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Re: Inf/Inf = NaN?
From: |
Craig Earls |
Subject: |
Re: Inf/Inf = NaN? |
Date: |
Thu, 17 Apr 1997 07:18:44 -0400 |
John Utz wrote:
>
> Hello everybody;
>
> On my fake pentium-100 running FreeBSD-2.2-GAMMA and octave-2.0.5 compiled
> with gcc-2.7.2 i get the above result ( Inf/Inf = NaN ). Shouldn't this
> equal 1?
No, if it is a value of a function, it can be evaluated using L'Hopitals
rule, i.e.
lim [f(x)/g(x)]=lim [ df/dx / dg/dx]
x->A x->A
where f(x)/g(x) =inf/inf as x->A, where A is any real Number.
--
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Craig P Earls address@hidden
LT US Navy, MIT Ocean Engineering address@hidden
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Craig P Earls address@hidden
LT US Navy, MIT Ocean Engineering address@hidden
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- Inf/Inf = NaN?, John Utz, 1997/04/17
- Re: Inf/Inf = NaN?,
Craig Earls <=