Re: Limit of function

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 01/10/05 

Date: Mon, 10 Jan 2005 10:37:53 +0000

ticbol wrote:

> Robert,
>
> lim ln(x^2 -2x +1) / ln(x^10 +3x +1)
> x->inf
>
> = (inf)/(inf)
> Indeterminate again.
>
> Use L'Hopital's Rule again.
>
> f(x) = ln(x^2 -2x +1)
> So, f'(x) = [1/(x^2 -2x +1)]*(2x -2)
> = (2x-2)/(x^2 -2x +1)
>
> g(x) = ln(x^10 +3x +1)
> So, g'(x) = [1/(x^10 +3x +1)]*(10x^9 +3}
> = (10x^9 +3)/(x^10 +3x +1)
>
> Then,
> lim ln(x^2 -2x +1) / ln(x^10 +3x +1)
> x->inf
>
> = lim [(2x-2)/(x^2 -2x +1)] / [(10x^9 +3)/(x^10 +3x +1)]
> x->inf
>
> = lim [(2x-2)(x^10 +3x +1)] / [(x^2 -2x +1)(10x^9 +3)]
> x->inf
>
> = lim [2x^11 +6x^2 +2x -2x^10 -6x -2] / [10x^11 -20x^10 +10x^9 +3x^2
> -6x +3]
> x->inf
>
> = lim [2x^11 -2x^10 +6x^2 -4x -2] / [10x^11 -20x^10 +10x^9 +3x^2 -6x
> +3]
> x->inf
>
> Divide both numerator and denominator by x^11,
>
> = lim [2 -2/x +6/x^9 -4/x^10 -2/x^11] / [10 -20/x +10/x^2 +3/x^9
> -6/x^10 +3/x^11]
> x->inf
>
> = [2] / [10]
>
> = 2/10 = 1/5 = 0.2 -----answer.

Typical "mathematics made difficult" :-(

log(x^2 -2x + 1) = log(x^2) + log(1 - 2/x + 1/x^2) = 2 log x + O(1/x)

similarly

log(x^10 + 3x +1) = 10 log x + O(1/x^9).

Hence
log(x^2 -2x + 1)/log(x^10 + 3x +1) = 2/10 + O(1/(x log x)) -> 1/5
as x -> infinity. 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_

Relevant Pages