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From: ahmed_abdelakher on 15 Mar 2010 15:50
I'm trying to evaluate the limit of:

F(x) = -1/x + e^(-x) /(1- e^(-x)) as x-->0

The correct answer is -1/2. I cannot get the correct answer. No
matter what I do, I get it equal to -1.

Thanks for your help.
From: Mû on 15 Mar 2010 15:58
ahmed_abdelakher a �crit :
> I'm trying to evaluate the limit of:
>
> F(x) = -1/x + e^(-x) /(1- e^(-x)) as x-->0
>
> The correct answer is -1/2. I cannot get the correct answer. No
> matter what I do, I get it equal to -1.

Using Taylor expansions at order 2 I get -1/2. Maybe you simply added
equivalents, which may lead to wrong results.

--
M�
From: Arturo Magidin on 15 Mar 2010 16:12
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
From: James Burns on 15 Mar 2010 16:26
ahmed_abdelakher wrote:
> I'm trying to evaluate the limit of:
>
> F(x) = -1/x + e^(-x) /(1- e^(-x)) as x-->0
>
> The correct answer is -1/2. I cannot get the correct answer. No
> matter what I do, I get it equal to -1.
>
> Thanks for your help.


It looks like you should use l'Hopital's rule twice.

Manipulate the two addends, -1/x and the other,
so that they have a common divisor, x*(1-exp(-x)).

Differentiate the divisor and (the sum of) the numerators
until you do NOT get 0/0 when you evaluate at x=0.


By l'Hopital's rule, IF g(a) = h(a) =0, then
limit(x->a) g(x)/h(x) = limit(x->a) g'(x)/h'(x)

But IF g'(a) = h'(a) =0, then
limit(x->a) g'(x)/h'(x) = limit(x->a) g''(x)/h''(x)
and so on.

Jim Burns
From: Arturo Magidin on 15 Mar 2010 16:41
On Mar 15, 3:26 pm, James Burns <burns...(a)osu.edu> wrote:
> ahmed_abdelakher wrote:
> > I'm trying to evaluate the limit of:
>
> > F(x) = -1/x + e^(-x) /(1- e^(-x))   as x-->0
>
> > The correct answer is -1/2.  I cannot get the correct answer.  No
> > matter what I do, I get it equal to -1.
>
> > Thanks for your help.
>
> It looks like you should use l'Hopital's rule twice.
>
> Manipulate the two addends, -1/x and the other,
> so that they have a common divisor, x*(1-exp(-x)).
>
> Differentiate the divisor and (the sum of) the numerators
> until you do NOT get 0/0 when you evaluate at x=0.
>
> By l'Hopital's rule, IF g(a) = h(a) =0, then
> limit(x->a) g(x)/h(x) = limit(x->a) g'(x)/h'(x)

I believe that you also need to assume that lim(x->a) g'(x)/h'(x)
exists; otherwise, the result may not hold.

I don't have a ready example for a 0/0 indeterminate, but an example
for an oo/oo indeterminate take lim(x->0) (x + sin(x))/(2x+sin(x)).
This limit equals 1/2 (divide numerator and denominator by 2), but if
you use L'Hopital's Rule, you get lim(x->0)(1 + cos(x))/(2 + cos(x)),
and that limit does not exist.

--
Arturo Magidin
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