PlanetMath: example of jump discontinuity

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example of jump discontinuity

(Example)

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

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

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 expansions of the function gotten by performing the divisions: for

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

and for

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

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

and thus we have the one-sided limits $\displaystyle \lim_{x\to 0\pm}f'(x) = 0$ (see growth of exponential function).

**Figure:** Graph of the function with jump discontinuity
$\includegraphics{e1xjump.eps}$

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Cross-references: growth of exponential function, one-sided limits, derivative, divisions, function, series, origin, jump discontinuity, real function
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This is version 13 of example of jump discontinuity, born on 2006-11-18, modified 2007-09-05.
Object id is 8567, canonical name is ExampleOfJumpDiscontinuity.
Accessed 3511 times total.

Classification:

AMS MSC:	54C05 (General topology :: Maps and general types of spaces defined by maps :: Continuous maps)
	26A15 (Real functions :: Functions of one variable :: Continuity and related questions )

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