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Limit As n Approaches Infinity


Date: 9/7/96 at 0:33:58
From: Shane Eaton
Subject: Limit as n tends to infinity

Dear Dr. Math,

Please help me with the following limit:

    lim             (sqrt(2n^2+1))/(2n-1)
n-> -infinity


I can get the answer when n tends to + infinity but I can't solve
this variation of the question. I know it has something to do with
using absolute value but I'm not quite sure.

Thanks,

Shane Eaton


Date: 9/7/96 at 4:17:34
From: Doctor Mike
Subject: Re: Limit as n tends to infinity
  
Hello Shane,

Here is one way to solve it. There may be others. If you let m = -n 
then having n tend to - infinity is the same as having m tend to 
+ infinity.  Your - infinity limit then becomes : 
  
   lim         (sqrt(2(-m)^2+1))/(2(-m)-1) 
   m->+infinity
  
=  lim         (sqrt(2m^2+1))/(2m+1)*(-1)
   m->+infinity
  
=  (-1) * lim         (sqrt(2m^2+1))/(2m+1)
          m->+infinity
  
Now this last limit is essentially the same problem as the + infinity 
limit you said you can do. In fact, having (2m-1) or (2m+1) or 2m in 
the denominator results in the same + infinity limit. (Of course, 
calling the variable m instead of n has no effect.)  So, your problem 
limit is just the negative of the + infinity limit you have already 
found.
  
I hope this helps. If you need any further details on this just write 
back.  

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

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