\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \)
deepdream of
          a sidewalk
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \)
Math and science::Analysis::Tao::09. Continuous functions on R

Left and right limits and discontinuities

Let \( X \subseteq \mathbb{R} \) and let \( x_0 \in X \). Let \( f: X \to \mathbb{R} \) be a function.

If \( f(x_0-) \) and \( f(x_0+) \) both exist and are both equal to \( f(x_0) \), then \( f \) is continuous at \( x_0 \).

\( x_0 \) is an element of \( X \) by definition, so \( f(x_0) \) must exist and be defined as \( f \)'s range includes all of \( X \).

\( x_0 \) must be an adherent point of both \( X \cap (-\infty, x_0) \) and \( X \cap (x_0, \infty) \), as these are requirements for the left and right limits to exist.

Discontinuities

Note that we require both ① limits to exist and ② be equal to each other and ③ to \( f(x_0) \), and ④ that the statement is only an implication, not a bi-implication. Discontinuities can be found by looking at these properties. Four discontinuities that arise at (at least):

  1. Asymptotic discontinuity and oscillatory discontinuity
  2. Jump discontinuity
  3. Removable discontinuity
  4. Domain break discontinuity, with function continuity maintained (I made up this name). Not 100% if it exists.

These are discussed further on the back side.


Example

1. A left or right limit doesn't exist

Two examples of discontinuites that arise from a limit not existing:

Asymptotic discontinuity
One or both limits do not exist as, informally, \( f(x) \) tends to infinity as \( x \to x_0 \). Example: \( f: \mathbb{R} \setminus \{0\}, f(x) := \frac{1}{x} \) has a discontinuity at 0.
Oscillatory discontinuity
A function can remain bounded but still not have a left or right limit near a point. For example: \( f : \mathbb{R} \to \mathbb{R} \) where \( f(x) \) is \( 1 \) if \( x \in \mathbb{Q} \) or \( 0 \) otherwise.

2. Jump discontinuity

Where both limits exist but are not equal, we have a jump discontinuity. At least one must not be equal \( f(x_0) \). For example, \( f: \mathbb{R} \to \mathbb{R} \) defined by:

\[ f(x) := \begin{cases} 
0 \quad \text{if } x \le 0 \\
1 \quad \text{if } x > 0 \\
\end{cases} \]

3. Removable discontinuity

Let an right limits can exist and be equal to \( f(x_0) \), yet \( f \) still might not be continuous. For example, \( f: \mathbb{R} \to \mathbb{R} \) defined by:

\[ f(x) := \begin{cases} 
1 \quad \text{if } x = 0 \\
0 \quad \text{if } x \neq 0 \\
\end{cases} \]

Then both left and right limits exist and we have \( f(x_0-) = f(x_0+) \neq f(x_0) \). This is called a removable discontinuity.

4. (Not sure what called?) Domain break discontinuity

If \( X = [0, 1] \cap [3, 4] \) and \( f : X \to \mathbb{R} \) is given by \( f(x):= 2x \) (or some other function continuous on \( \mathbb{R} \)), then surprisingly, according to how Tao defines continuity, \( f \) is continuous at all endpoints \( x = 0 \), \( x = 1 \), \( x = 3 \) and \( x = 4 \). This is so as for any \( \varepsilon > 0 \) we can find a \( \delta > 0 \) such that \( |f(i) - f(a)| < \varepsilon \) for all \( i \in \{x \in X \cap (a- \delta, a + \delta)\} \).

Considering this example, continuity might best be thought of as the question of whether for all sequences \( (a_n)_{n=0}^{\infty} \) that converge to \( x_0 \) it can be said that \( (f(a_n))_{n=0}^{\infty} \) converges to \( f(x_0) \). In this view, we are forced to ignore the parts of the domain that are not connected to \( x_0 \) (what is the techical term for not connected?).


Source

p234