Re: 0^0

>> f(x) = 0^x is undefined for all x < 0. There is obviously a
>> discrepancy in the behaviour about x = 0 for this rule, no
>> matter what you do with 0^0.
>>
>> f(x) = x^0 has the value 1 on either side of x = 0.
>>
>> It seems to me that defining 0^0 makes the latter function
>> continuous and constant everywhere, and doesn't add
>> any problems to the former.

Those are not the only two place of discontinuity. If you were to look
at the 3-D graph f(x,y) = x^y, you will see that there are an infinite
number of approaches to 0^0 that will yield you literally *ANY* value
you desire (like with 0/0).

You can see this better by choosing functions f, g such that lim x->0
f(x) = 0 and lim x-> 0 g(x) = 0, but lim x->0 f(x)^g(x) can be anything
you'd like. x^0 is simply the special case of f(x) = x and g(x) = 0,
and 0^x is the special case of f(x) = 0 and g(x) = x. But there are an
infinite number of choices of f(x) and g(x) in which lim x->0 f(x)^g(x)
gives you an answer different from 0 or 1.

As a trivial example, let f(x) = e^-1/x^2 (and f(0)=0), lim x->0 g(x) =
x^2. Both functions are continuous and infinitely differentiable.
Yet, lim x->0 f(x)^g(x) = e.

Jonathan Hoyle
Eastman Kodak

.

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