Analisi matematica 1 (1/6)

Lezione di Analisi 1 - 23/09/2021

Limiti Notevoli:

lim_x→0 log(1+2x) / log(1+x) = 0/0EI

lim_x→0 log(7+x) / x = 1log7x · xlog7x : xx² : x²

lim_x→0 log(1+2x) / 2x · x / log(1+x) = 1 · 2 = 2 //

lim_x→0 tgx · tg²x / x(1-cosx) = 0/0l.i.

lim_x→0 -senx / x²

lim_x→0 tgx / x · tg²x / 1-cosx = 1 · tg²x / x² · x² / 1-cosx = 2 //

lim_x→0 log(3x) / x = 1

lim_x→0 log(1+x) / x = 1 , lim_x→0 senx/x = 1

lim_x→0 log(6/3x) = (0/0) F.I

lim_x→0 log(1+6x) / x = (3x/3x) = 2

lim_x→0 ( x (4 - cos x) + 2 lnx ) / (4x - 3 lnx) = (0/0) F.I

lim_x→0 (x - x cos x + 2 lnx) / (4x - 3 lnx)

lim_x→0 (1 - cos x + 4 cos x = 4)

Forma Impotenziante

([2],θ) ⇵ z₁z₂=[Ref]+(Im)=2

V(|a|²|z|^x(-1)|z|) = √[1:1/16]

V( z/4 ) = V(z√2/2)

• z[<√z/2] ⇵ θ, x<0, y<0
• θ ⇵ arg( y/x )=π = arg( √√/(-1/2) )

θ = arg( -z/2 )-γ ⇒ θ=π/16π-π-3/4π

2¹ = ( √z/2 )^1/ -3/14

∨ ∋ 4⧫( √z/2 )^4

∙{cos4Uθ⊥3NM4Uθ}

2⧫(?:4)

×[ _ ]

4∧( √z/2 )^4

∙( cos4π±λNM3π )∨(4-3/4π)

2⧫ = ( √z/2 )^4/

∙{ cos3π±λNM3π ∨( √z/2 )^4 3(cosπ±λNMπ) }

( __ )=(x)

ADMB=-1

AIMFT=0

<m

calcolo del restante

lim _m→∞

(a_m+1)/(a_m) = l

1<l diverge

l=1 Non va bene il criterio del rapporto(inde).

(m+1)¹=(m+1)=(m¹)

lim _m→∞

(a_m+1+1)/2)(m¹)/2

lim _m→∞

(a_m+1)/1+0

m²

^2m

Σ (m!)² convergc

m≥1 3^2m