AD1 CI 2012 2 gabarito

Respostas AD01 - 2/2012 - CÁLCULO I
QUESTÃO 01. [3, 0 pontos]
Calcule os seguintes limites de funções: x2 + 2x − 3 x2 − 3x + 2

(a) lim x→1 (b) lim x→4 1 − cos (x + 1) x2 + 2x + 1

(c) lim x→−1 (d) lim x→3 Solução:

2x sen (x − 4) x2 − 3x − 4

x−3
√
−4 + x2 + 7

x2 + 2x − 3
(x − 1)(x + 3) x+3 4
(a) lim 2
= lim
= lim
=
= −4. x→1 x − 3x + 2 x→1 (x − 1)(x − 2) x→1 x − 2
−1
(b) lim x→4 = lim x→4 2x sen (x − 4)
2x sen(x − 4)
2x sen(x − 4)
= lim
= lim
=
2 x→4 x→4 x − 3x − 4
(x + 1)(x − 4) x+1 x−4

sen(x − 4)
8
8
2x
lim
= (1) = x + 1 x→4 x − 4
5
5

(c) lim x→−1 = lim x→−1 = lim x→−1 1 − cos (x + 1)
= lim x→−1 x2 + 2x + 1

1 − cos (x + 1) 1 + cos (x + 1)
=
x2 + 2x + 1 1 + cos (x + 1)

1 − cos2 (x + 1) sen2 (x + 1)
=
lim
=
(x2 + 2x + 1)(1 + cos (x + 1)) x→−1 (x2 + 2x + 1)(1 + cos (x + 1))
1
sen2 (x + 1)
1
lim lim =
2
1 + cos (x + 1) x→−1 (x + 1)
2 x→−1

sen (x + 1) x+1 2

=

1 sen (x + 1) sen (x + 1) sen (x + 1) sen (x + 1)
1
= lim lim
=
x→−1 x→−1 x+1 x+1 2 x+1 x+1
2
√ x−3 4 + x2 + 7 x−3 √
√
√
=
= lim
(d) lim x→3 −4 + x2 + 7 x→3 −4 + x2 + 7 4 + x2 + 7
√
√
(x − 3)(4 + x2 + 7)
(x − 3)(4 + x2 + 7)
= lim
= lim
=
x→3 x→3 (x2 + 7) − 16 x2 − 9
√
√
(x − 3)(4 + x2 + 7)
4 + x2 + 7
8
4
= lim
= lim
= = x→3 x→3
(x − 3)(x + 3) x+3 6
3
=

1 lim 2 x→−1

QUESTÃO 02. [2, 0 pontos]
Seja f a função definida por:

f (x) =









4 |1 − x| ,

se x < −1

6x2 − 3x − 1, se −1 ≤ x < 3
√
se x ≥ 3
2 x + 1,

1

(a) Calcule lim + f (x) e x→−1 lim f (x). O que você pode concluir do lim f (x)? x→−1 x→−1−

(b) Calcule lim+ f (x) e lim− f (x). O que você pode concluir do lim f (x)? x→3 x→3

x→3

Solução:
(a) Temos que: lim f (x) = lim + 6x2 − 3x − 1 = 8 e

x→−1+

x→−1

lim f (x) = lim − 4 |1 − x| = lim − 4 (1 − x) = 8.

x→−1−

x→−1

x→−1

Assim, como lim + f (x) = lim − f (x) = 8, concluimos que lim f (x) = 8. x→−1 x→−1

x→−1

(b) Temos que: lim+ f (x) = lim+ 2

x→3

√

x+1=4 e

x→3

lim f (x) = lim− 6x2 − 3x − 1 = 44.

x→3−

x→3

Assim, como lim+ f (x) = 4 = 44 = lim− f (x),