Re: help, epsilon - N proof of a limit of a sequence
- From: Tim Smith <reply_in_group@xxxxxxxxxxxxxxxx>
- Date: Wed, 17 Sep 2008 18:52:50 -0700
In article <a3e7a$48d13cdf$c650990a$15723@xxxxxxxxxxxxxxxxxx>,
TheGist <thegist@xxxxxxxxxx> wrote:
Tim Smith wrote:
In article <thegist-0E51F7.23083014092008@xxxxxxxxxxxxxxxxxx>,ok, one more question...
TheGist <thegist@xxxxxxxxxx> wrote:
AArgh, these radical expressions are a giant pain.
Here is another attempt at one...
Show lim n / (sqrt(n^2 + 3)) = 1
Solution:
to find N such that |(n / (sqrt(n^2 +3 )) - 1 | < epsilon
(n / (sqrt(n^2 + 3)) - ((sqrt(n^2 + 3) / (sqrt(n^2 + 3))
= (n - sqrt(n^2 + 3)) / (sqrt(n^2 + 3)
= (n(sqrt(n^2 +3 )) - n^2 + 3) / (n^2 + 3)
This last expression is <= 1/n
That is,
n(sqrt(n^2 + 3)) - n^2 + 3) / (n^2 + 3) < 1/n
So, if N > 1/epsilon, then for all N > n
|(n / (sqrt(n^2 + 3)) -1| < epsilon < 1/N < 1/n < epsilon
Does this look right?
I think I am getting the hang of this... ;)
This gives a tighter bound, and seems clearer to me:
abs( 1 - n/sqrt(n^2+3) )
Why did you do this instead of
abs(n/sqrt(n^2+3) -1) ?
I think I might be missing something very critical in my understanding
my not seeing why...
abs(x-y) = abs(y-x) for all x, y, so either order gives the same result.
--
--Tim Smith
.
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- From: Tim Smith
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