Esercitazione guidata Analisi Matematica 1 - seconda parte

Dimostrazioni, equazioni, disequazioni e limiti

2ⁿ ≥ n² ∀n∈ℕ: n≥4

n=4

2⁴ ≥ 4²

Supponiamo vero l'asserto per n e dimostriamolo per n+1:

2ⁿ⁺¹ ≥ (n+1)²

2ⁿ⁺¹ = 2ⁿ·2 ≥ n²·2

(n+1)² = n² + 2n + 1 ≤ 2ⁿ + 2n + 1

n²·2 = n² + n² ≥ n² + 2n + 1 = (n+1)²

n² ≥ 2n + 1

4² ≥ 8 + 1

(n+1)² ≥ 2(n+1) + 1

n² + 2n + 1 ≥ 2n + 2 + 1

n² + 1 ≥ 3 => n² ≥ 2

(2n)! / (n!)² ≤ 2²ⁿ ∀n≥1

n=1

Supponiamo vero per n e dimostriamolo per n+1, cioè

2ⁿ ≥ n² ∀n∈ℕ : n ≥ 4

n = 4

2⁴ ≥ 4²

Supponiamo vero l'asserto per n e dimostriamolo per n+1:

2ⁿ⁺¹ ≥ (n+1)²

2ⁿ⁺¹ = 2ⁿ · 2 ≥ n² · 2

(n+1)² = n² + 2n + 1

n² · 2 = n² + n² ≥ n² + 2n + 1 = (n+1)²

n² ≥ 2n + 1

4² ≥ 8 + 1

(n+1)² ≥ 2(n+1) + 1

n² + 2n + 1 ≥ 2n + 3

n² + 1 ≥ 3 => n² ≥ 2

(2n)! / (n!)² ≤ 2²ⁿ ∀n ≥ 1

→ (2n)! ≤ 2²ⁿ · (n!)²

n = 1

Supponiamo vero per n e dimostriamolo per n+1, cioè

^(2(n+1))! / _((n+1)!)² ≤ 2²⁽ⁿ⁺¹⁾ = 2²ⁿ 2²

^(2n+2)! / _((n+1)!)² = ^{(2n)! 2n+1 (2n+2)} / _{(n!)² (n+1)²}

≤ ^{22n (n!)3} / _(n!)² · ^(2n+1)(2n+2) / _(n+1)²

= 2²ⁿ · ^{(2n+1) · 2 · (2n+2)} / _(n+1)²

= 2²ⁿ · 2^22(2n+1) / _n+1

≤ 2 · 2(n+1)2n+1 / n+1 ≤ 2 ⇔ 2n+1 ≤ 2(n+1) ⇔ 2n+1 ≤ 2n+2

≤ 2²ⁿ 2^{2(2n)! 2n+1 (2n+2)} / _{(n!)² (n+1)²}

≤ ²²ⁿ 2^{2(2n+1)(2n+2)} / _(n+1)² ≤ 2²

Equazioni e disequazioni

ax+b=0      x=-^b/_a, a≠0

ax+b>0      x > -^b/_a, a≠0

ax²+bx+c=0

a≠0      c=0

bx≠0

ax²+bx=0

x(ax+b)=0

x=0 ∨ ax+b=0

c≠0

ax²+bx+c=0

Δ=b²-4ac

Δ>0   x_1,2=^-b±√Δ/_2a

2 soluzioni reali distinte

Δ=0   x_1,2=-^b/_2a

2 soluzioni reali coincidenti

Δ<0

x²+3x-4=0

x_1,2=-^3±5/₂

Δ=9+16=25>0

2x²+bx+c≥0

x₁<x₂

a≥0

Δ>0

x≤x_n ∪ x≥x₂

1) (2x²-3x+1)/(x²-1) ≤ 0

x²-1 ≠ 0 → x≠±1

2x²-3x+1>0

x²-1>0

2x²-3x+1>0 → x≤½ ∪ x≥1

2x²-3x+1=0

Δ=9-8=1>0

x^1,2= (3±√1)÷4 < ½

x²-1>0 ⇔ x<-1 ∪ x>1

-1<x≤½

1)n√p(x) > q(x)

n pari

q(x) ≥ 0

p(x) > [q(x)]ⁿ

∪q(x)

p(x) ≥ 0

n dispari

p(x) > [q(x)]ⁿ

2)n√p(x)

n pari

p(x) ≥ 0

q(x) > 0

p(x)

n dispari

p(x)

Limiti

1)\(\lim_{{n \to \infty}} \frac{\sqrt[3]{n}\sqrt{n} + 1}{\sqrt[3]{n}-\sqrt{n}}\)

= \(\lim_{{n \to \infty}} \frac{\sqrt[3]{n} \left[ 1 - \frac{\sqrt{n}}{\sqrt[3]{n}} + \frac{1}{n^{1/3}} \right]}{\sqrt[3]{n} \left[ 1 - \frac{\sqrt{n}}{\sqrt[3]{n}} \right]}\)

= \(\lim_{{n \to \infty}} \sqrt[3]{n} \left[ 1 - \frac{1}{n^{5/2}} + \frac{1}{n^3} \right]\)

\(\sqrt{n} \approx \) ... \( n^{1/2}\)

\(\sqrt[3]{n} = n^{1/3}\) ...\(\)

= \(\lim_{{n \to \infty}} \sqrt[3]{n} \left[ 1 - \frac{1}{n^{1/12}} \right]\) = \(\to \infty\)

2)\(\lim_{{n \to \infty}} \frac{\cos\left(\frac{2}{n}\right)-1}{\log\left(1 - \frac{2}{n^3}\right)}\)

\((1 - \cos x_n) \approx\)

\(\Rightarrow \frac{1-\cos x_n}{x_n^2} \rightarrow \frac{1}{2}\)

\(\approx \frac{x^2_n}{2}\)

\(\log (1 + x_n)\)

\(\approx\) ... \(\)log ⁡(1 + x_n) ≃ x_n

cos ​( ²⁄_n) - 1 ⁄ log ​(1 - ²⁄_n³) ≃ - ¹⁄₂ ( ²⁄_n )² ⁄ - ²⁄_n³ =¹⁄₂ . ⁄_n² . n³⁄₂ = n → &plus;∞

N = cos ​( ²⁄_n) - 1 ≃ - ¹⁄₂ ( ²⁄_n )² = - ¹⁄₂ ⁄_n²

D = log ​(1 - ²⁄_n³) ≃ - ²⁄_n³

3) lim ‿n→∞∛cos ​( ²⁄_√n) - 1 ⁄ log ‿_n⁄⁄_√n e^⁄⁄_√n - 1

x_n → 0

(1 &plus; x_n)^α ⁄ x_n → α, (1 &plus; x_n)^α - 1 ≃ α x_n

∛(cos ​( ²⁄_√n) - 1 = ∛ (1 &plus; (-1 &plus; cos ​ ²⁄_√n)) - 1

x_n= [1 &plus; (-1 &plus; cos ​ ²⁄_√n)]^1⁄3 - 1 ≃ ¹⁄₃ (-1 &plus; cos ​ ²⁄_√n)

≃ ¹⁄₃ (-¹⁄₂) ( ²⁄_√n )² = - ⁴⁄_6n = - ²⁄_3n

x_n → 0

e^x_n -1 ⁄ x_n → 1 , e^x_n -1 ≃ x_n

e^tg_1/n - 1 ∼ tg_1/n ∼ 1/n

x_n → 0     ^tg x_n⁄_xn → 1 ,     tg x_n ∼ x_n

^{3√cos 2 ⁄_√n} ⁄_e^1/ n - 1   ∼   ^- 2 ⁄_3n  ∼   ¹ ⁄_n   =   - ² ⁄₃.

4)   lim_n → ∞^{log ( cos 1⁄_2n )  .    (√cos 1⁄_2n - 1)}⁄_{e^sin1⁄4n  -  1}

N = log ( cos ¹⁄_2ⁿ )  .    ( cos ¹⁄_2ⁿ - 1) = log