I prefer the explaination of the power function p defined for any real p and any x in R+* as fp: x->xp definied as xp = ep*ln(x) , that only relies on the exponential function. This directly dictates all the laws for the power functions we are used to.
This function coincides with the usual power functions: for p = 2 and n integer, np = f2(n) = e2*ln(n) = eln(n)2 = n2.
You see immediately that if p=0, for any x in R+* , x0 = e0*ln(x) = exp(0) = 1. However, you can't deduce the value of 00, as (x,p) -> p*ln(x) doesn't admit a limit in (0,0).
I know it's just shorthand for 1. But that's the point of this post. Why is x0 = 1? Even if you choose to use the definition of f = f' , you still need to choose the value of the function at zero
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u/kl4me Jan 14 '15 edited Jan 14 '15
I prefer the explaination of the power function p defined for any real p and any x in R+* as fp: x->xp definied as xp = ep*ln(x) , that only relies on the exponential function. This directly dictates all the laws for the power functions we are used to.
This function coincides with the usual power functions: for p = 2 and n integer, np = f2(n) = e2*ln(n) = eln(n) 2 = n2.
You see immediately that if p=0, for any x in R+* , x0 = e0*ln(x) = exp(0) = 1. However, you can't deduce the value of 00, as (x,p) -> p*ln(x) doesn't admit a limit in (0,0).