Esercizi svolti, Analisi Matematica I

12 SERIE

SUP. E INF.

p.25

Aₙ = {n-1/n ; n ∈ ℕ}

a₁ = 0

a₂ = 1/2

a₃ = 2/3

a₄ = 3/4

Aₙ è crescente

lim_n→∞(n-1)/n = 1 e quindi

1 > n-1/n

0 ≤ 1 ∀ n ok

inf(A) = min(A) = 0

sup(A) = 1

A = {2n/n²+1 ; n ∈ ℤ}

a₁ = -1

a₂ = 4/5

a₃ = 6/10 = 3/5

lim_n→+∞2n/n²+1 = 1/+∞ = 0⁺

lim_n→-∞2n/n²+1 = -∞

quindi A è limitato

-1 < 2n/n²+1 < 1

- (n²+1) < 2n < (n²+1)

- (n²-2n+1) < 0 < (n²-2n+1)

- (n-1)² ≤ 0 < (n+1)² ∀ n

sup(A) = max(A) = 1

inf(A) = min(A) = -1

1.39

• A = {_n+1/n ; n ∈ ℕ}

  a₁ = 2, a₂ = 3/2, a₃ = 4/3

  decrease

  lim_{n→∞ n+1}/n = 1

  1 < _n+1/n < 2

  ∀ n ∈ ℕ

  0 < 1 < n, m ∈ ℕ

  ⇒ ∀ n ∈ A

  sup(A) = max(A) = 2

  inf(A) = 1

• B = {_3n+2/n ; n ∈ ℕ}

  a₁ = 5, a₂ = 4, a₃ = 11/3

  decrease

  lim_{n→∞ 3n+2}/n = 3

  3 < _3n+2/n < 5

  3n + 3n + 2 ≤ 5n

  0 < l < 2n

  ∀ n ∈ A

  sup(B) max(B) = 5

  inf(B) = 3

• C = {^{⟨(-1)ⁿ⟩}/_n ; n ∈ ℕ}

  a₁ = -1, a₂ = 1/2, a₃ = 1/3

  se n è pari:

  a₂ = 1/2, a₄ = 1/4 decrease

  lim_n→∞ 1/n = 0

  se n è dispari:

  a₂ = 1, a₃ = -1/3 decrease

  lim_n→∞ n = 0

  sup(C) = max(C) = 1/2

  inf(C) = min(C) = -1

2)

limx → -3|x + 1| = 2

|x + 1| - 2 |ε2 - ε | x + 1 | 2 + ε

(2 - ε) | x + 1 | (2 + ε)

2 - ε | x + 1 | (2 + ε) - 1

6)

limx → -∞(x⁵ + 5) = 4

| x + 5 - 4 | < ε1 < x³4 < ε

∀ ε, ε ∝ x ∝ 1 + ε

7)

limx → 1x² + 3x - 4 = 5

x² + 7x - 3x - 4 < εx - 1x - 1

x² + 7x - 3x - 4 - 5x + 5 < ε

xx

x² - 2k + r | εx - 1

(x - 1) | ꬖ ε

ε ∝ x ∝ 1 + ε

1 - ε ∝ x ∝ 1 + ε

Ok!

8)

limx → 32x² - 5x - 3 = 7

2x² - 5x - 3x - 32

2x² - 5x - 3 ∝ 7x - 3

2(x² - gak + 9)x - 33 - Σε

2 < ε

No!

Non trovo intorni!

9)

limx → 3x - k - 6 = 5x + 3

x² + x - 6 + 5 = εx + 3

x² + x - 6 + 5x + 5 = εx + 3

x² - 3x + x - 5< ε8 < ε

No!

10)

limx → ∞x³ + 1 = 3x + 1

x³ + x - 3x + 13 - ε

f(x) x - 3x + 1f x¹

f(f(x))

x + 1[ (x + x + 1) - 3] < εx(x3)

x² - x + x + 3 | ε

| x² - x - 2 | < ε

Σε ∝ (x + 2) | (x - 1) < ε

No!

7.18

(a) \(\lim_{{n \to +\infty}} \sqrt{n+1} - n = \lim_{{n \to +\infty}} \frac{n+1 - n^2}{\sqrt{n + 1} + n} = \lim_{{n \to +\infty}} \frac{1 - n}{\sqrt{1 + \frac{1}{n}} + 1} = 0\)

(b) \(\lim_{{n \to +\infty}} \frac{1}{n+1} - \lim_{{n \to +\infty}} \frac{b^2}{n(n+1)} = \lim_{{n \to +\infty}} \frac{b^2}{n(n+1)}\)

750

(a) lim_n→∞ (n² + n) / (n² + n + 2)ⁿ

= lim_n→∞ [(n² + n - 2 + 2) / (n² + n + 2)]ⁿ

= lim_n→∞ [(n² + 2n - n + 2 - 2) / (n² + n + 2)]ⁿ

= lim_n→∞ [(1 + 1 / (n² + n + 2))]^{n2 + n - 2}

= lim_n→∞ [1 + 1 / (n² + n + 2)]^{n2 + n + 2}

= e

(b) lim_n→∞ [(n² + n) / (n² + n + 1)]ⁿ

= lim_n→∞ [(n² + n - n) / (n² + n + 1)]ⁿ

= lim_n→∞ [(1 + 1 / (n² + n))]^-n2

= e^-1

751

(a) lim_n→∞ [(n² + n) / (n² - n + 1)]ⁿ²

= lim_n→∞ [(n² + 2n - n + 1 - 1) / (n² + n + 2)]ⁿ²

= lim_n→∞ [(n² + b + 1) / (n²)]^-n2

= e^-∞

7.66

lim _n->+∞ 3n + ln n / n

2/n + 3n/n + ln n / n

↓

0 1 0

0 0 0

7.67 (a) lim _n->+∞ 2 + cos n / ln^2(n / n)

0 < ln^2(1/n) ≤ 1

1 <= cos n ≤ 1

quindi.

2/ln^2(1/n)

+∞

7/ln^2(1/n)

+∞

2 + cos n / ln^2(1/n)

quindi

+∞ +∞ +∞

(b) lim _n->+∞ 2n + ln n / log n

trovo 2n / log n = 2n + log n / n = 2n + log n

lim _n->+∞ 2n / n + log n = lim _n->+∞ 2n log / n + log = 2

lim _n->+∞ log / +∞ + 2

quindi.

lim _n->+∞ 2n + ln n / n

trovo 2

7.68

lim _n->+∞ cos n log (√n^2 + 1) - log √n^2-1

trovo lim _n->+∞ cos n log (√n^2-1/√n^2)

lim _n->+∞ log n(1-√n^2-1/√n^2)=0

quindi = lim _n->+∞ cos n log (√) = 0

Limiti di Funzioni

8.15. _x→∞ lim log _α/x = 0 ( con α > 1 )

_x→∞ lim log_α/x = _x→∞ lim log_α(1^x/x) = 0

8.16. _n→∞ lim (1 + ^b/_x)^x = e^b

8.17. uguale

_x→∞ lim (1 + ¹/_x/b)^x/b b = e^b

8.18. ( a ) _x→∞ lim ^x log (1 + ¹/_x)

8.19. _x→∞ lim log (1+x)

_x→0 lim ^{log (1+x)}/_x = b_n = 1/_x

_bn→∞ lim log (1 + ¹/_bn)^b_n = 1

8.20. _x→0 lim ^{αx - 1}/_x ; b_n = e^x →

_bn→∞ lim ^{b_n - 1}/x→0 log (b_n + 1) [log(b_n + 1)b_n]

_bn→∞ lim ¹/_loge = 1

8.35

a) lim_x→0⁺ x log x

e^{x log x} = e⁰ - 1

(b) lim_x→0⁺ log x + lim_x→0⁺ e^{x log x} = e^{(log x)2}

8.36

lim_x→∞ x log ( x + 1/x ) = lim_x→∞ log [(1 + 1/x)^x] - x - 1

= log (1/e ) = -1

8.37

lim_x→∞ x² [log (x+2) - 2 log x -1 ]

= lim_x→∞ x² log ( x²/x² ) +lim_x→∞ log[(1 + 2/x)^x2] =

= log e² = 2

8.38

(a) lim_x→0 2^3x - 1 / x

2^3x - 1 = y

y + 1 = 2^3x

3x = log₂ (y+1)

x = log₂ (y+1) / 3

lim_y→0 3y / log₂ (y+1)

= log₂ (y+1) - log (y+1) / log₂

= lim_x→∞ log₂ 3y / log₂ - lim_x→∞ log₂

= ^{log 2}/_{log e³} = -3 log 2