Esercitazione guidata Analisi Matematica 1 - quinta parte

Esercizi su serie e limiti

1) Serie con parametro α

\[ \sum_{n=1}^{+\infty} \frac{\alpha^n}{\sqrt{1+\frac{2}{3^n}}-\cos(\frac{1}{2^n})} \]

\[ \lim_{n \to +\infty} \frac{\alpha^n}{\sqrt{1+\frac{2}{3^n}}-\cos(\frac{1}{2^n})} \]

\[ D = \sqrt{1+\frac{2}{3^n}} - \cos(\frac{1}{2^n}) = 2^n \]

\[ =\left (1+\frac{1}{2}\left(\frac{2}{3^n}\right)+o\left(\frac{2}{3^n}\right) \right)-\left[ 1-\frac{1}{2}\left(\frac{1}{2^n}\right)^2+o\left(\left(\frac{1}{2^n}\right)^2\right) \right] \]

\[ =\cancel{1} + \frac{1}{3^n}+o(\frac{1}{3^n}) -\cancel{1} + \frac{1}{2}\left(\frac{1}{4^n}\right)+o\left(\frac{1}{4^n}\right)\]

\[ =\frac{1}{3^n}+o\left(\frac{1}{3^n}\right) \]

\[\sim \frac{1}{3^n} \]

\[\frac{1}{3^n} = o \left(\frac{1}{4^n}\right) \]

\[\frac{1}{4^n} = o \left(\frac{1}{3^n}\right) \]

\[\frac{1}{4^n}{1}{3^n} = \left(\frac{3}{4}\right)^n \to 0 \]

\[\sum_{n=1}^{+\infty} \frac{\alpha^n}{2^n} \sim \sum_{n=1}^{+\infty} (3\alpha)^n \]

\[\lim_{n \to +\infty} \frac{\alpha^n}{\frac{1}{3^n}+o\left(\frac{1}{3^n}\right)} = \lim_{n \to +\infty} (3\alpha)^n = \begin{cases} +\infty & 3\alpha > 1 \\ 0 & -1 1)

∑_n=1^∞ (αⁿ) / (√(1 + 2/3ⁿ) - cos(1/2ⁿ)) lim_n→∞ (αⁿ) / (√(1 + 2/3ⁿ) - cos(1/2ⁿ)) D = (1 + 2/3ⁿ) - cos(1/2ⁿ) = 2ⁿ

= [(1 + 1/2 (2/3ⁿ) + o(2/3ⁿ))] - [(1 - 1/2 (1/2ⁿ)² + o((1/2ⁿ)²))]

= 1 + 1/3ⁿ + o(1/3ⁿ) - 1 + 1/4ⁿ + o(1/4ⁿ) = (1/3ⁿ) + o(1/3ⁿ) ~ 1/3ⁿ

(1/3ⁿ) = o(1/4ⁿ) 1/4ⁿ = o(1/3ⁿ) 1/4ⁿ / 1/3ⁿ = (3/4)ⁿ → 0

lim_n→∞ (αⁿ) / (1/3ⁿ + o(1/3ⁿ)) = lim_n→∞ (3α)ⁿ = 3α > 1 ∞ -1 < 3α < 1 0 3α = 1 1 3α ≤ -1 0

∑_n=1^∞ αⁿ / 2ⁿ ~ ∑_n=1^∞ (3α)ⁿ ∑_n=1^∞ (αⁿ)/(3ⁿ) ~ ∑_n=1^∞ (αⁿ)/(1/3)ⁿ = ∑_n=1^∞ (3α)ⁿ

conv. per | 3α | < 1 div. per 3α > 1

1/(3ⁿ) + O(1/(3ⁿ)) ~ 1/(3ⁿ) ? lim_{n -> ∞} 1/(3ⁿ) + O(1/(3ⁿ)))/(1/(3ⁿ)) = lim_{n -> ∞} (1/(3ⁿ))/(1/(3ⁿ)) + O(1/(3ⁿ))/(1/(3ⁿ)) = 1

lim_x->0 o(x)/x = 0

2) Limite con seno, esponenziale e radici

2) lim_x->0 (sin² x - e^x + 1)/(cos √x - √(1-x))

sin x = x + o(x) e^x = 1 + x + o(x) cos x = 1 - 1/2 x² + o(x²) (1 + x)² = 1 + 2x + o(x)

N = sin² x - e^x + 1   = (x + o(x))² - (1 + x + o(x)) + 1   = x² + 2x o(x) + (o(x))² - 1 - 2x + o(2x²) + 1

→ x \, o(x) = o(x²) 2x \, o(x) = 2o(x²) = o(x²) − (o(x)²) = o(x) \cdot o(x) = o(x²) o(2x²) = o(x²)

= x² + o(x²) + o(x²) - 2x² + o(x²) = -x² + o(x²)

D = cos √x - √(1-x) = (1 - 1/2 (√x)² + o((√x)²)) - (1 + 1/3 (-x) + o(-x)) = 1 - 1/2 x + o(x) - / + 1/3 x + o(x) = -1/6 x + o(x)

lim ... x → 0 = 06 \, x/x = x → 0

3) Limite con radici cubiche

3) \lim......= \sqrt[3]{1 + \frac{1}{3}(2x²) + o(2x²) - [\sqrt[3]{1 + \frac{1}{3}(x² - x³) + o(x² - x³)]

= x + ²/₃ x² + o(x²) - ¹/₃ x² + ¹/₃ x³ + o(x²) = ¹/₃ x² + ¹/₃ x³ + o(x²)

(o(x + x) = o(x)) = min(, ) x³ = o(x²) ^x3/_x² = 0 = ¹/₃ x² + o(x²)

D = sin x · log(1+x) + tg x = (x + o(x)) · (x + o(x)) + (x + o(x))

sin x = x + o(x) log(1+ x) = x + o(x) tg x = x + o(x)

= x² + 2xo(x) + (o(x))² = x + o(x) = x + o(x)

x² = o(x) x²/_x → 0 2xo(x) = o(x²) = o(o(x)) = o(x) o(x¹)² = o(x²) = o(o(x)) = o(x)

lim ( ¹/₃ x² + o(x²) / x + o(x) ) = 0

4) Limite con parametro α

4) lim ( ^ex/_{sin x - tg² x} - ^√1-x/₃ ) ∈ ℝ

N = e^αx - ³√(1 - x)   = αx + o(x) - [- ¹/₃ x + o(x)]   = αx + ¹/₃ x + o(x)   = (α + ¹/₃) x + o(x)

D = sin x - ∫₀^x ₂   = x + o(x) - x² + o(x²)   = x + o(x)

lim_x→0 (α + ¹/₃) x + o(x)       x + o(x)

• α + ¹/₃ = 0 lim_x→0 o(x) = 0
• X + o(x) o(x) = o(x) x + o(x) = x
• α + ¹/₃ ≠ 0 lim_x→0 (α + ¹/₃) x = α + ¹/₃      x

5) Limite destro con parametro α

5) lim_x→0⁺ α √cos x - e^x² = l , α ∈ R        √1 + x - √1 - x

N = α √cos x - e^x² = α √1 + (-1 + cos x) - e^x²

= α [1 + ¹/₂(-1+cosx) + o(-1+cosx)] - [1 + x² + o(x²)]

(1+x)^α = 1 + αx + o(x), e^x = 1 + x + o(x)

= α [1 + ¹/₂(-¹/₂x² + o(x³)) + o(-¹/₂x² + o(x²))] - [1 + x² + o(x²)]

cos x = 1 - ¹/₂x² + o(x²) - 1 + cos x = -¹/₂x² + o(x²)

= α [1 - ¹/₄x² + o(x³)] - [1 + x² + o(x²)] = (α-1) - (^α/₄ + 1)x² + o(x²)

D = √(1 + x) - √(1 - x) = x + o(x)

^lim/_x→0 (α-1) - (^α/₄ + 1)x² + o(x²) α - 1 = 0 α - 1 ≠ 0 x² = o(x)