Olá2)-1/x <= sen(x^1000)/x <= 1/xqdo x -> +inf.. -1/x e 1/x tendem para 0.. pelo teorema do confronto (sanduiche), o limite de sen(x^1000)/x -> 0 quando x-> 0.abraços,Salhab----- Original Message -----From: Klaus FerrazSent: Sunday, May 21, 2006 10:37 AMSubject: [obm-l] LIMITES1)Determine lim(n->+inf) (1+1/2)*(1+1/2^2)*(1+1/2^3)*...*(1+1/2^n).2)Determine lim(x-->+inf) sen(x^1000)/xGrato.
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