Editing Continuous function (section)

===Examples of discontinuous functions===
[[File:Discontinuity of the sign function at 0.svg|thumb|300px|Plot of the signum function. It shows that <math>\lim_{n\to\infty} \sgn\left(\tfrac 1 n\right) \neq \sgn\left(\lim_{n\to\infty} \tfrac 1 n\right)</math>. Thus, the signum function is discontinuous at 0 (see [[#Definition in terms of limits of sequences|section 2.1.3]]).]]
An example of a discontinuous function is the [[Heaviside step function]] <math>H</math>, defined by
<math display="block">H(x) = \begin{cases}
1 & \text{ if } x \ge 0\\
0 & \text{ if } x < 0
\end{cases}
</math>

Pick for instance <math>\varepsilon = 1/2</math>. Then there is no {{nowrap|<math>\delta</math>-neighborhood}} around <math>x = 0</math>, i.e. no open interval <math>(-\delta,\;\delta)</math> with <math>\delta > 0,</math> that will force all the <math>H(x)</math> values to be within the {{nowrap|<math>\varepsilon</math>-neighborhood}} of <math>H(0)</math>, i.e. within <math>(1/2,\;3/2)</math>. Intuitively, we can think of this type of discontinuity as a sudden [[Jump discontinuity|jump]] in function values.

Similarly, the [[Sign function|signum]] or sign function
<math display="block">
\sgn(x) = \begin{cases}
\;\;\ 1 & \text{ if }x > 0\\
\;\;\ 0 & \text{ if }x = 0\\
-1 & \text{ if }x < 0
\end{cases}
</math>
is discontinuous at <math>x = 0</math> but continuous everywhere else. Yet another example: the function
<math display="block">f(x) = \begin{cases}
  \sin\left(x^{-2}\right)&\text{ if }x \neq 0\\
   0&\text{ if }x = 0
\end{cases}</math>
is continuous everywhere apart from <math>x = 0</math>.

[[File:Thomae function (0,1).svg|200px|right|thumb|Point plot of Thomae's function on the interval (0,1). The topmost point in the middle shows f(1/2) = 1/2.]]
Besides plausible continuities and discontinuities like above, there are also functions with a behavior, often coined [[Pathological (mathematics)|pathological]], for example, [[Thomae's function]],
<math display="block">f(x)=\begin{cases}
1   &\text{ if } x=0\\
\frac{1}{q}&\text{ if } x = \frac{p}{q} \text{(in lowest terms) is a rational number}\\
  0&\text{ if }x\text{ is irrational}.
\end{cases}</math>
is continuous at all irrational numbers and discontinuous at all rational numbers. In a similar vein, [[Dirichlet's function]], the [[indicator function]] for the set of rational numbers,
<math display="block">D(x)=\begin{cases}
  0&\text{ if }x\text{  is irrational } (\in \R \setminus \Q)\\
  1&\text{ if }x\text{ is rational } (\in \Q)
\end{cases}</math>
is nowhere continuous.