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Infinity in a Fraction


Date: 05/13/97 at 21:44:58
From: Tovah Kerr
Subject: Pre-Calculus

I have a question on an answer to a problem. Here is the problem:     

   limit    (n+1)^.5
   n-->oo  ----------
            (n-1)^.5

According to my text, if you substitute infinity for n, the whole 
thing becomes:
                                                      
   infinity
   ---------  = 1
   infinity                                                                     

But infinity is a really large number, so infinity divided by itself 
should be undefined or no limit. This is because infinity is not a 
designated number and an unknown number over an unkown number is an 
unknown, right?  This is my logic, if you can understand it because it 
is hard to explain over e-mail.
                       
Thanks for your attention,
Tovah


Date: 05/14/97 at 11:29:54
From: Doctor Wilkinson
Subject: Re: Pre-Calculus

Dear Tovah,

You're right to notice that your text's method isn't very good. 
Sometimes you have to do something trickier than just substituting 
infinity for n. You can't really substitute infinity anyway, because 
infinity isn't a number. Remember that for an expression to tend to a 
limit as n tends to infinity really means that if you make n large 
enough, the expression will be as close as you want to the limiting 
value. Now the expression you have is a quotient of things that both 
get big when n gets big. That doesn't tell you much, as you noticed.  

Let's see if we can simplify things a little. We have the same 
exponent in the numerator and the denominator, so let's forget about 
it at first. If we can find the limit of (n+1)/(n-1), then we can 
raise it the the .5 power after we're done.

Now let's see if we can guess the answer. If we can guess it, maybe 
we can prove it.  Let's try a few values of n and make a table:

      n  (n+1)/(n-1)
      -  -----------

      2     3/1    3
      3     5/2    2.5
      4     5/3    1.66
      5     6/4    1.5
      6     7/5    1.4
     21    22/20   1.1
    101   102/100  1.02

It certainly looks like the values are getting close to 1.

So maybe we can write (n+1)/(n-1) as 1 + (something that gets small).

Well: (n+1)/(n-1) - 1 = (n+1)/(n-1) - (n-1)/(n-1) = 2/(n-1)

Now we have a fixed numerator and a denominator that gets big, so it
certainly tends to 0, so the original expression tends to 1.

I suggest you study this example carefully, since it's quite typical.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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