RE: 0^0 = ?

[j . e . drews]

I know that by L'Htpital's Rule you should get:

ln(y)=x*ln(x) = ln(x)/(1/x)  so

Lim x->0+  ln(x)/(1/x) = ( 1/x )/( -1/( x^2)) =  

Lim x->0+ (-x) = 0
so  ln(y) = 0 and then y=1. 

Maybe this is the reason for the behavior?
> In octave 2.1.50 -->
> >> 0^0
> ans = 1
> 
> How about 0^0 in Octave? It is convenient in my program if 0^0 = 1.
> 
> Cong



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