Continuous Functions

A function is continuous when its graph is a single unbroken curve ...

pencil ... that you could draw without lifting your pen from the paper.

That is not a formal definition, but it helps you understand the idea.

Here is a continuous function:

x^4-2x^2+x

Examples

So what is not continuous (also called discontinuous) ?

Look out for holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).


Not Continuous	Not Continuous	Not Continuous
(hole)	(jump)	(vertical asymptote)

Try these different functions so you get the idea:

Domain

A function has a Domain.

In its simplest form the domain is all the values that go into a function.

A function might be continuous or not, depending on its Domain!

Example: 1/(x-1)

At x=1 we have:

1/(1-1) = 1/0 = undefined

So there is a "discontinuity" at x=1


f(x) = 1/(x-1) over all Real Numbers		g(x) = 1/(x-1) for x>1
NOT continuous		Continuous

g(x) does not include the value x=1, so it is continuous.

So when a function is continuous within its Domain, it is a continuous function.

More Formally !

We can define continuous using Limits (it helps to read that page first):

A function f is continuous when, for every value c in its Domain:

f(c) is defined, and:

"the limit of f(x) as x approaches c equals f(c)"

The limit says:

"as x gets closer and closer to c
then f(x) gets closer and closer to f(c)"

And you have to check from both directions:

as x approaches c (from left) then f(x) approaches f(c)

AND as x approaches c (from right) then f(x) approaches f(c)

If you get different values from left and right (a "jump"), then the limit does not exist!

How to Use:

Make sure that, for all x values:

f(x) is defined
and the limit at x equals f(x)

Here are some examples:

Example: f(x) = (x²-1)/(x-1) for all Real Numbers

The function is undefined when x=1:

(x²-1)/(x-1) = (1²-1)/(1-1) = 0/0

So it is not a continuous function

Let us change the domain:

Example: g(x) = (x²-1)/(x-1) over the interval x<1

Almost the same function, but now it is over an interval that does not include x=1.

So now it is a continuous function (does not include the "hole")

Example: How about this piecewise function:

which looks like:

It is defined at x=1, because h(1)=2 (no "hole")

But at x=1 you can't say what the limit is, because there are two competing answers:

"2" from the left, and
"1" from the right

so in fact the limit does not exist at x=1 (there is a "jump")

And so the function is not continuous.

But:

Example: How about the piecewise function absolute value:

which looks like:

At x=0 it has a very pointy change!

But it is still defined at x=0, because f(0)=0 (so no "hole"),

And the limit as you approach x=0 (from either side) is also 0 (so no "jump"),

So it is in fact continuous.