RE: [obm-l] Limite

["Paulo Santa Rita" <p_ssr@xxxxxxxxxxx>]

Se a^X=1 + (1/Y) entao Y -> INF implica X -> 0.

a^X=1 + 1/Y implica : a^X - 1 = 1/Y mas implica tambem : X*Ln(a) = Ln(1 + 
1/Y) =>
1/X = Ln(a)/Ln(1 + 1/Y). Portanto : (a^X - 1)/X =[ (1/Y)*Ln(a) ]/Ln(1 + 1/y) 
=
Ln(a)/Ln[(1+1/Y)^Y].

lim (x->0) { (a^X - 1)/X } = lim (y->INF) { Ln(a)/Ln[(1+1/Y)^Y] } = 
Ln(a)/Ln(e) = Ln(a)



>From: "leonardo mattos" <leonar_matt@hotmail.com>
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>To: obm-l@mat.puc-rio.br
>Subject: [obm-l] Limite
>Date: Wed, 18 Feb 2004 14:16:00 +0000
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>Ola pessoal,
>
>Como calcular o limite de [(a^X)-1]/x quando x tende a zero sem utilizar a 
>regra de L´Hopital?
>
>Um abraço,
>Leonardo
>
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