Why Are 1^infinity, infinity^0, and 0^0 Indeterminate Forms?

Date: 05/08/98 at 11:44:56
From: Tim Hanna
Subject: Indeterminates

I am a senior in Advanced Placement Calculus, and my class is having a 
hard time understanding some indeterminate forms. We know that these 
are indeterminate forms:

   0/0
   infinity/infinity
   infinity - infinity,

But why are these indeterminate forms?

   1^infinity
   infinity^0
   0^0 

We feel that 1^infinity = 1, infinity^0 = 1, and 0^0 = 1. Part of 
these conclusions come from the fact that 0^infinity = 0 and 0^(-
infinity) = infinity. Could you please explain these determinate and 
indeterminate forms?

Date: 05/08/98 at 15:31:02
From: Doctor Rob
Subject: Re: Indeterminates

These forms are called indeterminate because if you replace 1, 0, and
infinity by functions the limits of which are 1, 0, and infinity as 
x -> 0, then the limit of the compound function does not exist, in the 
sense that the limit depends on which functions you choose.

For an example of this discrepancy for 1^infinity, on the one hand, 
take f(x) = 1 and g(x) = 1/x. Then:

   lim f(x)^g(x) = lim 1^(1/x) = lim 1 = 1

On the other hand, if we take f(x) = 1 + x and g(x) = 1/x, then:

   lim f(x)^g(x) = lim (1 + x)^(1/x) = e = 2.718281828459... > 1

For an example of 0^0, on the one hand, take f(x) = 0, g(x) = x. Then:

   lim f(x)^g(x) = lim 0^x = lim 0 = 0

On the other hand, if we take f(x) = x, and g(x) = 0, then:

   lim f(x)^g(x) = lim x^0 = lim 1 = 1 > 0

All of the seven indeterminate forms are the same:

   0/0                  [(k*x)/x -> k, for any real k]
   0*infinity           [(k*x)*(1/x) -> k, for any real k]
   infinity/infinity    [(k/x)/(1/x) -> k for any real k]
   infinity - infinity  [(k+1/x)-(1/x) -> k for any real k]
   1^infinity           [(1+x)^(ln[k]/x) -> k for any positive real k]
   infinity^0           [(1/x)^(ln[k]*x/[1-x]) 
                                         -> k for any positive real k]
   0^0                  [x^(ln[k]/ln[x]) -> k for any positive real k]

It is a useful exercise to prove all these statements.

-Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/