Re: Limit of function

From: denis feldmann (denis.feldmann_at_wanadoo.fr)
Date: 01/10/05 

Date: Mon, 10 Jan 2005 13:49:37 +0100

ticbol a écrit :
> Robert,
>
> lim ln(x^2 -2x +1) / ln(x^10 +3x +1)
> x->inf
>
> = (inf)/(inf)
> Indeterminate again.
>
> Use L'Hopital's Rule again.

Really, you should learn some other trick. What would you do for lim
ln(x^2 -2x +sin(x*exp(x^2))) / ln(x^10 +3x +cos(x*exp(x^2))) ??
   x->inf

Approximations, use of Taylor's formula, factorisations (by largest
term), etc. are all general methods , with two big advantages : 1) the
student has a chance of understanding what is happening 2) he dosn't
have to memorize the exact conditions of application of a "magical" rule
(for Taylor, this may be debatable).

>
> 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.
>

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