Prove d'esame: integrali e Limiti svolti

1^a Lim _x→0 (cosx)^1/x = (e^ln(cosx))^1/x -→ e ^{ln (cosx) / x} -→ e⁰ = 1

2^a Lim _x→0 (cos²x) log(cosx) = Lim _x→0 log(cosx) / sec²x = 0 / 1 = 0

cos²x log(cosx) = Lim _x→0 (cos²x log(cosx))^' / (x²)^' -→ Lim _x→0 cos²x log(cosx) / (x²) -→

(-2(cosx)^-1senx) = (-4cosx + 1) secx -→ Lim _x→0 cos²x log(cosx) = Lim _x→0 log(cosx) x² - 1) / x ²

Div. Cambiar Strade

• cosx = 1-t² / 1 + t² x = x sen = 2t / 1+t²

Lim _x→0 (2t / 1 + t²) = Lim _x→0 lg((1-t²)(1+t²)) / 2t(1+ t²) = Lim _x→0 (((1-t²)

-2t / 1 + t²) = Lim _x→0 (2t) / (1 + t²)

Lim _x→0 2t / x²) = Lim _x→0 -t((1 + t²), 0

Lim _x→0 2⁰ = 1

1) lim

x→0

(cosx - senx²) / (√x (e^√x - 1))

=

lim

x→0

(-cosx . x² - secx . x / x² + x / x) / (√x (e^√x - 1))

=

lim

x→0

4 (1/2 x² - x

√x (e^√x - 1)

= lim

x→0

√x (1/2 x^-1)

√x (e^√x - 1)

=

lim

x→0

4 (1/2 x²) - 1

√x

= lim

x→0

√x (1/2 x^-1)

A . √x

-∞

lim

x→0

√x² (1/2 (x - 1) = -∞

lim_x→0 ^{x√(1 + sin x) - 1}/_{x⋅x - 1}

lim_x→0 ^{1 + sin x}/_{sen x} -> lim_x→0 ^{(x + sen x) - 1}/_{sen x}

lim_x→0 f(x) = 0

lim_x→0 C² x - 6 x² + x² - x²

lim_x→0 ^{C x⋅senx⋅x - 6 x2 - xx - x2 - x}/_x² → lim_x→0 6x² + x²/_x² → ∞