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From: ahmed_abdelakher on 15 Mar 2010 16:43
Thanks you for replying. I got it.

Ahmed

From: Axel Vogt on 15 Mar 2010 17:17
Arturo Magidin wrote:
> On Mar 15, 2:50 pm, ahmed_abdelakher <ahmed.abdelak...(a)rogers.com>
> wrote:
>> I'm trying to evaluate the limit of:
>>
>> F(x) = -1/x + e^(-x) /(1- e^(-x)) as x-->0
>
> I'm having a hard time figuring out what the function *is*.
>
> I assume it is:
>
> (-1/x) + [ (e^(-x))/(1-e^(-x))].
>
> If that is the case, then simplifying the second fraction we get
>
> [ (1/e^x) / (1 - (1/e^x)) ] = [(1/e^x)/((e^x-1)/e^x)] = 1/(e^x-1)
>
> so the function would be
>
> F(x) = (1/(e^x-1)) - (1/x) = (x - e^x + 1)/(xe^x - x).
>
> If you know L'Hopital's Rule, then two applications will readily give
> the answer.
>
>> The correct answer is -1/2. I cannot get the correct answer. No
>> matter what I do, I get it equal to -1.
>
> Perhaps you can show us one of the things you did to get -1, so we can
> see where the error lies? I get -1/2 when I do L'Hopital's Rule after
> the simplification.
>
> --
> Arturo Magidin

One can "produce" -1 like that: x ---> 0, hence 1st term = 0
and using x=ln(t) makes exp(-x)/(1-exp(-x)) = ... = 1/(t-1)
and t ---> 0 gives it (though t falling to 1 is need and then
l'Hospital for a difference)
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