Re: Limit
- From: OPPT <oppt@xxxxxxxxxxxxxx>
- Date: Sun, 14 Oct 2007 07:11:26 -0400
On Sun, 14 Oct 2007 02:45:04 -0700, "Boen S. Liong"
<mr_bean_curdy@xxxxxxxxx> wrote:
On 14 Okt, 16:03, Thomas Nordhaus <thnord2...@xxxxxxxx> wrote:
Boen S. Liong schrieb:
Can somebody help with the limit?
How to derive:
limit(u->0) (1-u)^(-1/2) to be (1+0.5u)?
This surely isn't correct. The limit is equal to 1. (1+0.5*u) is the
first order Taylor-approximation however. Just compute it using the
definitions.
Boen S. Liong.
--
Thomas Nordhaus
It is correct. I quote from a math book. My hunch is it is from Taylor
series expansion. Use numerical for u, and you will see. Please check
before you say it is wrong.
(1 - 0)^(-1/2) = 1 and the limit you ask for is obviously 1.
The first three terms of the Maclaurin series for (1-u)^(-1/2) are
(1-u)^(-1/2) = 1 + 1/2 u + 3/4 u^2 / 2! + ...
The binomial series gives the same result. See (10) at
<http://mathworld.wolfram.com/BinomialSeries.html>
For a value of u close to 0,
(1-u)^(-1/2) is approximately equal to 1 + u/2
.
Regards,
Boen S. Liong
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