How do you prove a function is continuous at a point?

Show that f(a) is defined, that lim_x→ af(x) exists, and that the two are equal. For piecewise functions, compare the left-hand and right-hand limits with the value at the join.

Is every function continuous?

No. Fractions are discontinuous where the bottom is 0, piecewise functions can jump where the formula changes, and functions like sinfrac1x oscillate. But polynomials, sin, cos and e^x are continuous everywhere.

What is a removable discontinuity?

A single hole or misplaced point: the limit exists, but f(a) is undefined or has a different value. Redefining f(a) as the limit removes it.

Does the Intermediate Value Theorem tell you where the solution is?

No, only that at least one exists between a and b. To estimate it, apply the theorem repeatedly with the bisection method, or use other tools like Newton's method.

Can the Intermediate Value Theorem show there is exactly one solution?

Not on its own; it guarantees at least one. Combined with knowing the function is always increasing or always decreasing on the interval (from its derivative), it shows there is exactly one.

Continuity and the Intermediate Value Theorem Explained

What does it mean for a function to be continuous?

Informally, a function is continuous if you can draw its graph without lifting your pencil: no holes, no jumps, no places where the graph shoots off to infinity. Most quantities in the real world change this way. The temperature doesn't skip from $10^{\circ}$ to $25^{\circ}$ without passing through the temperatures in between.

To check continuity with algebra instead of a drawing, we turn "no break at $x = a$ " into a statement about limits. The graph has no break at $a$ exactly when the outputs near $a$ head to the output at $a$ :

Definition: continuous at a point

A function $f$ is continuous at $x = a$ if $lim_{x \to a} f (x) =$ $f (a)$ . This packs in three conditions: (1) $f (a)$ is defined, (2) $lim_{x \to a} f (x)$ exists, and (3) the two are equal.

A function is continuous on an interval if it is continuous at every point of the interval. A point where any of the three conditions fails is called a discontinuity.

What are the types of discontinuity?

Each kind of discontinuity breaks a different one of the three conditions:

Kind	What the graph does	Which condition fails	Example at the point
Removable (a hole)	A single missing or misplaced point	(1) or (3): the limit exists but $f (a)$ is missing or different	$\frac{x^{2} - 4}{x - 2}$ at $x = 2$
Jump	Leaps from one height to another	(2): the one-sided limits differ	a piecewise function at its join
Infinite	Shoots up or down a vertical asymptote	(2): the outputs grow without bound	$\frac{1}{(x - 6)^{2}}$ at $x = 6$
Oscillating	Wiggles faster and faster forever	(2): the outputs never settle	$\sin \frac{1}{x}$ at $x = 0$

Three discontinuities: a hole at

x = - 2

(the value sits elsewhere), a jump at

x = 3

, and an infinite discontinuity at

x = 6

A removable discontinuity is called removable because redefining one value would fix it: if the limit is $L$ , setting $f (a) = L$ fills the hole. Jumps and infinite discontinuities can't be fixed by changing one value.

How do you check whether a function is continuous at a point?

Check the three conditions in order. For a piecewise function, the only places to check are where the formula changes; each piece on its own is usually a polynomial, which is continuous.

Worked example: Choosing a value that makes a function continuous

Find the number $k$ that makes $f (x) = {\begin{cases} k x + 1 & x < 2 \\ x^{2} - k & x \geq 2 \end{cases}$ continuous at $x = 2$ .

(1) $f (2)$ uses the second formula: $f (2) = 2^{2} - k = 4 - k$ .
(2) For the limit to exist, the one-sided limits must agree. From the left: $lim_{x \to 2^{-}} (k x + 1) =$ $2 k + 1$ . From the right: $lim_{x \to 2^{+}} (x^{2} - k) =$ $4 - k$ .
Set them equal: $2 k + 1 = 4 - k$ . Add $k$ to both sides: $3 k + 1 = 4$ . Subtract $1$ : $3 k = 3$ , so $k = 1$ .
(3) With $k = 1$ the limit is $2 (1) + 1 = 3$ and $f (2) = 4 - 1 = 3$ : equal, so $f$ is continuous at $2$ .

Answer: $k = 1$

Common mistake

Checking only that $f (a)$ exists. A function can be defined at $a$ and still be discontinuous there, like the hole graph above where the dot sits away from where the outputs are heading. All three conditions must hold.

Which functions are continuous?

You rarely need the three conditions for familiar functions, because the limit laws show these are continuous wherever they are defined:

Polynomials, like $3 x^{4} - x + 7$ : continuous for every $x$ .
Fractions of polynomials: continuous everywhere except where the bottom is $0$ .
Roots, like $\sqrt{x}$ : continuous wherever the inside is allowed (here, $x \geq 0$ ).
$\sin x$ , $\cos x$ , $e^{x}$ : continuous everywhere; $\ln x$ for $x > 0$ ; $\tan x$ except where $\cos x = 0$ .
Sums, products, quotients and compositions of continuous functions, wherever they're defined.

This is exactly why direct substitution works: for a continuous function, the limit equals the value.

What does the Intermediate Value Theorem say?

If the temperature was $10^{\circ}$ at 6 am and $25^{\circ}$ at noon, it must have been exactly $18^{\circ}$ at some moment in between. The reason is continuity: a graph without breaks can't get from one height to another without passing through every height in between. Intermediate means "in between", and this rule is called the Intermediate Value Theorem (often shortened to IVT).

The Intermediate Value Theorem

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

between

a

and

b

with

f (c) = N

Why it works

Picture the graph as an unbroken curve from the point $(a, f (a))$ to the point $(b, f (b))$ .
The horizontal line $y = N$ lies between the heights of the two end points, so one end is below the line and the other is above it.
An unbroken curve can't get from below the line to above it without crossing it, and the crossing point has height $N$ : that's $f (c) = N$ .

The most common use is showing that an equation has a solution. If $f (a)$ and $f (b)$ have opposite signs, then $0$ is between them, so $f (c) = 0$ for some $c$ between $a$ and $b$ .

Worked example: Showing an equation has a solution

Show that $x^{3} - 4 x + 1 = 0$ has a solution between $0$ and $1$ .

Let $f (x) = x^{3} - 4 x + 1$ . It's a polynomial, so it is continuous on $[0, 1]$ .
$f (0) = 0 - 0 + 1 = 1$ , which is positive.
$f (1) = 1 - 4 + 1 = - 2$ , which is negative.
$0$ is between $- 2$ and $1$ , so by the Intermediate Value Theorem there is a $c$ between $0$ and $1$ with $f (c) = 0$ .

Answer: Such a solution exists (it is about $0.254$ ).

f (x) = x^{3} - 4 x + 1

is above the axis at

x = 0

and below it at

x = 1

, so its unbroken graph must cross

0

in between.

How do you narrow down where the solution is?

The theorem only says a solution exists, but using it repeatedly pins the solution down. Check the midpoint of the interval and keep the half where the sign changes; this is called the bisection method.

Worked example: One round of bisection

Starting from $[0, 1]$ for $f (x) = x^{3} - 4 x + 1$ , find a smaller interval containing the solution.

Midpoint $0.5$ : $f (0.5) =$ $0.125 - 2 + 1 =$ $- 0.875$ , negative. Since $f (0) > 0$ , the sign change is in $[0, 0.5]$ .
Midpoint $0.25$ : $f (0.25) =$ $0.015625 - 1 + 1 =$ $0.015625$ , positive. Since $f (0.5) < 0$ , the sign change is in $[0.25, 0.5]$ .

Answer: The solution is between $0.25$ and $0.5$ .

Common mistake

Using the theorem on a function that isn't continuous. $f (x) = \frac{1}{x}$ is $- 1$ at $x = - 1$ and $1$ at $x = 1$ , but it is never $0$ : it isn't continuous at $x = 0$ , so the theorem doesn't apply.

Tip

Continuity also matters later: the Mean Value Theorem and the rule that a continuous function on a closed interval has a largest and smallest value both need it.

Continuity and the Intermediate Value Theorem, explained simply