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Square Roots and Limits

Date: 03/12/2003 at 21:39:19
From: Jonathan Coveney
Subject: Square Roots

Given a value of x, find the value of the expression sqrt(x sqrt(x
sqrt(x... and so on, and explain why.

Date: 03/13/2003 at 03:41:32
From: Doctor Ian
Subject: Re: Square Roots

Hi Jonathan,

The easiest way to deal with this is to make use of the fact that




are just two ways of saying the same thing. If that doesn't seem
familiar to you, take a look at 

   Properties of Exponents


   |     __          1/2 1/2     
  \| x \| x  = (x * x   )         

                 3/2 1/2
             = (x   )

                (3/2 * 1/2)
             = x 

             = x 
   |     _______   
   |    |     __         3/4 1/2
  \| x \| x \| x = (x * x   )

                     7/4 1/2
                 = (x   )

                    (7/4 * 1/2)
                 = x   

                 = x   

At each step, the exponent changes from 



  ((k + (k-1))/k)/2 = (2k - 1)/(2k)

More generally, for n square roots, the exponent is 

  (2^n - 1)/(2^n)
So you're right that as n approaches infinity, the exponent approaches
1, and the value of the expression approaches x.  

However, note that the denominator of the exponent will always be a
power of 2, so x^(19/20) isn't really part of the sequence.  

Now for a bit of terminology. The way that we express this situation
is to say

  Limit    (2^n - 1)/(2^n)  = 1

  "The limit, as n goes to infinity, of [the expression] is
   equal to 1."

This is somewhat subtle.  Note that we _don't_ say that the expression

  (2^n - 1)/(2^n)

is ever actually equal to 1, because that's never the case. And we
_don't_ say that n ever _becomes_ 'infinity', because infinity isn't a
number, so we can't use it in arithmetical expressions.  

We're saying that we can make the value of the expression as close to
1 as we wish, by choosing a suitably high value for n.  

You'll learn more about limits when you get into calculus.  

Does this help? 

- Doctor Ian, The Math Forum

Date: 03/13/2003 at 08:29:19
From: Jonathan Coveney
Subject: Thank you (Square Roots)

Thanks, I wondered what limits were, exactly. But now it makes plain 
Associated Topics:
High School Calculus
High School Polynomials
High School Square & Cube Roots

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