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- From: bs@gauss.mitre.org (Robert D. Silverman)
- Subject: Re: Peculiar limit?
- Message-ID: <1992Dec17.212725.21906@linus.mitre.org>
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- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- References: <1992Dec17.202531.20920@infonode.ingr.com>
- Date: Thu, 17 Dec 1992 21:27:25 GMT
- Lines: 38
-
- In article <1992Dec17.202531.20920@infonode.ingr.com> tdj@infonode.ingr.com (Ted Johnson) writes:
- :Is the following true?
- :
- : g g g g
- : (1/b) - (1/a) (1/b) - (1/a)
- : lim ------------------ = lim ------------------
- : g->+inf. g g g->-inf. g g
- : (1/c) - (1/a) (1/c) - (1/a)
- :
- :
- : ln(1/b) - ln(1/a)
- : = -----------------
- : ln(1/c) - ln(1/a)
- :
-
- Unless I've made a mistake somewhere, this is wrong.
-
- Take the original expression and multiply it by 1 in the form a^g/a^g.
-
- This yields (a/b)^g - 1
- -----------
- (a/c)^g - 1
-
- If a = b this is 0. If a < b and a < c, then both the numerator
- and denominator go to -1 as g --> infinity. Thus, the limit is -1.
- [note that since (a/b) < 1, that (a/b)^g goes to zero. ditto for (a/c)]
-
- :If necessary:
- : 0 < a <= b <= c <= 1;
-
- As is easily seen, if a = b the limit is 0, unless a = c as well,
- but you stated a < c.
-
- --
- Bob Silverman
- These are my opinions and not MITRE's.
- Mitre Corporation, Bedford, MA 01730
- "You can lead a horse's ass to knowledge, but you can't make him think"
-