Folha de exercícios nº 3

Departamento de Matemática e Engenharias
MATEMÁTICA GERAL
Licenciatura em Biologia
2o Semestre 2004/2005
Resoluções da folha de exercícios no 3
Continuidade. Diferenciabilidade. Limites
1. Expressão da derivada de ordem n de funções:
1.1 f (x) = (2x)eπ ⇒ f ′ (x) = 2eπ (2x)eπ−1 ⇒ f ′′ (x) = 22 eπ (eπ − 1) (2x)eπ−2 ⇒
⇒ f ′′′ (x) = 23 eπ (eπ − 1) (eπ − 2) (2x)eπ−3 = 2eπ eπ (eπ − 1) (eπ − 2) xeπ−3
Concluimos, assim, que f (n) (x) = 23 eπ (eπ − 1) . . . (eπ − n + 2) (eπ − n + 1) (2x)eπ−n
1.2 h(x) = (1 + x)k ⇒ h′ (x) = k (1 + x)k−1 ⇒ h′′ (x) = k (k − 1) (1 + x)k−2 ⇒
⇒ h′′′ (x) = k (k − 1) (k − 2) (1 + x)k−3
Logo h(n) (x) = k (k − 1) . . . (k − n + 1) (1 + x)k−n
2
2
2
2
1.3 g(x) = e−e x+b ⇒ g′ (x) = −e2 e−e x+b ⇒ g ′′ (x) = −e2 e−e x+b ⇒
3
2
⇒ g′′′ (x) = −e2 e−e x+b
n
2
Portanto g(n) (x) = −e2 e−e x+b
1.4 j(x) = log x ⇒ j ′ (x) =
1
x
⇒ j ′′ (x) = − x12 ⇒ j ′′′ (x) =
Concluímos que j (n) (x) = (−1)n−1
1.5 i(x) =
(n−1)!
xn ,
2
x3
⇒ j (4) (x) = − 2·3
x4
para n ∈ N.
π
π
π·2
π·2·3
⇒ i′ (x) = −
⇒ i′′ (x) =
⇒ i′′′ (x) =
2
3
x−π
(x − π)
(x − π)
(x − π)4
Assim, i(n) (x) = (−1)n
1.6 m(x) =
π · n!
(x − π)n+1
2
3
e−1
e (e − 1)
′′ (x) = e (e − 1) 2 ⇒ m′′′ (x) = − e (e − 1) 2 · 3
⇒
m
⇒ m′ (x) = −
ex
(ex)2
(ex)3
(ex)4
Neste caso m(n) (x) = −
en (e − 1) n!
(ex)n+1
1
2. Calculo de limites:
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2x
2
lim
= −2
= lim
2
x→0 x − x
R.C. x→0 2x − 1
√
√2
√
√
√
2x + 1 − 3
2 2x+1
4√ x−2
2 2
4√2
√
= lim
=
lim
=
lim √
=
3
2 9
x→4 2 2x+1
x→4
x − 2 − 2 R.C. x→4 2√1x−2
sin (x + π)
= −1
lim
−1 = 1
cos 2x
x→ π2
√
x2
x
x2
= lim =
√
= lim √ lim
x→+∞
x→+∞ 2 + x x
x→+∞
2
2
√
x x x√
+
1
+
1
x
x x
+∞
0
= +∞
x
x 1 − 23
3x 1 − 23
x
x
= lim
=
= lim x lim
x→+∞ 3 + 2
x→+∞ 3 3 + 2
x→+∞
2−2
2−2
3
3
(2x + 1)2
2x + 1
4x2 + 4x + 1
√
√
lim
=
= lim
= lim
x→+∞
x→+∞
x→+∞
x2 + 1
x2 + 1
x2 + 1
4x2 + 4x + 1
4x2
4
=
=
=2
=
lim
lim
lim
2
2
x→+∞
x→+∞
x→+∞
x +1
x
1
sin x
lim
= lim cos x = 1
x→0
R.C. x→0 1
x
3 cos 3x
sin 3x
= lim
= 23
lim
2
x→0 1 − (1 − x)
R.C. x→0 2 (1 − x)
4 cos x
−4 sin x
=2
limπ
= limπ
π − 2x R.C. x→ 2 −2
x→ 2
2.10 lim
x→0
3x − 2x
3x+1 + 2x−2
sin x
1 − cos x
= lim
R.C. x→0
3.
cos x
=∞
sin x

kex+2




 x5 − 4x3
g(x) =
x+2


0



x sin x + 3 cos x
3.1 Temos lim g(x) = lim
x→0−
x→0−
x ≤ −2
−2 < x < 0
x=0
x>0
x5 − 4x3
=0
x+2
lim g(x) = lim (x sin x + 3 cos x) = 3
x→0+
x→0+
Como lim g(x) = lim g(x) concluímos que g não é contínua em x = 0.
x→0−
x→0+
2
1
3
3.2 Queremos que
Temos
lim g(x) = lim g(x) = g (−2)
x→−2+
x→−2−
lim g(x) = lim kex+2 = k = g (−2)
x→−2−
x→−2−
lim g(x) = lim
x→−2+
x→−2+
x5 − 4x3
5x4 − 12x2
= lim
= 32
x + 2 R.C. x→−2+
1
Para que g seja contínua em x = −2 é necessário que se tenha k = 32.
3.3 Temos:
g ′ (0− ) = lim
g(x)−g(0)
x−0
x5 − 4x3
4
2
x
x
−
4x
x4 − 4x2
2 = lim
= lim x +
=
lim
=0
x
x (x + 2)
x→0−
x→0−
x→0− x + 2
g ′ (0+ ) = lim
g(x)−g(0)
x−0
= lim
x→0−
x→0+
x→0+
x sin x+3 cos x
= lim sin x+x cos1 x−3 sin x
x
R.C. x→0+
=0
Como g ′ (0− ) = g ′ (0+ ) = 0 concluímos que g′ (0) = 0.
3.4
lim g(x) = lim kex+2 = 0.
x→−∞
x→−∞
4. Utilizando a folha de apoio 1, resolva 1 exercício de cada grupo.
1. Indeterminações do tipo: 00
5
5x4
x −1
=5
= lim
1.1 lim
x→1
x − 1 R.C. x→1 1
x − tg x
−2 sec2 x tg x
1 − sec2 x
1.2 lim
= lim
=
= lim
x→0 x − sin x
R.C.
R.C. x→0 1 − cos x R.C. x→0
sin x
−2(2 sec2 x tg2 x+sec4 x)
cos x
R.C. x→0
= lim

=
−2
1
= −2
−2
x2

x+2
x+2
 log

x
x = lim
 = lim
1.3 lim 
x→+∞
x→+∞ 
x − 2  R.C. x→+∞ x22
log
x
−
2
x
x
−x(x−2)
x→+∞ x(x+2)
= lim
1.4
−2 x
x2 x+2
2 x
x2 x−2
=
=
(x−2)
= lim − (x+2)
= −1
x→+∞
− π2 x cos π2 x
− π2 cos π2 x
1 − sin π2 x
=
lim
lim
=
lim
x→1
x→1
R.C. x→1
2 log x
(logx)2
2logx x
π
π
π
π
− cos 2 x − x 2 sin 2 x
−πx cos π2 x − x π2
= lim 2
=
lim
2
x→1
x→1
4
x
π2
−2
x(x+2)
2
x→+∞
x(x−2)
= lim
8
3
=
sin π2 x
=
−π (− π2 )
4
=
2. Indeterminações do tipo: ∞
∞
x
e
ex
ex
ex
=
lim
= +∞
=
lim
=
lim
2.1 lim
x→+∞ x3
R.C. x→+∞ 3x2 R.C. x→+∞ 6x R.C. x→+∞ 6
1 1 1 1
− x12 e x
ex
ex
ex
= lim
= lim
2.2 lim
= lim 2 =
−1
x2
x
cotg x R.C. x→0+
x→0+
x→0+
x→0+
2
sin2 x
sin x
sin x
+∞
1
= +∞
x
x + cos x
1 cos x
cos x 2.3 lim
= lim
+
+
= lim
=
x→+∞
x→+∞ 3
x→+∞ 3x
3x
3x
3x
=
1
1
1
1
+ lim
cos x = + 0 =
3 x→+∞ 3x
3
3
1
log x
x
= lim
2.4 lim
=
lim
x→+∞
x→+∞ 2x2 + log (x + 1)
R.C. x→+∞ 4x + 1
x+1
=
= lim
x→+∞
4x3
1
x
4x2 +4x+1
x+1
=
x+1
x
1
= lim
= lim
=0
2
3
x→+∞
x→+∞
+ 4x + x
4x
4x2
3. Indeterminações do tipo: 0.∞ e ∞ − ∞
3
3.1 lim x3 e−x = lim xex = lim
3.2 lim
x→0
3x2
= lim 6xx = lim 6x
x
R.C. x→+∞ e R.C. x→+∞ e R.C. x→+∞ e
x→+∞
x→+∞
1
3
− 2
sin x x
=0
2x − 3 cos x
x2 − 3 sin x
= lim
=∞
2
x→0
x sin x R.C. x→0 2x sin x + x2 cos x
= lim
4. Indeterminações do tipo: 00 , ∞0 e 1∞
log x
lim
sin x
1
4.1 lim xsin x = lim elog x
= lim esin x log x = lim e sin x = ex→0+
x→0+
lim
= e
x→0+
1
x
− cos x
sin2 x
sin 2x
lim − cos x+x
sin x
= ex→0+
R.C.
1
lim (x + 1) log x = lim elog(x+1)
x→+∞
= e
lim
x→+∞
4.3 lim |cos x|
1
x+1
1
x
1
x−π
4.4 lim (cos 3x)
lim
1
x2
= lim e
=
R.C.
= e0 = 1
log(x+1)
log x
x→+∞
lim
= ex→+∞
log(x+1)
log x
=
R.C.
x
= ex→+∞ x+1 = e
= lim e
log|cos x|
x−π
x→π
x→π
=e
1
log x
x→+∞
R.C.
x→0
x→0+
2
x
lim − xsin
cos x
= ex→0+
R.C.
4.2
x→0+
x→0+
log x
1
sin x
= lim e
log(cos 3x)
x2
x→0
−9 cos 3x
lim
x→0 2(cos 3x−3x sin 3x)
=e
lim
x→π
=e
lim
log|cos x|
x−π
x→0
9
= e− 2
4
=e
log(cos 3x)
x2
lim
x→π
=e
sin x
cos x
1
lim
x→0
= e0 = 1
−3 sin 3x
cos 3x
2x
lim
−3 sin 3x
= ex→0 2x cos 3x =
5. Calcule cada um dos seguintes limites:
e2x − 1
2e2x
1
=
lim
=−
x→0 1 − ex − 3x R.C. x→0 −ex − 3
2
5.1 lim
1
1
log x − 1
= lim x =
x→e x − e
R.C. x→e 1
e
5.2 lim
sin x − x
cos x − 1
= lim
=0
2
x→0 tg (2x) − x R.C. x→0 2 sec2 (2x) − 2x
1
log x
1
x−1
x
=
+
+
= lim
5.4 lim
x→1 x3 − 1
tg (x − 1) R.C. x→1 3x2 sec2 (x − 1)
5.3 lim
5
1
3
+1=
4
3