Limiti esercizi svolti

ANALISI 1

Limiti svolti

ANALISI 1

Limiti svolti

ESERCIZIO SVOLTO

^lim_x→3 x-3 / √x - √3 = 6 / 0

^lim_x→3 (x-3) / (√x - √3) * (√x + √3) / (√x + √3) = ^lim_x→3 (x - x ). (√x+√3) = 2√3

a² - b² = (a-b)(a+b)

^lim_x→∞ ^{ln (1+5x)} / _3x - ^5x / _5x = 5/3

^lim_x→0 ^{ln (1+g(x))} / _f(x) = 1

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim_x→1 lnx / (x²−1) → (x−1)(x+1)

x−1 = y

x→1 → y→0

x = y+1

lim_y→0 ln(y+1) / y(y+1+1)

= 1/2

lim_x→0 ln(1+4x) / (e^5x−1)

4x = ln(1+4x)

4x / 4x ➝ 5x / 5x

= 4/5

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim_x→0 (1 - cos2x) / x(e²x - 1)= 2x / 2x = 1/2ln / log_e/Log

lim_x→0 log₁₀(1 + 3x) / x² + 2x

lim_x→0 3 / 3_{log10(1 + 3x)} / x(x + 2)= 3/2 * 1 / ln10

lim_x→0 log_e(1 + x) / x = 1 / ln(e)

lim_x→∞ x² log³ x + x + log⁴ x / 1 + x³ = lim_x→∞ x⁷ log³ x (1 + 1/x log⁴ x) / x³ (1 + 1 / x³)

lim_x→∞ log³ x / x = 0

ESERCIZIO SVOLTO

lim_x→0 (1+√x)^1/√x = lim_x→0 e^{ln(1+√x)1/√x} =

e^ln(x) = x

log_ba² = 2 log_ba

lim_x→0 e^{1/√x . ln(1+√x)} = lim_x→0 e^e

lim_x→0 √x / x = e^∞ = ∞

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim_x→∞ (3x + 1 / 3x – 5)^{x – 2}

lim_x→∞ (3x + 1 + 5 – 5 / 3x – 5)^{x – 2} = lim_x→∞ (3x – 5 / 3x – 5 + 6 / 3x – 5)^{x – 2}

= lim_x→∞ (1 + 1 / (3x – 5 / 6))^{(x – 2) * (3x – 5 / 6)})

= lim_x→∞ e^{6(x – 2) / 3x – 5} = lim_x→∞ e^{6x – 12 / 3x – 5} = e²

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim_{x → 0} (e^x² - cos x) / x² = lim_{x → 0} (e^x² - 1 - cos x + 1) / x² =

(e^x² - 1) / x² + (1 - cos x) / x² = ³/₂

lim_{x → ∞} ((x + 8) / (x - 2))^x [→] 1^∞

lim_{x → ∞} ((x + x + 2 - 2) / (x - 2))^x = lim_{x → ∞} ((x - 2) / (x - 2) + (8 + 2) / (x - 2))^x

lim_{x → ∞} (1 + ¹⁰/_{x - 2})^x = lim_{x → ∞} (1 + ¹/_{x - 2}/₁₀ (x - 2) /₁₀)^x

lim_{x → ∞} ^10x/_{x - 2} = e⁰

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZI SVOLTI

lim_x→1

lim_y→0

cos( ^π⁄₂ y)

^{y = x - 1}⁄_{x = y + 1}

cos( ^π⁄₂ (y + 1) )⁄(1 - √(y + 1) )

=lim_y→0

cos( ^π⁄₂ y + ^π⁄₂ ) ⁄(1 - √(y + 1) )

lim_y→0

sin( ^π⁄_2y ) ⁄(√(y + 1) - 1)

S → ^π⁄₂ y

=^π⁄₂ lim_y→0(^√(y+1)+1⁄_y)

^y⁄(^{√(y + 1) + 1}⁄(^{√(y + 1) - 1}⁄)

^{√(y + 1) + 1}⁄_{√(y + 1) + 1}

=^π⁄₂ * 2 = π

VINCENZA SCIORTINO, ANALISI 1 2021

DE L'HOSPITAL ESERCIZI

^lim_x→0 ^senx⁄_x = 1   →   ^lim_x→0 ^cosx⁄₁ = 1

^lim_x→1 ^{cos(π⁄2 x)}⁄_1−√x = [⁰⁄₀] = = ^lim_x→1 ^{−sen(π⁄2 x) (π⁄2)}⁄_1⁄2√x = ^−π⁄2⁄_−1⁄2 = π

^lim_x→2 ^x−2⁄_ln(x⁄2) = [⁰⁄₀]   &box;  y = x−2 x→2 → y→0

^lim_y→0 ^y⁄_ln(1+y⁄2) = ^lim_y→0 ^y⁄_ln(1+y⁄2) = ^lim_y→0 ^2(y⁄2)⁄_(y⁄2+1) = 2

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZI SVOLTI

lim_{x → 2} ^{x - 2}/ln(^x/₂) = ⁰/₀

y = x - 2

x → 2 ⇒ y → 0

lim_{y → 0} ^y/ln(^{y + 2}/₂) = lim_{y → 0} ^y/ln(1 + ^y/₂) = lim_{y → 0}

^y/₂/ln(1 + ^y/₂) = 2

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim _{x→ ∞} √^{x3 - 1} / _{x² - x} = ∞

y = ∞

lim _{x→ ∞} √^{x3 - 1} / _{x² - x} = [∞ / ∞] = ∞

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCITAZIONE

LIMITI

lim_x→3 ^ln(x)/_{3 ln(x)-1} = lim_x→3 ^ln(x)/_{ln(x)(3 - ¹/ln(x))} = lim_x→3 ¹/_{3 - ¹/ln(x)} = ¹/₃

lim_x→3e ^ln(x)/_{3 ln(x)-1} = ∞

lim_x→3e ^ln(3e)/_{3 ln e^2/3-1} = lim_x→3e ^ln((3)e)/_{3/3 ln(e)-1}

lim_x→∞ ^ln(x/1)/_{3ln(x) - 1} = lim_x→∞ ^ln(x/1)/_{ln(x)(3 - ¹/ln(x))} = ¹/₃

ESERCIZIO SVOLTO

lim_x→+∞ 3x - 4√1-x² = [∞ - ∞] → INDETERMINATO

lim_x→+∞ 3x - 4√1-x² (3x + 4√1-x²) / (3x + 4√1-x²) = lim_x→+∞ (9x²-(16(1-x²))) / (3x + 4√1-x²)

Somma x Differenza

e²-b² = (e-b)(e+b)

Razionalizzazione

1/√2 . √2/√2 = √2/2

lim_x→+∞ (9x² - 16 + 16x²) / (3x + 4√1-x²) = lim_x→+∞ (25x² - 16) / (3x + 4√1-x²) = ∞

ESERCIZIO SVOLTO

lim_x→1 √^x³-1/_x²-x = lim_x→1 ^{√(x-1)(x²+x+1)}/_x(x-1) = √3

x³-1

x³-b³ = (a-b)⎡

(x-1)

(x²+x+1)¹(x-1)

1 0 -1 1 1 1 1 1 1 0

ϑ ^(x²-x)2 ━━━━━ ^(x²-x)2 √(x²-x)

√^(x²-x)³/_x² / x = √^(x²-x)³

^(x²-x)2-1/2 = (x²-x)^3/2

2⁽¹⁾ . ¹/_2⁺³ = z^(1-3) = z^-2

^{ac . qb} = e^{c+b a} / _q^b = e^c-b

VINCENZA SCIORTINO, ANALISI 1 2021

ESERCIZIO SVOLTO

lim_x→0

sin(sen(x)) sin x

x sen x

lim_x→0 ^{sen x}/_x = 1 → lim_x→0 ^{sen f(x)}/_f(x) = 1

VINCENZA SCIORTINO, ANALISI 1 2021