# example of jump discontinuity

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

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

has a jump discontinuity at the origin, since

 $\lim_{x\to 0-}f(x)=1\quad\mathrm{and}\quad\lim_{x\to 0+}f(x)=0.$

Indeed,

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

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

These results can be seen also from the series of the function gotten by performing the divisions:  for  $x<0$  we obtain the converging (http://planetmath.org/Converge) alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries)

 $\displaystyle 1:(1+e^{\frac{1}{x}})=\sum_{k=0}^{\infty}(-1)^{k}e^{\frac{k}{x}}% =1-e^{\frac{1}{x}}+e^{\frac{2}{x}}-e^{\frac{3}{x}}+-\ldots$

and for  $x>0$  the series

 $\displaystyle 1:(e^{\frac{1}{x}}+1)=\sum_{k=1}^{\infty}(-1)^{k+1}e^{-\frac{k}{% x}}=e^{-\frac{1}{x}}-e^{-\frac{2}{x}}+e^{-\frac{3}{x}}-+\ldots$

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$\displaystyle\lim_{x\to 0\pm}f^{\prime}(x)=0$ (see growth of exponential function).

Title example of jump discontinuity ExampleOfJumpDiscontinuity 2013-03-22 16:25:02 2013-03-22 16:25:02 pahio (2872) pahio (2872) 16 pahio (2872) Example msc 26A15 msc 54C05 ExponentialFunction ImproperLimits