Matematica
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6 páginas
Cálculo IProfº. Marcello Santos Chaves
LIMITES E CONTINUIDADES
1)
Lim f ( x ) = x →1
x −1
3
x −1
x+e − x e 2) Lim f ( x) = e→0 Solução :
Solução:
Faça → u 3 = x
x +e − x x+e + x
Lim f ( x) = Lim
×
e→0 e→0 e
x+e + x
u3 −1
Lim f (u ) = Lim
u3 −1 u3 −1
Lim f (u ) = Lim u →1 u →1 u − 1
(u − 1) ⋅ u 2 + u + 1
Lim f (u ) = Lim u →1 u →1
(u − 1) u →1
u →1 3
(
(
)
)
Lim f (u ) = Lim u 2 + u + 1 u →1
u →1
Lim f (u ) = 12 + 1 + 1
) ( x)
2
x+e − e→0 e→0 e × x+e× x+e− x
Lim f ( x) = Lim e→0 e→0 e × x+e × e Lim f ( x) = Lim e→0 e→0 e × x+e ×
1
Lim f ( x) = Lim e→0 e→0 x+e× x
1
Lim f ( x) = Lim e→0 e→0 x+0× x
1
Lim f ( x) = e→0 x× x
(
2
x
Lim f (u ) = 1
)
(
x
)
(
x
)
(
(
Lim f ( x) =
)
(
u →1 u →1
(
Lim f ( x) = Lim
)
)
1
e→0
2 x
Lim f ( x) =
⇒ Lim f ( x) =
2 x
2 x
⇒ Lim f ( x) =
⇒
e→0
4x
2 x 2 x
⋅
x
2x
e→0
1
Marcello Santos Chaves
Instituto Federal de Educação, Ciência e Tecnologia (IFPA)
e→0
1
Belém-PA, Abril de 2011
Cálculo I
Profº. Marcello Santos Chaves
3) Lim f ( x ) = Lim x →1
x →1 3
x −1 x− x
4)
Lim f ( x ) = Lim x→0 3
x→0
x +1 −1 x Solução :
Solução :
Faça → x = u 3
Faça → u = 3 x + 1 ⇔ x = u 3 − 1
Lim f (u ) = Lim
u3 −1
Lim f (u ) = Lim
u− u u3 −1
Lim f (u ) = Lim u →1 u →1 u − u u u →1
u →1
Lim f (u ) = Lim u →1
u →1
u →0
3
Lim f (u ) = Lim
)(
−1 ⋅ u + u u
3
Lim f (u ) = Lim
u →1
)
u2 − u u
2
)(
(
(u
)(
+ u +1 ⋅ u + u u u →1 u →1
− u2
12 + 1 + 1 ⋅ 1 + 1 1
Lim f (u ) = u →1
− 12
3× 2
Lim f (u ) = − u →1
1
Lim f (u ) = −6
Lim f (u ) = Lim
(
2
)(
u →0
(
2
)
2
u →0
u2 − u3
(u − 1) ⋅ u 2 + u + 1 ⋅ u + u u
Lim f (u ) = Lim u →1 u →1
− u 2 ⋅ (u − 1) u →1
u →0