Re: how to solve this limit?


David C. Ullrich wrote:
On 28 May 2006 07:39:26 -0700, "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?

It looks like you only read half of my post. It
wasn't that long.

yah, you're right.

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


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David C. Ullrich

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