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What is 1^infinity?


Date: 12/10/98 at 00:39:32
From: Jonathan Hiscock
Subject: What is 1^infinity?

My calculus class recently came across this problem while working on 
l'Hopital's rule for limits. Often using a direct substitution method, 
we would get an answer with 1^infinity. We were told that this is an 
indeterminate form and to try to rewrite the problem to get an 
appropriate form. Our calculus teacher showed us a proof using a 
problem similar to:

   f(x) = lim 1^x
   x->infinity
  
And then by using natural logs to evaluate. This method did not seem 
to work to me or the rest of our class. Our reasoning is that 
1 = 1 * 1 = 1 * 1 * 1 = ..., and so the answer will always be one. How 
can I prove that 1^infinity is one or indeterminate?
  
Thank you for your time,
Jonathan Hiscock


Date: 12/10/98 at 03:15:05
From: Doctor Schwa
Subject: Re: What is 1^infinity?

When you have something like "infinity," you have to realize that it's
not a number. Usually what you mean is some kind of limiting process.
So if you have "1^infinity" what you really have is some kind of limit: 
the base isn't really 1, but is getting closer and closer to 1 perhaps 
while the exponent is getting bigger and bigger, like maybe (x+1)^(1/x) 
as x->0+.

The question is, which is happening faster, the base getting close to 
1 or the exponent getting big? To find out, let's call:

   L = lim x->0 of (x+1)^(1/x)
 
Then:

   ln L = lim x->0 of (1/x) ln (x+1) = lim x->0 of ln(x+1) / x

So what's that? As x->0 it's of 0/0 form, so take the derivative of the
top and bottom. Then we get lim x->0 of 1/(x+1) / 1, which = 1.
So ln L = 1, and L = e. Cool!

Is it really true? Try plugging in a big value of x. Or recognize this 
limit as a variation of the definition of e. Either way, it's true. The 
limit is of the 1^infinity form, but in this case it's e, not 1. Try 
repeating the work with (2/x) in the exponent, or with (1/x^2), or with 
1/(sqrt(x)), and see how that changes the answer.

That's why we call it indeterminate - all those different versions of
the limit approach 1^infinity, but the final answer could be any 
number, such as 1, or infinity, or undefined. You need to do more work 
to determine the answer, so 1^infinity by itself is not determined yet. 
In other words, 1 is just one of the answers of 1^infinity.

I hope that helps clear things up!

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Calculus

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