = 2$ .

Common mistake

A function can be defined at a point and still be discontinuous there. Being defined is only the first of the three conditions. Always check that the limit exists and equals the value.

The Intermediate Value Theorem

Continuity has a striking consequence. If a continuous graph starts below a height and ends above it, it has to cross that height somewhere in between, because it can't jump over it.

Intermediate Value Theorem (IVT)

If $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there is at least one number $c$ in $(a, b)$ with $f(c) = N$ .

The most common use is locating zeros. If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $0$ is between them, so $f$ has a zero in $(a, b)$ .

Worked example: Locating a zero

Show that $f(x) = x^3 + x - 1$ has a zero between $0$ and $1$ .

Solution. $f$ is a polynomial, so it is continuous on $[0, 1]$ . Compute $f(0) = -1$ and $f(1) = 1 + 1 - 1 = 1$ . The value $0$ lies between $-1$ and $1$ , so by the IVT there is a number $c$ in $(0, 1)$ with $f(c) = 0$ .

Tip

The IVT tells you a solution exists, not where it is. To zoom in, split the interval in half and check the sign at the midpoint. Repeating this is how calculators hunt down zeros.

Practice

Practice 1

Which condition of continuity fails for $f(x) = \dfrac{x^2 - 9}{x - 3}$ at $x = 3$ ?

Practice 2

List every $x$ -value where $f(x) = \dfrac{x + 4}{x^2 - 16}$ is discontinuous.

Separate answers with commas, e.g. 2, -5

Practice 3

The function $f(x) = \dfrac{x^2 + x - 6}{x - 2}$ has a removable discontinuity at $x = 2$ . What value should $f(2)$ be given to make $f$ continuous there?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

What kind of discontinuity does $f(x) = \dfrac{|x|}{x}$ have at $x = 0$ ?

Practice 5

Find $k$ so that $f(x) = \begin{cases} x^2 + k, & x \le 3 \\ 2x + 5, & x > 3 \end{cases}$ is continuous everywhere.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find $c$ so that $f(x) = \begin{cases} cx^2, & x < 1 \\ 4x - c, & x \ge 1 \end{cases}$ is continuous everywhere.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Let $f(x) = x^3 - 4x + 1$ . On which interval does the Intermediate Value Theorem guarantee a zero of $f$ ?

Practice 8

Let $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \ne 2$ , and $f(2) = 3$ . Which statement is true at $x = 2$ ?