mathematical analysis
Phimath
lim(n*(exp(1/n)-1))^n = ?
I have spend much time on this problem, but even now I don't solve it.
Can you help me to solve it?
David Kastrup
Taylor series. exp(x) = 1 + x + x^2/2 + O(x^3). You can probably
prove the result with other methods, but getting it in the first place
will be most convenient using Taylor.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
William Elliot
> "Phimath" <phi...@gmail.com> writes:
>
> > Question:
> > lim(n*(exp(1/n)-1))^n = ?
>
> prove the result with other methods, but getting it in the first place
> will be most convenient using Taylor.
>
n.log(n(e^(1/n) - 1))
[e^(1/n) - 1] / (1/n)
e^(1/n) (-1/n^2) / (-1/n^2) = e^(1/n) -> 1
log(n(e^(1/n) - 1) / (1/n)
1/(n(e^(1/n) - 1) * [e^(1/n) - 1 + n.e^(1/n) (-1/n^2)] / (-1/n^2)
-n + e^(1/n) / (e^(1/n) - 1)
((1 - n)e^(1/n) + n) / (e^(1/n) - 1)
Yicks, does numerator -> 0 ?
Torsten Hennig
Hi,
(e^x-(1+x))/x^2 -> 1/2 (x->0) ---->
(e^(1/x)-(1+1/x))*x^2 = x^2*(e^(1/x)-1)-x -> 1/2 (x->oo)
---->
n^2*(e^(1/n)-1)-n -> 1/2 (n->oo) ---->
(1+(n^2*(e^(1/n)-1)-n)/n)^n -> e^(1/2) (n->oo)
The last step uses the fact that if a_n -> a (n->oo),
then (1+a_n/n)^n -> e^a (n->oo).
Best wishes
Torsten.
Dave L. Renfro
I'll assume you want the limit as n --> oo.
Let y = [ n*( exp(1/n) - 1 ) ]^n. We'll evaluate
limit as n --> oo of ln(y) and then use continuity
of the logarithm function to conclude that the
limit you want is exp[limit as n --> oo of ln(y)].
ln(y) = n*[ln(n) + ln( exp(1/n) - 1 )]
= [ln(n) + ln( exp(1/n) - 1 )] / (1/n)
Now make the variable change m = 1/n, and note that
limit as n --> oo of ln(y) is equivalent to
limit as m -->0+ of ln(y). Then we have
ln(y) = [-ln(m) + ln( e^m - 1 )] / m
At this point, apply L'Hopital's rule by differentiating
in the appropriate manner. You'll get
(e^m)/(e^m - 1) - (1/m)
= [m*e^m - e^m + 1] / [m*(e^m - 1)].
Now use the standard series expansion for e^m,
do some algebra, keep only the dominate terms
in the numerator and the denominator, and
you'll be left with limit as m --> 0+ of
[m^2 - (1/2)*m^2] / m^2, which is 1/2. Hence,
the limit you want is exp(1/2).
A few years ago I posted several elementary ways
of evaluating limits in sci.math, and you might
find that post to be a useful reference. The post
doesn't seem to be in google's sci.math archive,
but it's in the Math Forum sci.math archive at
http://mathforum.org/kb/message.jspa?messageID=225974
Dave L. Renfro
Ignacio Larrosa Cañestro
William Elliot <ma...@hevanet.remove.com> escribió:
> On Thu, 3 Nov 2005, David Kastrup wrote:
>> "Phimath" <phi...@gmail.com> writes:
>>
>>> Question:
>>> lim(n*(exp(1/n)-1))^n = ?
>>
>> Taylor series. exp(x) = 1 + x + x^2/2 + O(x^3). You can probably
>> prove the result with other methods, but getting it in the first
>> place will be most convenient using Taylor.
>>
> sqr e ?
e^(1/n) = 1 + 1/n + 1/(2n^2) + O(1/n^3)
L = Lim(n(e^(1/n) - 1)^n, n, inf)
Ln(L) = Lim(n*Ln(n(1/n + 1/(2n^2) + O(1/n^2))), n, inf)
= Lim(n*Ln(1 + 1/(2n) + O(1/n^2)), n, inf)
(As L(1 + x) ~ x, when x --> 0)
= Lim(n*(1/(2n) + O(1/n^2)), n, inf)
= Lim(1/2 + O(1/n)), n, inf) = 1/2
Then, L = e^(1/2)
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com
William Elliot
> William Elliot <ma...@hevanet.remove.com> escribió:
> > On Thu, 3 Nov 2005, David Kastrup wrote:
> >> "Phimath" <phi...@gmail.com> writes:
> >>
> >>> Question:
> >>> lim(n*(exp(1/n)-1))^n = ?
> >>
> >> Taylor series. exp(x) = 1 + x + x^2/2 + O(x^3). You can probably
> >> prove the result with other methods, but getting it in the first
> >> place will be most convenient using Taylor.
> >>
> > sqr e ?
>
> e^(1/n) = 1 + 1/n + 1/(2n^2) + O(1/n^3)
>
>
> L = Lim(n(e^(1/n) - 1)^n, n, inf)
>
> Ln(L) = Lim(n*Ln(n(1/n + 1/(2n^2) + O(1/n^2))), n, inf)
>
> = Lim(n*Ln(1 + 1/(2n) + O(1/n^2)), n, inf)
>
Typo, Ln((1 + 1/(2n) + O(1/n^2)) ?
Star...@earthlink.net
can answer that question we must know what n approaches.
Robert Low
> You have the limit equation but what number does n approach? Before we
> can answer that question we must know what n approaches.
Eh?
Ignacio Larrosa Cañestro
William Elliot <ma...@hevanet.remove.com> escribió:
> On Thu, 3 Nov 2005, Ignacio Larrosa Cañestro wrote:
>> William Elliot <ma...@hevanet.remove.com> escribió:
>>> On Thu, 3 Nov 2005, David Kastrup wrote:
>>>> "Phimath" <phi...@gmail.com> writes:
>>>>
>>>>> Question:
>>>>> lim(n*(exp(1/n)-1))^n = ?
>>>>
>>>> Taylor series. exp(x) = 1 + x + x^2/2 + O(x^3). You can probably
>>>> prove the result with other methods, but getting it in the first
>>>> place will be most convenient using Taylor.
>>>>
>>> sqr e ?
>>
>> e^(1/n) = 1 + 1/n + 1/(2n^2) + O(1/n^3)
>>
>>
>> L = Lim(n(e^(1/n) - 1)^n, n, inf)
>>
>> Ln(L) = Lim(n*Ln(n(1/n + 1/(2n^2) + O(1/n^2))), n, inf)
>>
> Typo, O(1/n^3) ?
Yes, it must be O(1/n^3). I apologize.
>
>> = Lim(n*Ln(1 + 1/(2n) + O(1/n^2)), n, inf)
>>
> Typo, Ln((1 + 1/(2n) + O(1/n^2)) ?
No, it is correct
>> (As L(1 + x) ~ x, when x --> 0)
>>
>> = Lim(n*(1/(2n) + O(1/n^2)), n, inf)
>>
>> = Lim(1/2 + O(1/n)), n, inf) = 1/2
>>
>> Then, L = e^(1/2)
William Elliot
> You have the limit equation but what number does n approach? Before we
> can answer that question we must know what n approaches.
>
I've not the time nor the inclination for the clerical chore of back
tracking the thread to reconstruct the line of thought. If you want
intelligent and well thought answers, instead of sloppy inaccurate
answers from what I remember, then include the context pertinent to your
reply and to whom your are talking.
Please learn and use better math group manners as demonstrated
by other participants of this newsgroup and as described at
http://oakroadsystems.com/genl/unice.htm#quote
If you're posting from Mathforum or Google, it is requested you use
the quote feature. Many of us use different news browsers than you.
William Elliot
> Question:
> lim(n*(exp(1/n)-1))^n = ?
lim(x->oo) (x.e^(1/x) - 1)^x = lim(x->0) ((e^x - 1)/x)^(1/x)
So consider
lim(x->0) (1/x) log((e^x - 1)/x)
As
(e^x - 1)/x -> e^x / 1 -> 1 (by l'Hospital's rule)
log((e^x - 1)/x) / x is of the form 0/0.
Sketching use of l'Hospital's(sp?) rule:
x/(e^x - 1) * (e^x / x + (e^x - 1)/(-x^2)) / 1
e^x / (e^x - 1) - 1/x
((x - 1)e^x + 1)/x(e^x - 1)
Which also is of the from 0/0. Thus again l'Hospital's rule
(e^x + (x - 1)e^x) / (e^x - 1 + x.e^x)
x.e^x / ((x + 1)e^x - 1)
Another 0/0 form. So again:
(e^x + x.e^x) / (e^x + (x + 1)e^x)
(x + 1)/(x + 2) -> 1/2
Thus original limit = e^(1/2)
----