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Re: 0^0 = 1?
Hi, and Derive make a mistake, as expected. It is stuff from the elementary school that 0^0 is undefined because x^0 /; x!=0 is 1 but 0^x /; x!=0 is 0. Only if you have some sequences x[n] and y[n] with Limit[x[n],n->Infinity]==0 and Limit[y[n],n->Infinity]==0 you can take the limit Limit[x[n]^y[n],n->Infinity] and may get a defined result. Regards Jens ivflam at gmail.com wrote: > Mathematica says 0^0 = Indeterminate > Derive says 0^0 = 1 > > May I have any opinions? > > Bruno >