Lista 6 - Exercícios resolvidos de limites

Prévia do material em texto

1. Verifique se o limite existe ou não. Se o limite existir, calcule-o e se não existir justifique sua
resposta.
a. lim
x→3
√
x2 − 2x+ 6−
√
x2 + 2x− 6
x2 − 4x+ 3
;
Temos que
lim
x→3
√
x2 − 2x+ 6−
√
x2 + 2x− 6
x2 − 4x+ 3
= lim
x→3
(x2 − 2x+ 6)− (x2 + 2x− 6)
(x2 − 4x+ 3)(
√
x2 − 2x+ 6 +
√
x2 + 2x− 6)
=
= lim
x→3
−4(x− 3)
(x− 3)(x− 1)(
√
x2 − 2x+ 6 +
√
x2 + 2x− 6)
=
= lim
x→3
−4
(x− 1)(
√
x2 − 2x+ 6 +
√
x2 + 2x− 6)
=
=
limx→3−4
limx→3(x− 1)(
√
x2 − 2x+ 6 +
√
x2 + 2x− 6)
=
=
−4
(3− 1)(
√
32 − 2 · 3 + 6 +
√
32 + 2 · 3− 6)
=
=
−4
2 · (3 + 3)
=
−4
12
= −1
3
b. lim
x→−∞
2x+ 3
x+ 3
√
x
;
Temos que
lim
x→−∞
2x+ 3
x+ 3
√
x
= lim
x→−∞
2x+ 3
x+ x
1
3
= lim
x→−∞
2 + 3
x
1 + x
1
3
x
= lim
x→−∞
2 + 3
x
1 + 1
x·x−
1
3
= lim
x→−∞
2 + 3
x
1 + 1
x
2
3
=
2
1
= 2
c. lim
x→π
4
sen(x)− cos(x)
1− tan(x)
;
Temos que
lim
x→π
4
sen(x)− cos(x)
1− tan(x)
= lim
x→π
4
sen(x)− cos(x)
1− sen(x)
cos(x)
= lim
x→π
4
sen(x)− cos(x)
cos(x)−sen(x)
cos(x)
=
= lim
x→π
4
− cos(x) = − cos(π
4
) = −
√
2
2
d. lim
x→+∞
√
x√
x+
√
x+
√
x
.
Temos que, lim
x→+∞
1
x
+
√
1
x3
= lim
x→+∞
1
x
+
1
x
3
2
= 0.
Logo, lim
x→+∞
√
1
x
+
√
1
x3
=
√√√√ lim
x→+∞
(
1
x
+
√
1
x3
)
= 0
2
Exercícios resolvidos de cálculo 1 - Limite
Portanto, lim
x→+∞
1 +
√
1
x
+
√
1
x3
= 1. Desta forma,
lim
x→+∞
√
x√
x+
√
x+
√
x
= lim
x→+∞
√
x
x+
√
x+
√
x
= lim
x→+∞
√√√√√ 1
1 +
√
1
x
+
√
1
x3
=
=
√√√√√√√ limx→+∞
 1
1 +
√
1
x
+
√
1
x3
 = 11 = 1
3