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/
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