Analisi 1 esercizi svolti sui limiti di funzioni, prof Giovanni Dore

Limiti Parte 1

1. lim_x→5 (2x³ - 3x² + 5x) / (x - 5) = 0/0

   lim_x→5 (2x³ - 3x² + 5x) / (x - 5) = -5

2. lim_x→∞ (2x³ - 3x² + 5x) / x² = +∞/∞

   lim_x→∞ 2x³/x² - 3x²/x² + 5x/x² = +∞/∞

   lim_x→∞ (2x³ - 3x² + 5x) / x⁴ = 0

3. lim_x→0 (2x³ - 3x² + 5x) / x⁶ = +∞/∞

   lim_x→-∞ 2x³/ x⁶ - 3x²/x⁶ + 5x/x⁶ = 0

   lim_x→∞ x⁴/x⁶ = 0

   lim_x→∞ 2ᵪ/x² = 0

4. lim_x→2 (3x⁴ - 2x² + 5x - 2) / (x² - 4) = 0/0

   lim_x→2 x² - 4x² + 5x - 2 / (x² - 4) = Hopital

   lim_x→2 3x² - 8x + 5 / 2x = 4/4

5. lim_x→1 x² - 1 / x² - 3x + 2 = 0/0

   lim_x→1 Hopital = lim_x→1 x - 1 / 3x - 3 = -∞/∞

6. lim_x→0 x⁵ - sin(x) + 5x² / x² - x = 0/0

   lim_x→0 2x⁵ - sin(x) + 5x² / x² - x ∼ lim_x→0 -x/x = -1

7. lim_x→0 sin(x + x⁴) / x³ - x = 0/0

   lim_x→0 sin(x + x⁴) / lim_x→0 x + x⁴ / x³ - x ∼ lim_x→0 x/x = -1

8) _limx→0 ^sin(3x)/_tan(5x) = ⁰/₀

_limx→0 ^sin(3x)/_tan(5x) = _limx→0 ^3x/5x = ³/₅

9) _limx→0 ^{3x + x³}/_{2x² + x sin(2x³)} = ^3x²/_2x² = ³/₂

10) _limx→0 ^{√x² - 2}/_{x sin(9 - x)} = ⁰/₀Posto t = 4 - x → x = 4 - t

_limx→0 ^{√(4 + t) - 2}/_{(4-t) sin(t)} = _limt→0 ^{√(4+t) - 2}/_{(4-t) sin(t)} = _limt→0 ^{2(√(4+t) - 2)/t}/_{(4-t) (h)}

x = ^-t/_4+t ~ _limx→0 ^-t/_4t = ¹/₁₆

11) _limx→0 ^sin(x/2)/_cos(x) = ⁰/₀

Posto t = π/2 - x → x = π/2 - t

_limt→0 ^{sin(π - 2t)}/_{cos(π - t)} ~ _limt→0 ^{sin(t) cos(2t) - cos(t) sin(2t)}/_{cos(t) sin} ~ _limt→0 ^sin(t)/_sin(t) = 2

12) _limx→0 ^sin(x²)/_{sin(x) + x} = ⁰/₀

_limx→0 ^sin(x)/_{sin(x²) + x} ~ _limx→0 ^x²/_x ~ _limx→0 ^x/_x = 0

13) _limx→0⁺ ^{√(x + x² - √(x + 2x²)} = ⁰/₀

_limx→0⁺ ^{√(x + x²) - √(4+x)²}/_{√(x(1+x)) - √(1+2x)} = _limx→0⁺ ^{√(x(1+x)) - √(1+2x) =}

= _limx→0⁺ ^{1/² - x}/_{k(1 + ³/₂ x)} ~ _limx→0⁺ ^{-1/2 x}/_x = -1/2