Re: how to solve this limit?

On 28 May 2006 19:02:30 -0700, "Mystera" <yli116@xxxxxxxxx> wrote:


The World Wide Wade ???

In article
<1148827166.003270.252550@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Mystera" <yli116@xxxxxxxxx> wrote:

so what do you think if m is a positive integer, david?
since m is a positive integer which means that n(=1/m) is a rational,
we cannot use the binominal theorem. how can we do with this condition?

HInt: If f''(0) exists, then [f(x) - f(0) - f'(0)x]/x^2 ->
f''(0)/2 as x -> 0.

PS: Stop top-posting.

L'Hopital's rule does work, of course.

Golly. Accusing Wade of suggesting L'Hopital; this could be bloody.

He didn't say anything about L'Hopital, btw. What he said is just
Taylor's theorem.

i just wanna find a solution in
a classical way.
what about putting y = (1+x)^(1/m) - 1, then
expr = ( y - (1/m)*( (1+y)^m - 1 ) ) / ( (1+y)^m - 1)^2
=( (-(m-1)/2)*y^2+ ... ) / ( (m^2)*y^2 + ... ) \* by binominal
theorem
=(1-m) / 2*m^2

btw, anything wrong with top-posting?


David C. Ullrich wrote:
On 28 May 2006 02:12:25 -0700, "Larry Hammick"
<larryhammick@xxxxxxxxx> wrote:


Jean-Marc Gulliet wrote:
Mystera wrote:
lim ( (1+x)^(1/m) - 1 - x/m ) / x^2, x-->0.

thanks.

I just put 1/m = n, so we want the limit at 0 of
((1+x)^n - 1 - nx ) / x^2
which seems to be n-choose-2 or
n(n-1) / 2 = (m-1) / (2 m^2).

Or rather (1-m)/(2m^2).

I suppose this manoeuvre can be justified :)

If n happens to be a positive integer then it's
correct by the binomial theorem, which is
presumably where you got the "n choose 2" from
in the first place.

Otoh, if n is any real then it's correct by the
binomial theorem! A slightly fancier version...

LH


************************

David C. Ullrich


************************

David C. Ullrich
.

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