example of jump discontinuity

The elementary (http://planetmath.org/ElementaryFunction) real function

$$f:x\mapsto \frac{1}{1+e^{\frac{1}{x}}}$$

has a jump discontinuity at the origin, since

$$\underset{x\to 0-}{lim}f(x)=1\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}\underset{x\to 0+}{lim}f(x)=0.$$

Indeed,

• if  $x\to 0-$,  then  ${\displaystyle \frac{1}{x}}\to -\mathrm{\infty }$,  $e^{\frac{1}{x}}\to 0$,  ${\displaystyle \frac{1}{1+e^{\frac{1}{x}}}}\to 1$;

• if  $x\to 0+$,  then  ${\displaystyle \frac{1}{x}}\to \mathrm{\infty }$,  $e^{\frac{1}{x}}\to \mathrm{\infty }$,  ${\displaystyle \frac{1}{1+e^{\frac{1}{x}}}}\to 0$.

These results can be seen also from the series of the function gotten by performing the divisions:  for    we obtain the converging (http://planetmath.org/Converge) alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries)

$1:(1+e^{\frac{1}{x}})={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{(-1)}^k{e}^{\frac{k}{x}}=1-e^{\frac{1}{x}}+e^{\frac{2}{x}}-e^{\frac{3}{x}}+-\mathrm{\dots }$

and for  $x>0$  the series

$1:(e^{\frac{1}{x}}+1)={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{(-1)}^{k+1}{e}^{-\frac{k}{x}}=e^{-\frac{1}{x}}-e^{-\frac{2}{x}}+e^{-\frac{3}{x}}-+\mathrm{\dots }$

Note.  The derivative of the function may be written as

$$f^{\prime }(x)=\frac{1}{x^2(e^{-\frac{1}{x}}+1)(1+e^{\frac{1}{x}})},$$

and thus we have the one-sided limits  $\underset{x\to 0\pm }{lim}{f}^{\prime }(x)=0$ (see growth of exponential function).

Figure 1: Graph of the function $f$ with jump discontinuity

Title	example of jump discontinuity
Canonical name	ExampleOfJumpDiscontinuity
Date of creation	2013-03-22 16:25:02
Last modified on	2013-03-22 16:25:02
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	16
Author	pahio (2872)
Entry type	Example
Classification	msc 26A15
Classification	msc 54C05
Related topic	ExponentialFunction
Related topic	ImproperLimits