(November 23, 2022 at 11:15 pm)LinuxGal Wrote:(November 23, 2022 at 10:58 pm)polymath257 Wrote: More generally, if f(x) and g(x) are analytic functions with f(0)=g(0)=0, then the limit of f(x)^g(x) as x-->0 is 1. Your calculation was with f(x)=g(x)=x.
But if you go away from analytic functions, that may no longer be the case. So, if f(x)=e^(-1/x^2) and g(x)=x^2, then
as x-->0, both f(x) and g(x)-->0, but f(x)^g(x) =e^(-1) is not 1.
In many situations, it is convenient to have the *convention* that 0^0 =1 to make certain formulas correct when one variable is 0.
But, for limits, you should always regard 0^0 as indeterminate.
If you take the limit of 0^x from the right, you get 0. From the left, you get infinity. If you take the limit of x^0 you get 1. So it's a case of choose your own adventure.
Which is why it is said to be indeterminate.
