Temi d'esame svolti anni 2022-2023

13/11/23

A: (z−3)

A ∈ ℝ

B: z² + 4 < z

B: z² < z + 4 − z

C:

• z = x ± iy
• z = x + iy

(0−A)²+4∈z

(z+3).3.x

A ∩ B = r

equilibrio

1. A − (a, 0)
2. B − (z − 0)

z₀e²

equilibrio al centro

A ∩ B = equilibrio

a = equazione

1. a colpi
2. a sin(x)

a∗∗∗x)

x−0 < < -

1

• 3x
• −sin(x2x)

−−(−3) < x − 3³

−sin (x2x)=

y = 1

• = √−3
• a − sin (x3x)

= −

3/√

=

−1 − √x + − x = 0

a²

= x3

• −sin(x3x)=
• x < 0
• sin(x) = 0

3x =

=sin(x)1

2

=sinx

• x ≤ 0
• 1 − (x + y)
• cosin (1 − sinx) =

n

• = −x³ - x³ =
• x0
• =_xcosin(1−sinx)=

−1

n²cosin(1-sinx)

x

n⁵ 1+0

sin¹=

0

1. conoscenza bₙ = ∑ aₙ

nᵏ (1+n) = 0

devo verificare che bₙ non interpolisca e decrementi

bₙ = ∑ aₙlim bₙ

∫₁ₙ (1-e⁻ⁿ) dx = c(n-1) ∫₁_n (1-e⁻ⁿ) = 0 con 0 < t < 1

n→∞∫₁_n e⁻(x/y) dx = b = 0 c 0

e^x_n

6) f(x) = ^5x+1/_x

• D: x ≠ 0
• N: 5x + 1 = 0 → 5x = -1
• x = -1/5
• -∞;0
• 0; +∞
• ^{5x + 1}/_x = ⁺/_- - ⁺/₊
• 0,0, 5x = -1
• ψ = -1/5
• @F
• lim_{x→ 0^-} f(x) = +∞
• lim_{x→ 0⁺} f(x) = -∞
• lim_{x→ -1/5} f(x) = +-∞
• x - 0
• ψ = 0

• x ≠ 0

(@)^F

• x² + ¹/_{5x ²} = 2/3
• -^2x-Cx+5x+1/_x² x ≠ 0
• 3x - 5x+1 ≠ 2/3
• x - 0; x ≠ 0
• x = 0

• x ⋮ 5x = 0
• x = 0; x ≠ 0

(@)¹

f'(x) = (x+1)

K1-1

K+1 > 3

K2 < 1

= (K+1)

f(x) = ³_{(x+2)(x+1)(x-1)} x > 1

3(x+2)(x+1)

1 (x+1)(x-1)

K1 < 0

- (x-1)(x+1)

x < 1- (x-2)

3

K+2 = 0

x > 0

K+2 > 1

- f x 0

- f

3 ³_(x+2)(x+1)

12/12/13

1. Z_n=1((1/2³)^1/n+(1/4)^1/n)

(A:1) (B:1)

P3=3

Z(1/2)^1/n - (1/4)^1/n)

Z(1/2)^1/n (1/2)⁵ - (1/3)^1/n(1/1)⁵

(1/4)^1/3 - (1/2)^1/n(1/8)^1/n

(1/D)^1/3(1/2)⁵(1/32)_1/n⁵

Z=1/2 (1/2-1/4)(1/1) (1/2 -1/2)(1/2)

1/32⁵

2. lim Lim X 2 x(3x-1)(X -0) - 3lim

= lim X/1/X-1) X/x 3x² X - X/( -X/X )

= lim lim lim 3/4

X/ X + 3

- cos Lim lim

X (X+2.3-1)

lim lim/lim lim

X 0(1 - x 3) lim x(1/3)³1/4

Y= Y₀ x/sinx + cirl 3

log x/ (3 x (1/x))²

X₃

X(1) = X

lim

cos x (3(x 1 1)/lim (1+1-1)(2x)

(p2:3)(1)(3 3)

= - 1/

-1 cos

- 2 deg img cos (1+ sin^X)

= 3

ho pensato pare

X=8+8

X

1/

Z (

lim,

20/11/22

(w: figura 9

-8 ln1/16 =11 ln

e

.3e = 4x/3

x₂ e = -4x/3

e + 3/2 = 1.1.2.1/3

20 ln -20 ln - 0 ln ₂

2)

s (x,n) /

ln 16 = ln 1

e

(x+1) …

x+3/2

...x x