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Re: [obm-l]
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l]
- From: "Marcos Martinelli" <mffmartinelli@xxxxxxxxx>
- Date: Thu, 26 Apr 2007 23:10:25 -0300
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Se x->0^+, naturalmente x>0. Como 2/x-1<[2/x]<=2/x ->
2/3-x/3<x/3*[2/x]<=2/3. Pelo Teorema do Confronto temos, portanto,
lim(x->0^+)(x/3.[2/x])=2/3.
Mas no outro item, basta observar que 2/x.[x/3]=0 para qq 0<x<1. Logo
o limite procurado é nulo.
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