"Stavros Macrakis" <macrakis at alum.mit.edu> writes:
> On 3/11/07, Andrey G. Grozin <A.G.Grozin at inp.nsk.su> wrote:
>> lim x^y
>>
>> does not exist when both x and y tend to 0 (and are positive). The limit
>> depends on the path along which (x,y) approaches (0,0). By choosing an
>> appropriate path, one can easily obtain any non-negative value for this
>> limit.
So? What does pointing out that a limit doesn't exist have to do
with 0^0? Blindly plugging in values is not how to find limits; doing
it for x^y will not work whether or not 0^0 has a value.
> I agree, Andrey, but isn't the value limited to [0,1]?
>
> Jay, consider a^b where a=exp(-1/x^k)) (k>0) and b=x; both a and b go
> to 0 as x goes to 0.
> This is equivalent to exp(-1/x^k)^x = exp(-x^(1-k)), and its limit as
> x->0+ is 0 for k<1, 1/e for k=1, and 1 for k>1.
Saying that a function is not continuous at a point isn't the same as
saying that 0^0 = 1 gives the wrong value; your a^b is not continuous
at 0 whether or not 0^0 has a value.
I still don't see any way in which defining 0^0 = 1 gives a wrong
answer. The only reasons I've ever heard against it are along the
lines of "If someone doesn't know what they're doing and try to find
certain limits, then leaving 0^0 undefined will serve as a flag that
they're doing something wrong."
Jay