Discontinuity: Jump discontinuity

[samir]

I earlier posted about point discontinuity. It became overwhelming pretty quickly. Now I would like to focus this thread at jump discontinuity specifically, if you don't mind me posting multiple threads about discontinuity.

From what I understand, "jump discontinuity" occurs where the left-hand limit and right-hand limit for a given are not equal. Correct?

Assume we have the following functions.

$$f(x)=\begin{cases}x^2, & x\leq 1 \\ 2-x, & x\gt 1\end{cases}$$

$$g(x)=\begin{cases}x^2, & x\leq 1 \\ 6-x, & x\gt 1\end{cases}$$

Function $f$ is continuous, but function $g$ is discontinuous. Correct?

$$\lim_{{x}\to{1^{-}}}f(x)=1$$

$$\lim_{{x}\to{1^{+}}}f(x)=5$$

$$\lim_{{x}\to{1^{-}}}f(x) \neq \lim_{{x}\to{1^{+}}}f(x)$$

$$\lim_{{x}\to{1^{-}}}g(x)=1$$

$$\lim_{{x}\to{1^{+}}}g(x)=1$$

$$\lim_{{x}\to{1^{-}}}g(x) = \lim_{{x}\to{1^{+}}}g(x)$$

So far so good? If we were to examine the limits of a function in this manner for every point, we would be able to tell if it has any jump discontinuities?

Now, do only piece-wise functions have jump discontinuities? Are there any other kind of functions that have "jump" discontinuities?

Does the jump always occur in the vertical direction? As in jumping from $(1,1)$ to $(1,5)$ as opposed to $(1,1)$ to $(2,4)$?

 

[I like Serena]

So far so good? If we were to examine the limits of a function in this manner for every point, we would be able to tell if it has any jump discontinuities?

Now, do only piece-wise functions have jump discontinuities? Are there any other kind of functions that have "jump" discontinuities?

Does the jump always occur in the vertical direction? As in jumping from $(1,1)$ to $(1,5)$ as opposed to $(1,1)$ to $(2,4)$?

Hey samir! ;)

So far so good... just that $f$ and $g$ seem to have been mixed up.
And yes, examining the limits of each point of a function will tell us where the jump discontinuities are.

Which functions are you considering?
If we only look at standard continuous functions like $\sin x$, and at functions we compose with nested function calls and additions and such, we can't get jump discontinuities.
However, consider the standard unit step function, written as $H(x)$ or $u(x)$ or $\Theta(x)$, which is also known as the Heaviside step function.
It implicitly has a jump discontinuity.
Of course, to define what we mean with $\Theta(x)$, we'll need a piece-wise function definition.

Anyway, we can write your $g(x)$ as:
$$g(x) = x^2\Theta(1-x) + (6-x)\Theta(x-1)$$
The actual value of $\Theta(0)$ is a bit ambiguous though, and depends on which text you're reading.
It can be any of $0$, $\frac 12$, or $1$. If you consider it relevant, we should simply define what we mean by $\Theta$ before using it.
(For the record, I picked $\Theta$ as symbol to use, because it's the most exotic. )

We don't really have a "vertical" jump in a function - a function is supposed to be an abstract thing without geometrical meaning.
However, if we get a pencil and draw a graph of $y=f(x)$ where $y$ is considered to be vertical on the paper, then yes, a jump discontinuity corresponds to an abrupt change in $y$ from one $x$ value to the next - a vertical jump.