Esercitazione guidata Analisi Matematica 1 - settima parte

Funzioni, limiti e sviluppi

Continuità e derivabilità della funzione

f(x) = { ^{sin(αx) - x/_x-1} 0 ≤ x < 1 ∪ x > 1 e^-1/_x2 x < 0 } α ∈ ℝ

f(0) = 0

lim_x→0⁺ f(x) = 0

lim_x→0^- f(x) = 0 ⇒ f cont.

Per x ∈ (0,1) ∪ (1,+∞)

f’(x) = cos(αx) · α - ^{x - 1 - x}/_{(x - 1)²} = α cos(αx) + ¹/_{(x - 1)²}

Per x < 0 f’(x) = e^-1/_x2 · ²/_x³

In x₀ = 0

limx→0+ f(x) - f(0)/x - 0 = limx→0+ sin(αx) - x/x-1/x = limx→0+ [sin(αx)/x - x/x(x-1)] = α + 1 → α · sin(αx)/αx → α se α ≠ 0

limx→0- f(x) - f(0)/x - 0 = limx→0- e-1/x2/x

f(x) = { sin(αx) - ^x/_x-1     0 ≤ x < 1 ∪ x > 1 { e^-₁/_x2                x < 0                                α ∈ ℝ

f(0) = 0

lim_x→0⁺ f(x) = 0

lim_x→0^- f(x) = 0 => f cont.

Per x ∈ (0,1) ∪ (1,+∞)

f'^'(x) = cos(αx) ⋅ α - _x-1-x/_(x-1)² = α cos(αx) + ¹/_(x-1)²

Per x < 0

f'^'(x) = e^-₁/_x2 ⋅ ²/_x³

In x₀ = 0

limx→0+ f(x) - f(0)/x - 0 = limx→0+ sin(αx) - x/x-1/x == limx→0+ [sin(αx)/x - x/x(x-1)] = α + 1                  → α ⋅ sin(αx)/αx                  → α , se α ≠ 0

limx→0- f(x) - f(0)/x - 0 = limx→0- e-1/x2/x

y = ¹/_x

lim_y→∞ y e^−y2 = lim_y→∞ ^y/_e^y2 = 0

Se α + 1 = 0 (=> α = −1) e f è derivabile e f'^'(0) = 0.

1) Limite con seno, arctangente e logaritmo

1) lim_x→0 ^{(sen x)2 − x2}/_{(arctg x)² − (log (1 + x))²}

(sen x)² − x² = (x − ^x3/₆ + o(x³))² − x² = x² − ^x3/₃ + o(x⁵) − ¹/₃ x⁴ + 2 x o(x³) − ¹/₃ x³ o(x³) − x² = − ¹/₃ x⁴ + o(x⁴)

(arctg x)² − (log (1 + x))² = (x − ^x3/₃ + o(x⁴))² − (x − ^x2/₂ + ^x3/₃ + o(x³))² = [x² + ^x6/₉ + o(x⁵) − ²/₃ x³ + 2 x o(x³)] − [x² − ²/₃ x³ o(x³) + ³/₃ o(x³)] = [x² − ²/₃ x⁴ + o(x⁴)] − [x² − x³ + (²/₃ + ¹/₄) x⁴ + o(x⁴)] = x³ + x⁴ (⁻²/₃ − ²/₃ − ¹/₄) + o(x⁴) = x³ + o(x³)

\lim_x→0 \frac{- \frac{1}{3} x⁴ + o(x⁴)}{x³ + o(x³)} = \lim_x→0 \frac{- \frac{1}{3} x⁴}{x³} = 0.

2) Limite con esponenziale e radice

2) \lim_x→0 \frac{e^x \sin x - x \cdot \sqrt{1 + αx}}{e^x \log(1+x) - \sin x \cdot \sqrt{1 + 2x}}

α ∈ \mathbb{R}

e^x \sin x - x \sqrt{1 + αx} == (1 + x + \frac{x²}{2} + \frac{x³}{6} + o(x³)) \left(x - \frac{x³}{6} + o(x³)\right)- x \left(1 + \frac{1}{2}(αx) - \frac{1}{8}(αx)² + \frac{1}{16}(αx)³ + o(x³)\right) = \frac{x³}{6} + o(x³) + \frac{x²}{2} x - \frac{x³}{2} - \frac{1}{2}αx² + \frac{1}{8} α² x³ = (\frac{1}{2} - α)x² + \left(- \frac{1}{6} + \frac{1}{2} + \frac{1}{8} α²\right)x³ + o(x³)

e^x \log(1+x) - \sin x \cdot \sqrt{1 + 2x} = (1 + x + \frac{x²}{2} + \frac{x³}{6} + o(x³)) \left(x - \frac{x²}{2} + \frac{x³}{3} + o(x³)\right)- \left(x - \frac{x³}{6} + o(x³)\right) (1 + \frac{1}{2}(2x) - \frac{1}{8}(2x)² + \frac{1}{16}(2x)³ + α(x³)) = \bigg[x - \frac{x²}{2} + \frac{x³}{3} + o(x³)\bigg] + \bigg[\frac{x²}{2} - \frac{x³}{2}\bigg]- x \cdot \frac{x³}{6} + o(x³) + \frac{x²}{2}+ \frac{x³}{3} - \frac{1}{2}x³ = -\frac{x²}{2} + x³ \left(\frac{1}{3} + \frac{1}{6} -\frac{1}{2}\right) + o(x³) = -\frac{x^2}{2} + O(x^2)

\lim_{x \to 0} \frac{(1 - \frac{1}{2}\alpha)x^2 + (\frac{1}{3} + \frac{1}{8}\alpha^2)x^3 + O(x^3)}{-\frac{1}{2}x^2 + O(x^2)}. \ \ \text{se} \ \alpha = 2

\lim_{x \to 0} \frac{(\frac{1}{3} + \frac{1}{2})x^3 + O(x^3)}{-\frac{1}{2}x^2 + O(x^2)} = 0. \ \ \text{se} \ \alpha \neq 2

\lim_{x \to 0} \frac{(1 - \frac{1}{2}\alpha)x^2 + O(x^2)}{-\frac{1}{2}x^2 + O(x^2)} = 2(1 - \frac{1}{2}\alpha) = \alpha - 2.

3) Limite con coseno, seno e logaritmo

3) \ \lim_{x \to 0^+} \frac{\cos(x^\alpha) - \sqrt{1 - \sin x}}{-x \cdot \log(1 + \sin x)}, \ \ \alpha > 0

= \cos(x^\alpha) - \sqrt{1 - \sin x} = \left[-\frac{1}{2}(x^\alpha)^2 + \frac{1}{4!}(x^\alpha)^4 + O((x^\alpha)^4)\right]- \left[-\frac{1}{2}\sin x - \frac{1}{8}(\sin x)^2 + \frac{1}{16}(\sin x)^3 + O(\sin^3 x)\right] = -\frac{1}{2}x^{2\alpha} + \frac{1}{4!}x^{4\alpha} + O(x^{4\alpha}) + \frac{1}{2}\sin x + \frac{1}{8}\sin^2 x = -\frac{1}{2}x^{2\alpha} + \frac{1}{4!}x^{4\alpha} + O(x^{4\alpha}) + \frac{1}{2}\sin x + \frac{1}{8} \left(x - \frac{x^3}{6} + O(x^3)\right)^2 + \frac{1}{16} \left(x - \frac{x^3}{6} + O(x^3)\right)^3 + o((x - ^x3₆ + o(x³))³)

= -₁²x^2α + ₁²⁴x^4α + o(x^4α) ++ ₁²x - ₁¹²x³ + o(x³) + ₁⁸x² + ₁¹⁶x³ = -₁²x^2α + ₁²⁴x^4α + o(x^4α) ++ ₁²x + ₁⁸x² + (₁¹⁶ - ₁¹²)x³ + o(x³) = ...

se 2α = 1 (α = ₁²) ... = (₁⁸ + ₁²⁴)x² + o(x²) = ₁⁶x² + o(x²)

se 0 < α < ₁² ... = -₁²x^2α + o(x^2α)

se α > ₁² ... = ₁²x + o(x)

x - log (1 + sin x) == x - (sin x - 12sin2x + 13sin3x + o(sin3x)) = x - [(x - x36 + o(x3)) - 12(x - x36 + o(x3))2 ++ 13(x - x36 + o(x3))3 + o((x - x36 + o(x3))3)]