The function $f$ is defined as follows: $$f(x):=\sum_{j=1}^{\infty} \frac{x^j}{j!} e^{-x}$$
It's easy to see that $f(0)=0$. But I am interested in the value $$\lim_{x \rightarrow 0^+} f(x).$$
Even Wolfram Alpha does not help here. I tried to plot this function, but this doesn't work neither. And my calculator doesn't give a solution for concrete values of $x$, so I have no idea how to get on here.
$$ f(x) =\sum\limits_{j=1}^\infty \frac{x^j}{j!}e^{-x} = e^{-x}\sum\limits_{j=1}^\infty \frac{x^j}{j!} = e^{-x}\cdot (e^x-1) = 1-e^{-x}.$$
Here we have used that $$e^x = \sum\limits_{j=0}^\infty\frac{x^j}{j!} = 1+\sum\limits_{j=1}^{\infty}\frac{x^j}{j!} $$ So that $$ \sum\limits_{j=1}^{\infty}\frac{x^j}{j!} = e^x-1.$$
Hint: Note that $$f(x) = \sum_{j=1}^{\infty}\frac{x^j}{j!}e^{-x} = e^{-x}\sum_{j=1}^{\infty}\frac{x^j}{j!} = e^{-x}\left(\sum_{j=0}^{\infty}\frac{x^j}{j!}-1\right).$$
Wolfram Alpha can do this limit. "Direction->-1" means to approach the limit from larger values.
Notice that $e^x=\displaystyle\sum_{j=0}^{\infty}\displaystyle\frac{x^j}{j!}$.
So $\displaystyle\sum_{j=1}^{\infty} \frac{x^j}{j!} e^{-x}=\displaystyle\sum_{j=0}^{\infty} \frac{x^j}{j!} e^{-x}-e^{-x}=1-e^{-x}$
Thank you all very much!!
– user136457 Jun 11 '14 at 15:12