Continuity

A function is continuous at a point if its graph has no holes, jumps, or vertical breaks there. Continuity is a powerful property: continuous functions on closed intervals attain maximum and minimum values, and they satisfy the Intermediate Value Theorem.

Concept	Quick Reference
Continuous at $a$	$f (a)$ defined, $lim_{x \to a} f (x)$ exists, and $lim_{x \to a} f (x) = f (a)$
Discontinuous at $a$	if any of the three conditions above fails
Removable discontinuity	limit exists but $f (a)$ is undefined or $\neq$ limit
Jump discontinuity	left and right limits exist but differ
Infinite discontinuity	one or both one-sided limits are $\pm \infty$
Oscillating discontinuity	$f (x)$ neither approaches a number nor $\pm \infty$

Intuitive Introduction

Consider a function $f$ whose graph is shown below.

Graph of a function that is discontinuous at x=-3, -1, 3, and 6

Intuitively, $f$ is discontinuous at $x = - 3, - 1, 3$ , and $6$, we must lift the pen to draw the graph at those points. At all other points, $f$ is continuous.

Continuity at a Point

The function $f$ is continuous at $a$ if:

$f$ is defined at $a$ , and
$lim_{x \to a} f (x) = f (a)$ .

Condition (2) implicitly requires that the limit exists and equals the function value. So when $f$ is continuous at $a$ , all three of the following hold:

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

If any of these three conditions fails, then $f$ is discontinuous at $a$ .

Right and Left Continuity; Continuity on an Interval

Left-continuity: $f$ is left-continuous at $a$ if $lim_{x \to a^{-}} f (x) = f (a)$ .
Right-continuity: $f$ is right-continuous at $a$ if $lim_{x \to a^{+}} f (x) = f (a)$ .
Continuity at an endpoint: Since two-sided limits do not exist at endpoints of an interval, continuity at an endpoint always means one-sided continuity.
Continuity on an open interval $(a, b)$ : $f$ is continuous at every point in $(a, b)$ .
Continuity on a closed interval $[a, b]$ : $f$ is continuous on $(a, b)$ , right-continuous at $a$ , and left-continuous at $b$ . The graph including endpoints can be drawn without lifting the pen.

Types of Discontinuity

1. Removable Discontinuity

If $lim_{x \to a} f (x) = L$ exists but either $f (a)$ is not defined or $f (a) \neq L$ , the discontinuity is removable. There is a hole in the graph at $x = a$ .

Graph showing a removable discontinuity (hole) at x=a

To remove this discontinuity: simply define (or redefine) $f (a) = L$ .

2. Jump Discontinuity

$f$ has a jump discontinuity at $x = a$ if the right and left limits exist but are not equal:

lim_{x \to a^{-}} f (x) \neq lim_{x \to a^{+}} f (x) .

Graph showing a jump discontinuity at x=a

3. Infinite Discontinuity

$f$ has an infinite discontinuity at $x = a$ if one or both one-sided limits equal $+ \infty$ or $- \infty$ .

Graph showing an infinite discontinuity at x=a

4. Oscillating Discontinuity

This occurs when $f (x)$ does not approach a number or infinity as $x \to a$ . For example, $f (x) = \sin (1 / x)$ oscillates between $- 1$ and $1$ infinitely often as $x \to 0$ (the following figure), so its limit at $0$ does not exist.

Graph of y=sin(1/x) showing oscillating discontinuity at x=0

**A function discontinuous everywhere: the Dirichlet function.** Consider the Dirichlet function: D(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} This function is discontinuous at every point. To see why: suppose

D (x)

had a limit

L

at some point

a

. Taking

ϵ = 1 / 2

, there should exist

δ > 0

such that

0 < | x - a | < δ \Rightarrow | D (x) - L | < 1 / 2

. But every deleted neighborhood

0 < | x - a | < δ

contains a rational point

x_{1}

and an irrational point

x_{2}

. Then:

| D (x_{1}) - L | = | 1 - L | < \frac{1}{2} and | D (x_{2}) - L | = | 0 - L | < \frac{1}{2} .

Adding gives

1 = | 1 - 0 | \leq | D (x_{1}) - L | + | D (x_{2}) - L | < 1

, a contradiction. So

D (x)

cannot have a limit anywhere.

Investigate the discontinuity of $f (x) = 2^{1 / x}$ at $x = 0$ .

Solution

x \to 0^{+}

\frac{1}{x} \to + \infty

, so

f (x) = 2^{1 / x}

increases without bound. As

x \to 0^{-}

\frac{1}{x} \to - \infty

, so

f (x) = 2^{1 / x} \to 0

. There is an **infinite discontinuity** at

x = 0

. Also,

f

is not defined at

x = 0

Graph of f(x)=2^(1/x) showing infinite discontinuity at x=0

Elementary Continuous Functions

The following functions are continuous on their domains:

Polynomial functions are continuous everywhere: for any $a$ , $lim_{x \to a} P (x) = P (a)$ .
Rational functions $R (x) = P (x) / Q (x)$ are continuous at every $a$ where $Q (a) \neq 0$ .
$f (x) = \sqrt[n]{x}$ with $n$ a positive odd integer is continuous everywhere.
$f (x) = \sqrt[n]{x}$ with $n$ a positive even integer is continuous on $[0, \infty)$ .
Sine and cosine are continuous everywhere.
Tangent, cotangent, secant, cosecant are continuous on their domains. For example, $y = \tan x$ is continuous everywhere except where $\cos x = 0$ , i.e., when $x = \frac{π}{2} + k π$ for integer $k$ .
Inverse trigonometric functions are continuous on their domains. For example, $y = \arcsin x$ and $y = \arccos x$ are continuous on $[- 1, 1]$ , and $y = \arctan x$ is continuous on $(- \infty, + \infty)$ .
$f (x) = e^{x}$ is continuous on $(- \infty, + \infty)$ .
$f (x) = \ln x$ is continuous on $(0, + \infty)$ .

The floor function $f (x) = ⌊ x ⌋$ is an example of a function that is not continuous on its domain: it is discontinuous at every integer.
The sign function \text{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases} is defined at $0$ but is discontinuous there.

Algebraic Operations on Continuous Functions

If $f$ and $g$ are continuous at $x = a$ , then the following functions are also continuous at $x = a$ :

(a) $f (x) + g (x)$
(b) $f (x) - g (x)$
(c) $f (x) g (x)$
(d) $f (x) / g (x)$ , provided $g (a) \neq 0$

In (d), $g (a) \neq 0$ prevents division by zero.

It follows that functions such as $F (x) = \ln x + \cos^{2} x - \frac{3}{x^{2} + 1}$ and $G (x) = \tan x - \frac{3 x^{2} + 2 x - 1}{x^{2} - 1}$ are continuous on their domains.

A Useful Theorem for Calculating Limits

If $lim_{x \to a} g (x) = b$ and $f$ is continuous at the point $b$ , then

lim_{x \to a} f (g (x)) = f (lim_{x \to a} g (x)) = f (b) .

This theorem says: if $f$ is continuous, you can move the limit inside $f$ .

Properties of Continuous Functions

Continuous functions on closed intervals have important structural properties. **Bolzano's Theorem (special case of the Intermediate Value Theorem).**

f

is continuous on

[a, b]

and

f (a)

and

f (b)

have opposite signs, then there exists a point

c

[a, b]

such that

f (c) = 0

Geometrically: a continuous curve that starts below the

x

-axis and ends above it must cross the

x

-axis somewhere.

Bolzano's Theorem guarantees at least one zero, but there may be more.
If $f (a) f (b) > 0$ (same sign), the equation $f (x) = 0$ may still have a solution in $[a, b]$ .
If $f$ is discontinuous even at one point in $[a, b]$ , the theorem may fail.

**The Intermediate Value Theorem.**

Let

f

be continuous on the closed interval

[a, b]

. If

k

is any number between

f (a)

and

f (b)

, then there exists a point

c

[a, b]

such that

f (c) = k

In other words, a continuous function takes on every value between its values at the endpoints. Example application: a moving vehicle that goes from speed $0$ to speed $100$ km/h must at some moment have been going at exactly $50$ km/h (or any other speed in between). **The Extreme Value Theorem.**

f

is continuous on a closed interval

[a, b]

, then

f

attains both an absolute maximum and an absolute minimum on

[a, b]

Both conditions, closed interval and continuity, are needed. If either fails, the theorem may not hold.
The theorem guarantees the existence of extreme values but does not tell us how to find them (those methods come in the differentiation chapter).

Frequently Asked Questions

How do I check whether a piecewise function is continuous?

Check continuity at the "break points" where the formula changes. At each break point

x = c

, verify: (1) both one-sided limits exist, (2) they are equal, and (3)

f (c)

equals that common value. The function is continuous on the rest of its domain by whatever continuity properties the individual formulas have.

What is the difference between a removable and a jump discontinuity?

In a **removable discontinuity**, the two-sided limit exists (both one-sided limits agree), but either the function is undefined or takes a different value at that point. In a **jump discontinuity**, the left and right limits both exist but are different, so the function "jumps" between two values.

Is the composition of continuous functions always continuous?

Yes. If

g

is continuous at

a

and

f

is continuous at

g (a)

, then the composite function

f \circ g

is continuous at

a

. This follows from the Composite Function Limit Theorem.

Intuitive Introduction

Continuity at a Point

Right and Left Continuity; Continuity on an Interval

Types of Discontinuity

1. Removable Discontinuity

2. Jump Discontinuity

3. Infinite Discontinuity

4. Oscillating Discontinuity

Elementary Continuous Functions

Algebraic Operations on Continuous Functions

A Useful Theorem for Calculating Limits

Properties of Continuous Functions

Frequently Asked Questions

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