<Jer### [at] enstfr>  wrote:

>  Moreover your implication goes the wrong way (ie the second inequation
> implies the first, not the other way round).

Hmm, maybe you are forgetting some fundamental things here...

2n lg n   >   2n + n lg n    |   : n lg n

=>

n lg n  >  lim   2n

because n lg n can never be 0 in this case (so the division is legal).  Of
course I am just silently dropping the + 1 here and some other changes to
the function because of the division.

>  This doesn't make any mathematical sense since all four limits are
> infinity and therefore can't be compared. What can be said is that for a
> big enough n, you have:
> 2n lg(n) > 2n + n lg(n)
> and
> n lg(n) > 2n

Nope, you can do the following (as what I am up to is not a result, but only
a relative comparison of the growth rate):

 lim   2n lg n   >   lim   2n + n lg n
n->oo               n->oo

<=>

 lim   2n lg n - 2n + n lg n  >  0
n->oo

<=>

 lim   n lg n - 2n  >  0
n->oo


       Thorsten


____________________________________________________
Thorsten Froehlich, Duisburg, Germany
e-mail: tho### [at] trfde

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