Apostila CDI2 2013 01

http://slidepdf.com/reader/full/apostila-cdi2-2013-01 1/226
M i = M ax{f (x) : x ∈ [xi−1, xi]}.
a y c b


[a, b]

|[x0, x1]| = x1 − x0 = x1 |[x1, x2]| = x2 − x1 = x2
|[x2, x3]| = x3 − x2 = x3· · · |[xi−1, xi]| = xi − xi−1 = xi
· · · |[xn−1, xn]| = xn − xn−1 = xn.

[1, 12].
1 = x0 < 2 = x1 < 4 = x2 < 8 = x3 < 12 = x4.
[a, b]

P = {x0, x1, x2, · · · , xi, · · · , xn} Q = {x0, x1, x2, · · · , y0, · · · , xi, · · · , xn}
[a, b].
[1, 12]

M i
f
P

S (f, P ) = M 1(x1 − x0) + M 2(x2 − x1) + .. + M n(xn − xn−1) = n
i=1
[0, 2] .
mi
f

S (f, P ) = m1(x1 − x0) + m2(x2 − x1) + ... + mn(xn − xn−1) = n
i=1
[0, 2]
S (f, P )
n i=1
[a, b].
n i=1

k vezes
2
6

14 + 24 + 34 + ... + k4 = k (k + 1) (6k3 + 9k2 + k − 1)
30
y = 0
xi ∈ P

M i

S (f, P ) = M 1x + M 2x + M 3x + .... + M nx = f (x1)x + f (x2)x + f (x3)x + ... + f (xn)x
= f (x)x + f (2x)x + f (3x)x + ... + f (nx)x
= x[(x)2 + 1 + (2x)2 + 1 + (3x)2 + 1 + ... + (nx)2 + 1]
= x[1 + 1 + ... + 1 + (x)2 + 4(x)2 + 9(x)2 + ... + n2(x)2]
= x[n + x2(1 + 22 + 32 + ... + n2)]
= x

xi ∈ P
x0 = 1, x1 = 1 + x, x2 = 1 + 2x, x3 = 1 + 3x, · · · , xn = 1 + nx.
mi

S (f, P ) = m1x + m2x + m3x + .... + mnx
= f (x1)x + f (x2)x + f (x3)x + ... + f (xn)x
= f (1 + x)x + f (1 + 2x)x + f (1 + 3x)x + ... + f (1 + nx)x
= [16
− (1 + nx)2]x
= 16nx − [1 + 2x + (x)2 + 1 + 2 · 2x + (2x)2 + 1 + 2 · 3x + (3x)2 +
+ · · · + 1 + 2 · nx + (nx)2]x
= 16nx − nx − 2(1 + 2 + 3 + · · · + n)(x)2 − (12 + 22 + 32 + · · · + n2)(x)3
= 15nx − 2 · n(n + 1)
2 · (x)2 − n(n + 1)(2n + 1)
6 · (x)3
= 15n · 3
2n3

a
c

[
f
[−1, 1]

x0 = −1, x1 = −1 + x, x2 = −1 + 2x, x3 = −1 + 3x, · · · , xn = −1 + nx.
M i
[xi−1, xi].
f

S (f, P ) = M 1x + M 2x + M 3x + · · · + M nx
= f (x0)x + f (x1)x + f (x2)x + · · ·
+ f (xn − 1)x
= f (−1)x + f (−1 + x)x + f (−1 + 2x)x + · · · + f (−1 + (n − 1)x)x
= x{5 + [
+ [
+
+ · · · + [
(−1 + (n − 1)x)2 − 2(−1 + (n − 1)x) + 2 }
= x{5 + [
+ [
+
+ · · · + [
1 − 2(n − 1)x + (n − 1)2(x)2 + 2 − 2(n − 1)x + 2 }
= x{5 + [
+
+ · · · + [
= x [5n
6

x0 = 1, x1 = 1 + x, x2 = 1 + 2x, x3 = 1 + 3x, · · · , xn = 1 + nx.
M i,

S (f, Q) = M 1x + M 2x + M 3x + · · · + M nx
= f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x
= [f (1 + x) + f (1 + 2x) + f (1 + 3x) + · · · + f (1 + nx)]x
= {[(1 + x)2 − 2(1 + x) + 2] + [(1 + 2x)2 − 2(1 + 2x) + 2] +
+[(1 + 3x)2 − 2(1 + 3x) + 2] + · · · + [(1 + nx)2 − 2(1 + nx) + 2]}x
= {[1 + (x)2] + [1 + (2x)2] + [1 + (3x)2] + · · · + [1 + (nx)2]}x
= nx + (12 + 22 + 32 + · · · + n2)(x)3
= n
3 +
4
n +
4
3n2 +
4
3 +
1
2n +
1

P,
S (f, P ) = m1x + m2x + m3x + · · · + mnx
= f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x
= f (−1 + x)x + f (−1 + 2x)x + f (−1 + 3x)x + · · · + f (−1 + nx)x
= x { [
+ [
+
+ · · · + [(−1 + nx)2 − 2(−1 + nx) + 2 }= x{ [1 − 2x + (x)2 + 2 − 2x + 2+ [1 − 4x + 22(x)2 + 2 − 4x + 2+ + · · · +
[ 1 − 2nx + n2(x)2 + 2 − 2nx + 2
} = x
+ · · · + [
= x [
1 + 22 + · · · + n2
6
[1, 2]
Q,
S (f, Q) = m1x + m2x + m3x + · · · + mnx
= f (x0)x + f (x1)x + f (x2)x + · · · + f (xn−1)x
= f (1)x + f (1 + x)x + f (1 + 2x)x + · · · + f (1 + (n − 1)x)x
= x{1 + [
+ [
+
+ · · · + [
(1 + (n − 1)x)2 − 2(1 + (n − 1)x) + 2 }
= x{1 + [1 + (x)2] + [1 + (2x)2] + · · · + [1 + ((n − 1)x)2]} = nx + [12 + 22 + · · · + (n − 1)2](x)3
= n · 1 n
3 − 4
(x2 − 2x + 2)dx
x ≤ 0,

R1 : AR1
[−3, 0]

g(xi) = x2i .
= [(−3 + x)2 + (−3 + 2x)2 + · · · + (−3 + nx)2]x
= [(
+ · · · + (
x
= 27 − 54
2 2 + 3
[0, 4]

= [f (x0) + f (x1) + f (x2) + · · · f (xn−i)]x =
{−1 + [−(x)2 − 1] + [−(2x)2 − 1] + · · · + [−((n − 1)x)2 − 1] }
x
= −n · 4
3n2 = −4 − 64

f (t)
F (x)
x
a f (t) dt
− x) = f (c)x
f (c)
x → 0
c → x
x = f (x) .
G
f,
b
a
g (α) = a
g (β ) = b
f : [a, b]
g (α) = a
g (β ) = b

(F g)′ (t) = F ′ (g (t)) g′ (t) = f (g (t)) g′ (t)
t
f (g (t)) g′ (t)
β
α
f (g (t)) g′ (t) dt = F (g (β )) − F (g (α)) =
b
a
β 2 + 1 = 5 ⇒ β 2 = 4 ⇒ β = 2.

5 1
√ x − 1
x dx =
2 0
b
b

u = sin2 x ⇒ du = 2 sin x cos xdx dv = sin xdx ⇒ v =
∫ sin xdx = − cos x
π 3
= − sin2 x cos x
3 cos3 x)

a f (x) dx,
b→+∞ arctan b =
arctan a = − −π
c
a
b
c
(arctan b − arctan 0)
α
a
b
β
[−1, 1] ,
+ −1 − 1
⇒ 6 − x2 = 3 − 2x ⇒ x2 − 2x − 3 = 0 ⇒ x = −1
x = 3.
=
y = 2x + 8.

(2x + 6 + x2)dx = 38 3
,
3 +
38


x + 2y = 5,
,
x = 2 − y
x = y2 + 2
=
y = 2
y = −2x + 12 ⇒ (2, 8)

√ x
sin t(−2sin t)]dt
=
0 π 2
[α, β ]
θ1, θ2, θ3,..., θn
ri
θi

|θ| X
θ − 1 4
sin 2θ π
3 ⇒ θ = π
π 6

2
3y ⇒ x2 + (y − 5 √

5√ 3sin θ = 5 cos θ ⇒ √ 3tan θ = 1 ⇒ tan θ =
√ 3


xi − xi−1
f (xi) − f (xi−1) xi − xi−1
= f ′ (ξ i) .
|x|
AB.
n → ∞,
n∑i=1 1 + (f ′ (ξ i))
2
l =
b
a
l = 1
0
φ′ (t) .
= β
=
=

t ∈ [0, 2π].
r2(sin2 t + cos2 t)dt = 2π 0
rdt = rt|2π0 = 2πr.

ψ′ (t) = 9 sin2 t cos t,

= 36

3
2
t
r = f (θ)

φ′ (θ) = f ′ (θ)cos θ − f (θ)sin θ = r ′ cos θ − r sin θ

+ (r′senθ + r cos θ) 2

r′ = −a sin θ.
= 2a
π
0
= 2a π
8a u.c.
r = 2e2θ
θ ≥ 0
.

dV= r dx
dV= f(x) dx
n − cilindros
|θ|
n i=1
b
a
V 1
y = 2
1 9


5 x5 2
1 − y2

) 2
dy
V = π
π 2
−π 2
= 8π π2 −π
2
π 2
−π 2

(1 + 2x2 + x4)dx = 94

= π
= π
2 √
x
4
a
f (x) = x + 2
g (x) = x2+ x
f : [−2, 5] → R
f (x) = x2 + 2
f (x)dx.
(b) f
x ∈ R+.
f
g.
xe−xdx (n) 1 −1
1
0 π
(b) r = 1 + cos θ r = 1;
(c) r = sin θ r = 1 − cos θ;
(d) r2 = cos(2θ) r2 = sin(2θ);

r = 6.
r = 2.
y(t) =
t = 0
π 2
π 6 √

R
4 (e) 2e3 − 2 (f ) 0
(g) − 1 (h) 0, 027 (i) 4, 59
(a) 1 s − a
para s > 0
(a) 2 √

(d) 23 6
17
2
(π − 2) (d) 1 − √ 2 2
(e) 6π − 8 √
π 2
π 6
π3
l =
2
(y + 8 + 4√ y + 4)dy − π −3 −4
(y + 8 − 4√ y + 4)dy − π 0 −3
(y2 + 8y + 16)dy
(a) 134 189
r2 = x2 + y2
θ = π 2
a, b ∈ R

2n
n
r²=-a²cos(2 )θr²=a²sin(2 )θ
r²=-a²sin(2 )θ

r=eaθ θr= θr=-

T (x , y, z, w , t) = x + 0, 1y + 0, 5z + 2, 5w + 0, 6t.
T

T : R5 → R.
n−upla
f
(a) f : D ⊂ R2 → R
f (x, y) = 2x + 3y + 1. (b) f : D ⊂ R3 → R
f (x,y,z ) = x2 + y + z + 6.
(c) f : D ⊂ R4 → R

2

(a)
(b)
,

• x
y,
z = ±
+ z 2 32
z = 3.
D(f ) = {(x, y) ∈ R2/ 4x2 + y2 ≤ 4} = {(x, y) ∈ R2/ x2 +
y2
Gr(f ) = {(x,y,z ) ∈ R3/ (x, y) ∈ D(f ) z = f (x, y)}.
z =
f,

A (y1, y2, · · · , yn)
B (A, δ ) = {(x1, x2, · · · , xn) ∈ Rn; ||P − A|| < δ } .
δ = 1
A(1, 2)
}

lim (x,y)→(x0,y0)

0 <


2 |(x − 1)| + 3 |(y − 3)| < 2δ + 3δ = 5δ.
2 |(x − 1)| + 3 |(y − 3)| < ε 2 |(x − 1)| + 3 |(y − 3)| < 5δ
5δ = ε
0 < ||(x, y) − (1, 3)|| <
δ,

C 1
y→0 0 · y
C 2
(0, 0) .
x2 + (kx)2
(x,y)→ S2

3x2kx

(x,kx2)→(0,0) f (
(x,y)→ C 2
0 < ||(x, y) − (0, 0)|| < δ.

0 <
≤ 3 (x2 + y2) |y| x2 + y2
= 3 |y| < 3 √
x2 + y2 < 3δ.
lim (x,y)→
C 1
(0,1) 3x
4
(kx)
4
lim (x,y)→
C 2
3x4(kx2)4
(x − 3)(y − 1)(z + 5) .
C 1 = {
.

(x − 3)(y − 1)(z + 5) = lim
t→0 (3 + at + 2 + 2bt − 5 + ct)3
(at)(bt)(ct)
lim (x,y)→(x0,y0)

f (x, y) ± lim (x,y)→(x0,y0)
g (x, y) .
f (x, y)
(x,y)→(xo,yo) f (x, y) · lim
(x,y)→(x0,y0) g (x, y) .
lim (x,y)→(x0,y0)
lim (x,y)→(1,2)
x2 + 2xy + y2
x + y + 1
x + y + 1 = lim (x,y)→(2,−2)
x2 − y2
f (x, y) + ln
f (x, y) + ln
f (x, y) + ln
lim
2 .
4
D = {(x, y) ∈ R2/ x = 0
y > x2
x2 − y2
(x − y) √
g(x, y)

r 2
g(x, y)
lim (x,y)→(0,0)

0 se (x, y) = (0, 0)
(0, 0) .
0
[x2 − (y − 1)2][x2 + (y − 1)2]
x2 + (y − 1)2 = lim
f (x, y)
0 se (x, y) = (0, 0)
(0, 0) .
f (x, y)
b,
lim (x,y)→(0,0)
(x,kx)→(0,0) f (x,kx) = lim
lim (x,y)−→
C 2
(x,kx2)→(0,0) f (x,kx2) = lim

δ =
√ ,
lim

x2y2
2
· y2
lim(x,y)−→ C 1
(0,5) g(x, y) = lim(x,kx+5)→(0,5) g(x,kx + 5) = limx→0 x3(kx)2
2x7 + 3(kx)4 = lim
lim (x,y)−→
C 2
x→0
x→x0
g(x) − g(x0)
x − x0
x − x0
− f (x0, y0)
y
→ y0
y − y0
y
x
y .

x = lim x→0
x
x2y + 2xyx + y (x)2 + xy2 + y2x − x2y − xy2
x
x
x = lim
x →0

y = x2 + 2xy.

∂x = 6xyz 3t2 + 4xyz 3t2 cos x2yz 3t2.
x,z,t
y :
∂y = 3x2z 3t2 + 2x2z 3t2 cos x2yz 3t2.
x,y,t
z :
∂z = 9x2yz 2t2 + 6x2yz 2t2 cos x2yz 3t2.
x,y,z
∂t = 6x2yz 3t + 4x2yz 3t cos x2yz 3t.

t1 m = tgα.
C 1
C 1 : y = y0z = f (x, y0) t1 : y = y0
z − f (x0, y0) = ∂f (x0, y0)
∂x (x − x0)
∂y (y − y0)


π.

∂x (x − x0) = 2x0(x − x0),

y = 2.

∂y (y − y0) = 2y0(y − y0),

x = 1.

P (1, 2, 5),
P (1, 2, 5), −2x − 4y + z + 5 = 0.

∂x (x0, y0), −∂f
∂y (x0, y0), 1

P (x0, y0, z 0)
π
P 2(2, −1, 1)
π
−2x0x0 − 2y0y0 + x20 + y20 + d = 0 −2x0 − 2y0 + 1 + d = 0 −4x0 + 2y0 + 1 + d = 0
⇒ d = x20 + y20
⇒ d = 5y20
⇒ y0 = 1

P 2 ( 2 5
2

F
x
y

F (u(x, y), v (x, y)) = 2 [u(x, y)]2 + 3 [v (x, y)]
= 2(x2y + y)2 + 3(x + y2)
= 2x4y2 + 4x2y2 + 5y2 + 3x

∂z (x, y)

z
x
y.
= 8x3y2 + 8xy2 + 3
= 4x4y + 8x2y + 10y.
F
1
5 ,

∂ 2w
∂x2 +
∂ 2w
∂y2 =
∂ 2w
∂u2 +
∂ 2w
∂v2 .
L(x, y)
∂x (x − x0) + ∂f (x0, y0)
∂y (y − y0) + f (x0, y0)
z = f (x, y)
f (x, y)
f ≈ L(x0 + x, y0 + y) − L(x0, y0)
f ≈ ∂f (x0, y0)
V 1
V 2
100 = −0.04; dh = 2
−4
=
3 dh
= 2π · 4 · 2 · (−0, 04) + π · 16 · (0, 06) + 2π · 4 · 5
3 (−0, 04) +
2πrh
πr2
= 0, 81(10π + 8π)0.2 + 0, 81.(4π)0, 1 10, 17.

= 0, 81 (
2π(2, 2) · (5, 1) + 2π(2, 2)2 − 0, 81(20π + 8π) 10, 47.
R$10, 47,
14, 7%.


w = f (x,y,z ) =
dw =
∂w
dz = −0, 1.
w ·(0, 01)+
3

• ∂

y 4
z 5
y
∂ 3f
x
∂x∂y∂z∂t (x , y, z, t) = 120x2y3z 4t.
u

∂ 2u
∂u
∂ 2u

− (cos x) e y
+ (cos x) e y
− (sin y) e x
(x, y)
f (x, y) = 3x4 + 2y4;

⇒ x − x3 = 0 ⇒ x(1 − x2) = 0 ⇒ x = 0; x = ±1
P (0, 0), Q(1, 1)
R(−1, −1).
Θ(x, y) = ∂ 2f

Q(1, 1)
R(−1, −1)

V (x,y,z ) = xyz
y +
2V
x .
S.
x2

(a, b) = (6, 6)

∂ 2S

∂ 2S
40, 00
60, 00
L

⇒
Q
f,
2
− 2y
2
y = −x
x2 − 1 = 0 ⇒ x = 0

P 3(−1, 1) .
P 3(−1, 1).
− 16xy
− 16
∂x2 = 2 + 2y2.
f (x, y).
(−1, 1) = 16 > 0 Θ(−1, 1) = 4 > 0 ⇒ P 3(−1, 1)
f (x, y).

x2 + y2 − 9 = 0
F (x, y) = x2+ y2− 9


(2xy3 + 3x2y2y′) + (3x2y2 + 2x3yy ′) + (y + xy′) + 1 + y′ = 0 (3x2y2y′ + 2x3yy ′ + xy′ + y′) + (2xy3 + 3x2y2 + y + 1) = 0
(3x2y2 + 2x3y + x + 1) y′ = − (2xy3 + 3x2y2 + y + 1) .

3x2y2 + 2x3y + x + 1 . (
F,

f (x, y) =

(a) x2 + y2 + z 2 = 25; (b) x2 + y2 − z 2 = 25;

(2x − 4y) = −10.
x2y2
x3 + y3
x2 + y2
x2 + y3
x − y
x + y
(x − 3)5y2 + (x − 3)4y4
(x2 − 6x + 9 + y6)3
lim (x,y)→(0,0)
sin(x2 + y2)
x2 + y2 ;
lim (x,y)→(0,0)
z = f (x, y),
P (1, 1, 1),
g(x, y) = −x2 − y2.
k
∂ 2w
∂x2 +
∂ 2w
∂y2 +
∂ 2w

f
g
∂w
∂r
2 +
1
r2
∂w
∂θ
2 =
∂w
∂x
2 +
∂w
∂y
2 .

∂f
243π cm3.

K, M

P (1, 2, 2).


(a) esfera de raio 5 (b) hiperboloide de uma folha (c) plano (d) cone circular
2, 1 2

(a) nao existe (b) L = 0, com δ = √ ε (c) L = 0, com δ = ε 2
(d) nao existe (e) nao existe (f ) nao existe

(d) √ 2 2
lim (x,y)→(4,4)
(b) descontnua

∂z
∂y =
= 2x2 cos(2y)
(d) ∂z
∂z
∂y =
∂ 3f
∂ 3g
−24x + 24y − z = 36
2x + 4y − z − 3 = 0 (−1, −2, −5)
{(x, y) ∈ R2/ − 2 ≤ x
x ≥ 2, y2 ≥ 25x2 4 − 25} ∪ {(x, y) ∈ R2/ − 2 ≤ x ≤
2, y };
w = z − x,
∂ 2f
1, 28
dC = 616, 38
dV = 100, 4

y2
x + y = 1},

x = 7 3
x = 2 3
x = 1000, y = 2000
z = x − 1
= 74.
0
= −2
R
xy
n
i−
Ai
n i=1

⇒ x = −3, y = 9 x = 2, y = 4

x =
y = x2 y = 1 y = x2
y = 6 − x y = 6 − x y = 6 − x

+ 28 3
= 39 2

x = y2
x


θi.
Ri
Ri
f (r, θ)
|P | Ri

α r2

r2
drdθ

y = x2 ⇒ r sin θ = r2 cos2 θ ⇒ r = sin θ
cos2 θ = tan θ sec θ
x = 3 ⇒ r cos θ = 3 ⇒ r = 3
cos θ = 3 sec θ.

xy,
x = r cos θ, y = r sin θ
dxdy = rdrdθ,
I,
π
0
y2
0
3 + 12
2 − 2
y = 16
3 − 4x

e12 − 13
8 (eln

− 2π ln 6 (e)
I =
2 − 1 2
I =
(x∗
i
S,
i−
mi = f (x∗i , y∗i , z ∗i ) xiyiz i
S
m ≈ n
i=1
f (x∗i , y∗i , z ∗i ) xiyiz i.
|N | S,
n

y = y2(x)
(x, y) ,
z = 0, y = 0, x = 0

2x− 22 − 1
⇒ V =
⇒ V =
⇒ V =
z = 3,

∪

⇒ V = 3 0
∪ x ∈ [0, √
⇒ V =
y = 5.
y = 5
0dy = 0.

x2 + y − 16 = 0, x + y − 4 = 0, y = 2x + 13, z = 0
z = 10,
z = 0 z = 0
z = 10 z = 10
3
0
dzdxdy
0 ≤ z ≤ 2 ⇒ I =
0
0 ≤ z ≤ √ 16−y2
0 √ 16−4z2 0
r1
r2
θ
θ ∈ [θ1, θ2] .

z = 0
π
0
=
π
0

r ⇒ r2 = 2x ⇒
xy
x2 + z 2 =
y = yx2 + z 2 = r2
tan θ = x z
⇒ x2 + z 2 = x
r = √ 3cos θ
r = sin θ
6cos2 θdθ

x = ρ cos θ sin φ, y = ρ sin θ sin φ, z = ρ cos φ, ρ2 = x2 + y2 + z 2, tan φ =
√ x2 + y2
ρ1 0 ≤ θ0 < θ1 ≤ 2π, 0 ≤ φ0 < φ1 ≤ π
0 ≤ ρ0 < ρ1.
f (x,y,z )
θ1 φ2
φ1 ρ2
dV (x,y,z ) = dxdydz
T (ρ + dρ, θ, φ) .
x
z
y
x2 + y2 + x2 + y2 = 4 x2 + y2 + 3x2 + 3y2 = 4

z 2 = x2 + y2
z 2 = 3x2 + 3y2.
z =
3ρ sin φ ⇒ tan φ = √ 3 3
⇒ φ = π 6
z = √
x2 + y2 ⇒ ρ cos φ = ρ sin φ ⇒ tan φ = 1 ⇒ φ = π 4
φ ∈ [ π 6

.
z = 4
z = 0.
0
dzdydx
x + 2y = 6,
f (x,y,z ) = 12z

x2+ y2 = 2z.

z = x2 + y2.

d(x,y,z ) =
dzdydx.
z = 0
x2 + y2

x2 +
y2 + z 2 = 4z z = 1 + 1 2 √ x2 + y2, f (x,y,z ) =
(x2 + y2)z 2
z = 2−
√ x2 + y2
x + y + z .

V = 32a
I = √ 12 0
dzdydx−
V =
0 π 2
2x2+2y2 dzdydx
0
M =
cos(ρ2) dρdφdθ
1−√ 1−r2 e
I =
0
eρ2
+
eρ2

R.

1
R
− 1 = n − n − 13 n + 13
= 13 n + 13
|a − b| = |a − un + un − b| = |−(un − a) − (un − b)| ≤ |un − a| + |un − b| < ε
2 + ε 2
unk
k1 ≤ un ≤ k2 n ∈ N∗.
un : N∗ → R
{u1, u2, u3, · · · , uK }
M = max u1, u2, · · · , uK , {u1, u2, u3, · · · , un−1, un, · · · } ⊂ B(a, 1)∪B(0, M ).
un
•

= n+2 n2+2n+3
.
n2 + 2
⇔ n3 + 2n2 + 2n + 4 ≤ n3 + 3n2 + 5n + 3
⇔ 1 ≤ n 2
un,
n−

S 3 = u1 + u2 + u3 = S 2 + u3
· · · S k = S k−1 + uk

3 ,

B =
− 20000.


ε < k.
N 0 =
20000 − ε
S k
2 5
5k

+ uα.
k→∞ S k = lim
k→∞ S α + lim
k→∞ S k−α,

· · · + uk +
S
S ′,

2 |S k−1 − S | < ε
2 .
S k = S k−1 + uk,
uk = S k − S k−1
|uk − 0| = |S k − S k−1 − 0| = |S k − S + S − S k−1|
= |(S k − S ) + (S − S k−1)| = |S k − S | + |S − S k−1| ≤ |S k − S | + |S k−1 − S | <
ε
2 +
ε
= 2 3 = 0.
· · ·
S 21 = S 2 = 1 + 1
2 >
1
2 +
1
2 =
2
2
3 +
1
5 +
1
6 +
1
7 +
1
9 +
1
10 +
1
11 +
1
12 +
1
13 +
1
14 +
1
15 +
1
16

a1q n−1 = a1 + a1q + aq 2 + · · · + a1q n−1 + · · ·
n−
+ a1q n−1.

qS n − S n = (a1q + a1q 2 + a1q 3 + · · · + a1q n) − (a1 + a1q + aq 2 + · · · + a1q n−1) ,
(q − 1)S n = a1q n − a1 = a1(q n − 1),
S n = a1(q n − 1) (q − 1)
.

.

x > 1,
x = 1
= 2
0 ≤ yn ≤ un

n > K.
yn ≥ un
y1 + y2 + y3 + · · · + yk + · · ·
n(n + 1)
n ∈ N∗.
n3 + n2 + n + 1 ≤ 1
n(n + 1) ⇔ n2(n + 1) ≤ n3 + n2 + n + 1 ⇔ n3 + n2 ≤ n3 + n2 + n + 1 ⇔ 0 ≤ n + 1
n. ∞∑ n=1
un+1
un+1 < unq un+2 < un+1q < unqq < unq 2
un+3 < un+2q < unq 2q < unq 3
· · · un+k < un+(k


u2 + u3 −

n p + · · ·

un = 0.

S 2n+1 = S 2n + u2n+1

S 2n + lim n→∞

ε > 0
n > K

n + 3
(n + 1) (n + 2) ⇔ (n + 2) (n + 1) (n + 2) > n (n + 1) (n + 3) ⇔ n3 + 5n2 + 8n + 4 > n3 + 4n2 + 3n ⇔ n2 + 5n + 4 > 0,
n

∞
n = 1 − 1
n
n =
(n + 1)3 + 4 ≤ n2
f (x) =
f i : R → R
f 0 (x) = 1, f 1 (x) = x, f 2 (x) = x2,
f 3 (x) = x3, f 4 (x) = x4, · · · , f n (x) = xn, · · · ,

· · · + xn +
3 + · · · + cnxn + · · · .
L ,
= lim
n→∞
R = 5 3
• x = −5
n
lim n→∞
n2 + 3
( (n + 1)2 + 3
n→∞ n2 + 3
1 2n−1
5 − x 1 3+ x1
7 − x 1 5+ · · · + x 12n+1 − x 12n−1
S n (x) = −x + x
1
−x + x
= cos x + 22 cos(24x) + 32 cos(34x) + 42 cos(44x) + · · · + n2 cos(n4x) + · · ·
x = 0,
S ′ (0) = cos 0 + 22 cos 0 + 32 cos 0 + 42 cos 0 + · · · + n2 cos 0 + · · · = 12 + 22 + 32 + 42 + · · · + n2 + · · ·
x = 0,
x = 0.
n=0
n=1
∞∑ n=2
C = K + ac0,
(x − a)2
2 + c2
(x − a)3
3 + · · · = C +
∞ n=0
xn.
∞ n=1
nxn−1.
ln(1 − x) = − ∞
n=1

f (x) = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · · + cn (x − a)n + · · ·
c0, c1, c2, · · ·
x = a

c1,
f ′ (x) = c1 + 2c2 (x − a) + 3c3 (x − a)2 + · · · + ncn (x − a)n−1 + · · ·
f ′ (a) = c1 + 2c2 (a − a) + 3c3 (a − a)2 + · · · + ncn (a − a)n−1 + · · ·
f ′ (a) = c1.
c2,
f ′′ (x) = 2c2 + 3 · 2c3 (x − a) + 4 · 3c4 (x − a)2 + · · · + n(n − 1)cn (x − a)n−2 + · · ·
f ′′ (a) = 2c2 + 3
c3.
f (3) (x) = 3·2c3+4·3·2c4 (x − a)+5·4·3c5 (x − a)2+· · ·+n(n−1)(n−2)cn (x − a)n−3+· · ·
f (3) (a) = 3·2c3+4·3·2c4 (a − a)+5·4·3c5 (a − a)2+· · ·+n(n−1)(n−2)cn (a − a)n−3+· · ·
f (3) (a) = 3 · 2c3
c3 = f (3) (a)
f ′′ (a)

f (x) = f (a) +f ′ (a) (x − a) + f ′′ (a)
2! (x − a)2+
f (3) (a)
3! (x − a)3+ · · ·+
f (n) (a)
n! (x − a)n + · · ·
a) −
3! (x − a)3 + · · ·
2n! (x
(2n + 1)! (x
f ′′ (0)
2! x2 +
sin x =
sin0 − sin0
f ′.
x
x2,
f ′(x) = (1 + x2)−1 = 1 − x2 + x4 − x6 + · · · + (−1)nx2n + · · ·
arctan x =
3 +
x5
3 +
1
3!
+ x5
1

k! (n − k)! =
k! ,
(a + b)n = an+nan−1b+ n (n − 1)
2! an−2b2+· · ·+n (n − 1) (n − 2) · · · (n − (k − 1))
k! an−kbk+· · ·+bn.
2! x2 + · · · +
k! xk + · · · + xn,
x2 + n (n − 1) (n − 2) 3!
x3 + · · · +
k! xk + · · ·
k! xk
|x| < 1.
n = −1
−1 (−1 − 1)
3! x3 + · · ·
k! xk + · · ·
k! xk + · · ·
∞
∞ n=0
(−1)n xn,
3! x3 + · · ·
+ −1 2
233! x3 + · · · + (−1)k 1 · 3 · 5 · ... · (2k − 1)
2kk! xk + · · ·
233!
(−x2n + · · · 1√
1 − x2 = 1 +
2nn! x2n + · · ·
233! x6dx + · · ·
2nn!
x2ndx + · · ·
2nn! (2n + 1) x2n+1 + · · · + C

2 n
(c) un = (−1)n√ n n+1
(d) un = 100n n 3 2 +4
(e) un = n+1√ n
(f ) un = lnn n
(h) un = n2
( j) un = arctan n (k) un = (
1 − 2 n
n ( p) un = 7−n3n−1

, · · ·
}
(a) un = n 2n−1 (b) un = n − 2n (c) un = ne−n (d) un = 5
n
2n2
3n
un
n + k
√ 10.
u1 = u2 = 1.

3 n
2 n+1
n 2
enn3n (b)
∞∑ n=1
n cos(nπ)
n2 + 5 (h)
1
x
x3

f (x) =

(a) f (x) = sin2 x (b) f (x) = x2 sin2x (c) f (x) = e3x (d) f (x) = e−x2
(e) f (x) = cos 2x (f ) f (x) = sin(x5)
x3 (g) f (x) =

2
2
x→0 ln(1 + x2) − 3 sin(2x2)
x2
ln(1 + x4)
e−x4 − cos(x2)
1√ 1 + x
− x2
x dx (f ) f (x) = e−x2dx
(g) f (x) = ln(1 + x)
x dx (h) f (x) = ln
1 + x 1 − x
(i) f (x) = arcsin x
( j) f (x) = arccos x (k) f (x) = arctan x (l) f (x) = 3 √
1 + x
(a) 1 4
(b) 0 (c) 0 (d) 0 (e) (f ) 0 (g) (h)
(i) ( j) π 2
(k) e−2 (l) 0 (m) 0 (n) (o) 1 ( p) 0
(a) un = 2 n−1
3n (b) un = (−1)
n2
(a) decrescente (b) decrescente (c) decrescente (d) decrescente (e) decrescente (f ) crescente (g) decrescente (h) nao
decrescente
L
(k + 1)2
(e) S k = 1
k + 1
(k + 2)!
1 2
k + 1 − 1

(a) F (b) F (c) F (d) F (e) V (f ) V (g) F (h) F (i) F ( j) F (k) V (l) V (m) V (n) V (o) V ( p) F
S k = 2 − 2


(a) R = 1, I = [−1, 1) (b) R = 1, I = [−1, 1] (c) R = ∞, I = (−∞, ∞) (d) R = 1
4 , I = (−1
4 , 1 4
, 1 2
] (f ) R = 4, I = (−4, 4] (g) R = 3, I = (−5, 1) (h) R = 1, I = (3, 5) (i) R = 2, I = (−4, 0] ( j) R = 0, I = {1
2 } (k) R = 3, I = [−3, 3] (l) R = 1
4 , I = [1, 3
− 4, 0), R = 2 (o) I = (1
− e, 1 + e), R = e

2x2
(1 − x)3

∞
∞∑ n=1
∞∑ n=0
xn+1
1.3.5. · · · .(2n − 1)x
∞∑ n=0
(−1)nx2n+1

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Apostila CDI2 2013 01