Associated Topics || Dr. Math Home || Search Dr. Math

### 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,

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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search