Continuity and the Intermediate Value Theorem

Q: What does it mean for a function to be continuous at a point?

f(a) is defined, the limit of f(x) as x approaches a exists, and that limit equals f(a).

Q: What are the types of discontinuity?

Removable (the limit exists but f(a) is missing or different), jump (the one-sided limits differ), infinite (a vertical asymptote) and oscillating (the values never settle).

Q: What is the Intermediate Value Theorem?

If f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c between a and b.

Q: How do I make a piecewise function continuous?

Set the one-sided limits at the break point equal to each other and to the function's value there, and solve for the unknown constant.

Q: Which functions are continuous?

Polynomials everywhere; rational, root, trigonometric, exponential and logarithmic functions on their domains; and sums, products, quotients and compositions of continuous functions where defined.

Quick answer

A function is continuous at a when f(a) is defined, the limit at a exists, and the two are equal. Discontinuities are removable (a hole or a misplaced point), jump, infinite or oscillating. Polynomials and rational, root, trigonometric, exponential and logarithmic functions are continuous on their domains. The Intermediate Value Theorem says a continuous function on [a, b] takes every value between f(a) and f(b), proving that equations have solutions without solving them.

What you'll learn

Test continuity at a point with the three-part definition
Classify discontinuities as removable, jump, infinite or oscillating
Choose a constant to make a piecewise function continuous
Use the Intermediate Value Theorem to show that an equation has a solution

The definition

A function $f$ is continuous at $a$ when all three of these hold:

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

In one line: $\lim_{x \to a} f(x) = f(a)$ . The first two conditions are what it takes for that equation to make sense.

A function is continuous on an interval when it is continuous at every point of it. At an endpoint, only the one-sided limit from inside the interval counts.

Four kinds of discontinuity

When continuity fails at $a$ , the graph shows how.

y = f(x)

Three discontinuities on one graph

Kind	What happens at $a$	In the graph
Removable	the limit exists, but $f(a)$ is missing or different	$x = -2$
Jump	the one-sided limits exist but differ	$x = 1$
Infinite	a one-sided limit is $\pm\infty$	$x = 3$
Oscillating	the values never settle, as $\sin\!\left(\tfrac{1}{x}\right)$ at $0$	—

A removable discontinuity is the only fixable kind: redefine $f(a)$ as the limit and it disappears. The other three have no limit to redefine to.

Why continuity means “limit equals value”

The limit is where the function is heading; the value is where it arrives. Continuity at $a$ says those are the same place, which is the precise meaning of drawing the graph through $a$ without lifting the pen.

It is also what makes direct substitution legitimate. For a continuous function, computing $\lim_{x \to a} f(x)$ comes down to computing $f(a)$ . That is why the long list of continuous functions matters:

polynomials are continuous everywhere;
rational, root, trigonometric, exponential and logarithmic functions are continuous at every point of their domains;
sums, differences, products, quotients (where the denominator is not $0$ ) and compositions of continuous functions are continuous.

Continuity is exactly the condition under which a limit can be found by plugging in.

The Intermediate Value Theorem

Intermediate Value Theorem

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 $c$ in $(a, b)$ with $f(c) = N$ .

A continuous graph cannot get from one height to another without passing through every height in between. The theorem is most often used with $N = 0$ : if $f(a)$ and $f(b)$ have opposite signs, $f$ has a root between $a$ and $b$ .

Worked examples

Common mistakes

Practice problems

Is $f(x) = \tfrac{x + 3}{x^2 - 9}$ continuous at $x = -3$ ? At $x = 3$ ? Classify any discontinuity.

Answer

Neither. Removable at $-3$ ; infinite at $3$ .

Full solution

$\tfrac{x + 3}{(x - 3)(x + 3)} = \tfrac{1}{x - 3}$ for $x \ne -3$ . At $-3$ the function is undefined but the limit is $-\tfrac{1}{6}$ : removable. At $3$ the values are unbounded: infinite.
Find $k$ so that $g(x) = \begin{cases} x^2 - k & x < 1 \\ 2x + 3 & x \ge 1 \end{cases}$ is continuous.

Answer

$k = -4$

Full solution

Left limit at $1$ : $1 - k$ . Right limit and value: $5$ . Setting $1 - k = 5$ gives $k = -4$ .
Classify the discontinuity of $f(x) = \tfrac{|x|}{x}$ at $0$ .

Answer

A jump

Full solution

For $x > 0$ the function is $1$ ; for $x < 0$ it is $-1$ . The one-sided limits exist and differ.
Where is $f(x) = \ln(x - 2)$ continuous?

Answer

On $(2, \infty)$

Full solution

A logarithm is continuous on its domain, and $\ln(x - 2)$ is defined exactly when $x - 2 > 0$ .
Show that $\cos x = x$ has a solution in $\left(0, \tfrac{\pi}{2}\right)$ .

Answer

Apply the Intermediate Value Theorem to $f(x) = \cos x - x$ .

Full solution

$f$ is continuous. $f(0) = 1 > 0$ and $f\!\left(\tfrac{\pi}{2}\right) = -\tfrac{\pi}{2} < 0$ . So $f(c) = 0$ for some $c$ in between, and there $\cos c = c$ .
What value should $f(0)$ have to make $f(x) = \tfrac{\sin x}{x}$ continuous at $0$ ?

Answer

$1$

Full solution

$\lim_{x \to 0} \tfrac{\sin x}{x} = 1$ , so defining $f(0) = 1$ makes the limit equal the value.
A continuous function has $f(1) = 5$ and $f(4) = -2$ . Must $f$ equal $3$ somewhere in $(1, 4)$ ? Must it equal $7$ ?

Answer

It must equal $3$ ; it need not equal $7$ .

Full solution

$3$ lies between $-2$ and $5$ , so the theorem applies. $7$ does not, so the theorem is silent: $f$ might reach $7$ or might not.
Find $a$ and $b$ so that $h(x) = \begin{cases} x + 1 & x < 1 \\ ax + b & 1 \le x < 3 \\ 3x & x \ge 3 \end{cases}$ is continuous.

Answer

$a = \tfrac{7}{2}$ , $b = -\tfrac{3}{2}$

Full solution

At $x = 1$ : $1 + 1 = a + b$ , so $a + b = 2$ . At $x = 3$ : $3a + b = 9$ .

Subtracting, $2a = 7$ , so $a = \tfrac{7}{2}$ and $b = 2 - \tfrac{7}{2} = -\tfrac{3}{2}$ .
Is $f(x) = \tfrac{x^2 + 1}{x^2 + 4}$ continuous everywhere?

Answer

Yes

Full solution

It is a quotient of polynomials, and the denominator $x^2 + 4$ is never $0$ . So it is continuous at every real number.
A student argues: ” $f(x) = \tan x$ has $f\!\left(\tfrac{\pi}{4}\right) = 1$ and $f\!\left(\tfrac{3\pi}{4}\right) = -1$ , so it has a zero between them.” What went wrong?

Hint

Is $\tan x$ continuous on $\left[\tfrac{\pi}{4}, \tfrac{3\pi}{4}\right]$ ?

Answer

$\tan x$ is not continuous there: it has a vertical asymptote at $\tfrac{\pi}{2}$ .

Full solution

The Intermediate Value Theorem needs continuity on the whole closed interval. $\tan x$ is undefined at $\tfrac{\pi}{2}$ , which lies in $\left[\tfrac{\pi}{4}, \tfrac{3\pi}{4}\right]$ .

In fact $\tan x$ has no zero in that interval: its zeros are at multiples of $\pi$ . The sign change comes from the asymptote, not from a root.

Frequently asked questions

What does it mean for a function to be continuous at a point?