Changing the domain of a function can change whether it's continuous or not.

Example

Is the function \(\dfrac{1}{x + 1}\) continuous?

It's function definition gives us a problem when \(x +1 = 0\), or when \(x = -1\):

However, if we modify the domain to exclude the gap, we get a continuous function as shown in the picture above.

Let's make those mathematicians happy

So far, we've only talked about the intuitive idea of continuity. It's time to come up with a definition involving symbols and limits that will make these ideas precise. It would be a good idea to read some of the articles on limits (if you haven't done so already) before reading on.

We say that a function \(f\) is continuous if, for every value of \(c\) in its domain,:

1. \(f(c)\) is defined, and
2. \(\displaystyle{\lim_{x \to c} f(x) = f(c)}\)

This last one's the killer. It's easy to check if \(f(c)\) is defined. For the limit to be defined, the function has to approach the same value as \(x\) approaches \(c\) from all directions. Fortunately, in the real numbers, "in all directions" translates into just two directions: from the left and from the right.

We need to check that as \(x\) approaches \(c\) from the LEFT, \(f(x)\) approaches \(f(c)\), AND

as \(x\) approaches \(c\) from the RIGHT, \(f(x)\) approaches \(f(c)\).

Example

Is the function \(f(x) = \dfrac{x^2 - 4}{x -2}\) continuous over the real numbers?

There is a problem when \(x -2 = 0\), that is, at \(x = 2\). The function is not defined there, so it can't be continuous.

If you check the limits of \(f(x)\) from either side of \(x = 2\) you'll see that they're both equal to \(4\), so we get a hole in the graph at \(x = 2\). This is called a removable discontinuity.

Note: If we change the domain to be the set of all real numbers \(x\) such that \(x > 2\), then the removable discontinuity is no longer in the domain, and the function is continuous.

Note: Another way to build a continuous function out of \(f(x)\) is to rebuild the function. Here's a new definition of a function that agrees with \(f(x)\) everywhere except at \(x = 2\) and is continuous over the real numbers:

\( g(x) = \begin{cases} \dfrac{x^2 -4}{x - 2} & x \neq 2\\ 4 & x = 2 \end{cases} \)

Note: We call removable discontinuities removable because we can "remove" them, just as we did in the example above, by redefining the function.

Piecewise Functions

You have to take extra care with piecewise functions to make sure their pieces match up at the end-points. The function

\( g(x) = \begin{cases} \dfrac{x^2 -4}{x - 2} & x \neq 2\\ 4 & x = 2 \end{cases} \)

defined in the example above is continuous everywhere, but this isn't always the case.

To check a piecewise defined function for continuity:

1. Check that each branch is continuous.
2. Check that the branches match up, so that there are no jumps in the graph.

Let's look at an example:

Example

Is the function \(f(x) = \begin{cases} 2 & \text{ if } x \leq 1\\ x &\text{ if } x > 1 \end{cases}\) continuous over the real numbers?

There is no problem for \(x\)-values inside the two branches. The functions \(g(x) = 2\) and \(h(x) = x\) are continuous everywhere.

The function is defined at \(x = 1\): \(f(x) = 2\)

So, the only problem is likely to occur at the end-points of the two branches. Let's see if \(\displaystyle{\lim_{x \to 1} f(x)}\) exists:

• From the left, the function values are all equal to \(2\), so the left-sided limit is 2.
• From the right, the limit is equal to \(\displaystyle{\lim_{x \to 1}x = 1}\).

Since the limits don't match up from the left and right, \(\displaystyle{\lim_{x \to 1} f(x)}\) does not exist, and the function is discontinuous at \(x = 1\). See the jump in the picture? We call this a jump discontinuity.

Here's an example of a piecewise-defined function that is continuous.

Another example:

This time, we want to look at the absolute value function, \(f(x) = |x|\), which is also piecewise-defined. Just a quick reminder...

\( |x| = \begin{cases} x &\text{ if } x \geq 0\\ -x &\text{ if } x and this function definition makes sense for all real numbers \(x\). So, the domain is all real numbers. In fact, the graph is continuous for all real numbers as the two branches match up at \(x = 0\): the limits from both left and right are equal to \(0\). So, there are no gaps, jumps or holes in the graph.

Note: Although the absolute value function is continuous at \(x = 0\), it isn't differentiable there. See the article on differentability for further information about differentiability.