Carissimo
Gabriel,
1)
lim
(x->3) (raiz(x^2+16) – 5) (raiz(x^2+16) +5)/((x^2-3x) (raiz(x^2+16) + 5)) =
lim(x->3) (x^2-9)/ )/((x^2-3x) (raiz(x^2+16) + 5)) =
=
lim(x->3)(x+3)(x-3)/x(x-3) raiz(x^2+16) + 5) = lim(x->3) (x+3)/x
raiz(x^2+16) + 5) = 6/3(5) = 3/5.
2)
lim(x->00)
raiz(x+1)/raiz(9x+1) = lim(x->00) raiz(x)*raiz(1+1/x) / raiz(x)*raiz(9+1/x)
= raiz(1+1/x)/raiz(9+1/x) = 1/9
3)
lim(x->0)
(1-cosx)(1+cosx)/x^2(1+cosx) = lim(x->0) (sinx)^2/(x^2*(1+cosx)) =
lim(x->0) ((sin x)/x)^2 * 1/(1+cosx) = ½
Acho que os
resultados sao esses caro amigo.
Leandro.
-----Original
Message-----
From:
owner-obm-l@sucuri.mat.puc-rio.br [mailto:owner-obm-l@sucuri.mat.puc-rio.br]
On Behalf Of Afemano
Sent: Tuesday, March 11,
2003 12:00
PM
To: obm-l@mat.puc-rio.br
Subject: [obm-l] AJUDA COM
LIMITES
Olá, alguém pode me ajudar com
esse problemas "simples" ??
1) lim(x-> -3)
(raiz(x^2 + 16) - 5 )/ ( x^2 - 3x )
2) lim(x-> +00) ( raiz(x
+ 1) ) / ( raiz(9x + 1) )
3) lim(x-> 0) ( 1 - cosx ) /
x^2