For more information on the definition of continuity see Continuity. For other types of discontinuities, see Removable Discontinuity

## Jump Discontinuity Example

Let's look at the unit step function, also called the Heaviside function. It was developed by Oliver Heaviside for use in telegraphs but is today used in biology and neuroscience to model binary cellular switches in response to chemical signals. This function is defined by the formula

$H\left(x\right)=\left\{\begin{array}{ll}0,& x<0\\ 1,& x\ge 0\end{array}\right.$,

and the graph of this function looks like this:

This function has a discontinuity at $x=0$, but it isn't a removable discontinuity or an infinite discontinuity. Instead, it has what is called a jump discontinuity.

## Jump Discontinuity Definition

Here is the formal definition of a jump discontinuity.

A function $f\left(x\right)$ has a **jump discontinuity** at $x=p$ if

- $\underset{x\to {p}^{+}}{\mathrm{lim}}f\left(x\right)=A,$
- $\underset{x\to {p}^{-}}{\mathrm{lim}}f\left(x\right)=B,$

where $A,B$ are real numbers, but $A\ne B$.

In other words, the limit from the left at the point and the limit from the right at the point both exist but aren't the same number.

For more information on limits from the left and right see One-Sided Limits

A jump discontinuity can't be an infinite discontinuity because the limit from the left and right are both real numbers. It also can't be a removable discontinuity because that requires the limit from the left and right to be the same number. So let's look at some more examples of functions with jump discontinuities.

## Jump Discontinuity Graph

The Heaviside function is nice because it is just a piecewise-defined function where all of the pieces are constants. But that doesn't need to be the case.

For the function in the graph below,

- Find an equation for the function
- Show that it has a jump discontinuity at $x=2$.

Answer:

When writing the equation of a function like this, it is easier to think of it in parts and then put all the parts together at the end.

- To the left of $x=2$, the function has a y-intercept at $\left(0,2\right)$ and the slope of the line is 1, so the equation of the line on the left in slope-intercept form is $y=x+\hspace{0.17em}2$.
- To the right of $x=2$, the function has an x-intercept at $(3,0)$ and the slope of the line is $-1$. So the equation of the line on the right is $y=-x+3$.
- At the point $x=2$ the function value is 1.
- Putting all of those parts together, you get the piecewise defined function

$f\left(x\right)=\left\{\begin{array}{ll}x+\hspace{0.17em}2,& x<\hspace{0.17em}2\\ -x+3,& x\ge 2\end{array}\right.$.

To show that it has a jump discontinuity, unfortunately it isn't sufficient to point to the spot on the graph and say "see, it jumps!". Instead you need to look at the limits from the left and right. Using the function definition you found,

$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{-}}{\mathrm{lim}}\left(x+2\right)=2+2=4$,

and

$\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}\left(-x+3\right)=-2+3=1$.

So you can see that both the limit from the left and the limit from the right are real numbers, but they certainly aren't the same number. That shows there is a jump discontinuity at $x=2$.

## Jump Discontinuity and Limits

You don't have to graph a function to see whether or not it has a jump discontinuity, you can just look at the limits.

Does the function

$f\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& x<1\\ 0,& x=1\\ 2-({x}^{2}-1),& x>\hspace{0.17em}1\end{array}\right.$

have a jump discontinuity at $x=1$?

Answer:

Let's check the limits:

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{-}}{\mathrm{lim}}{x}^{2}=1$

and

id="2752728" role="math" $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}(2-\left({x}^{2}-1\right)=2-\left({1}^{2}-1\right)=2-0=2$,

The left-handed limit is not equal to the right-handed limit at $x=1$, so you know that not only is the function discontinuous at $x=1$, it has a jump discontinuity there.

### Common Mistake!

Don't be fooled into thinking the function has a discontinuity just because the function is piecewise. Not all piecewise-defined functions are discontinuous where the function changes definition.

Does the function

$g\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& x<1\\ 2x-1,& x\ge 1\end{array}\right.$

have a jump discontinuity at $x=1$?Answer:

The idea is again to check the limits:

$\underset{x\to {1}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to {1}^{-}}{\mathrm{lim}}{x}^{2}=1$

and

$\underset{x\to {1}^{+}}{\mathrm{lim}}g\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}\left(2x-1\right)=1$The left-handed and right-handed limits are equal, and the function value at $x=1$ is also 1. Therefore, $g\left(x\right)$ is actually continuous there, and does not have a jump discontinuity.

## Jump Discontinuity Functions and Unexpected Results

You may wonder what happens if you multiply two functions that have jump discontinuities, or a function that has a jump discontinuity with one that doesn't have one.

Can the product of two functions that have a jump discontinuity at $x=3$ not have a jump discontinuity at $x=3$?

Answer:

The product might be, but doesn't have to be, continuous. Let's look at a couple of cases. For both of them, you will use the function

$f\left(x\right)=\left\{\begin{array}{ll}x+1,& x\le 3\\ x-1,& x>3\end{array}\right.$.

You can see by looking at the left and right limits at $x=3$ that $f\left(x\right)$ has a jump discontinuity there.

Now think about the function

$g\left(x\right)=\left\{\begin{array}{ll}x-1,& x<3\\ x+\hspace{0.17em}1,& x\ge 3\end{array}\right.$.

That function also has a jump discontinuity at $x=3.$ Then at $x=3$,

$f\left(3\right)\xb7g\left(3\right)=(3+1)(3+1)=16$.

Looking at the limit from the left and right,

$\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)\xb7g\left(x\right)=\underset{x\to {3}^{-}}{\mathrm{lim}}(x+1)(x-1)=(3+1)(3-1)=8$,

and

$\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)\xb7g\left(x\right)=\underset{x\to {3}^{+}}{\mathrm{lim}}(x-1)(x+1)=(3-1)(3+1)=8$.

So, we have a function where the left-handed limit is equal to the right-handed limit at $x=3$, but the function value here is not the same as the limit...this is a removable discontinuity, not a jump discontinuity!

Can the product of a function that has a jump discontinuity at $x=3$ and a continuous function be continuous at $x=3$?

Answer:

Sure! Take your continuous function to be the one that is identically zero, or in other words $g\left(x\right)=0$ for all values of $x$. From the previous example, you know that

$f\left(x\right)=\left\{\begin{array}{ll}x+1,& x\le 3\\ x-1,& x>3\end{array}\right.$

has a jump discontinuity at $x=3$, but

$f\left(x\right)\xb7g\left(x\right)=f\left(x\right)\xb70=0$The product of the two functions is always equal to 0, so the product is in fact continuous at$x=3$.

Can the product of two functions, both with jump discontinuities at $x=3$, be continuous at $x=3$?

Answer:

You bet! Take your functions to be defined by

$f\left(x\right)=\left\{\begin{array}{ll}1,& x\le 3\\ -1,& x>\hspace{0.17em}3\end{array}\right.$,

and

$g\left(x\right)=\left\{\begin{array}{ll}-1,& x\le 3\\ 1,& x>3\end{array}\right.$.

Both of these functions have a jump discontinuity at $x=3$, but their product is

$f\left(x\right)\xb7g\left(x\right)=\left\{\begin{array}{ll}1\xb7(-1),& x\le 3\\ (-1)\xb71,& x>3\end{array}\right.=\left\{\begin{array}{ll}-1,& x\le 3\\ -1,& x>\hspace{0.17em}3\end{array}\right.=-1$,

which is continuous at $x=3$.

## Jump Discontinuity - Key takeaways

- A function $f\left(x\right)$ has a jump discontinuity at $x=p$ if $\underset{x\to {p}^{+}}{\mathrm{lim}}f\left(x\right)=A,$ $\underset{x\to {p}^{-}}{\mathrm{lim}}f\left(x\right)=B,$ where $A,B$ are real numbers, and $A\ne B$.
- An example of a function with a jump discontinuity is the Heaviside function, which is also called the unit step function.
- Not all piecewise-defined functions are discontinuous where the function changes.
- Products of functions where one of them has a jump discontinuity can do strange things. For example, the product of two functions with jump discontinuities at the same point could be continuous or could end up having a removable discontinuity.

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##### Frequently Asked Questions about Jump Discontinuity

How do you know if something is a jump discontinuity?

You know it has a jump discontinuity if (a) it is a discontinuity, and (b) the function "jumps" there. An example is the Heaviside function, which has a jump discontinuity at x=0.

What is jump discontinuity in math?

It is when a function has a discontinuity at a point and the function jumps in value. An example is the Heaviside function, which has a jump discontinuity at x=0.

What is an example of jump discontinuity?

An example is the Heaviside function, which has a jump discontinuity at x=0.

Which function has a jump discontinuity?

An example is the Heaviside function, which has a jump discontinuity at x=0.

Is a function differentiable at a jump discontinuity?

A function is most definitely NOT differentiable where it has a jump discontinuity. Differentiable at a point implies continuous at that point.

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