RE: [obm-l] Limite

["leonardo mattos" <leonar_matt@xxxxxxxxxxx>]

Gostei do vodoo do a^x=1+1/y


>From: "Paulo Santa Rita" <p_ssr@hotmail.com>
>Reply-To: obm-l@mat.puc-rio.br
>To: obm-l@mat.puc-rio.br
>Subject: RE: [obm-l] Limite
>Date: Wed, 18 Feb 2004 15:03:43 +0000
>
>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>
>>Reply-To: obm-l@mat.puc-rio.br
>>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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