Re: Inf/Inf = NaN?

[Craig Earls]

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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