ll asi: Oblicz granice funkcji

	sin3x−3x
limx−−>0
	x3

undefined

Adamm:

	sin(3x)−3x		3cos(3x)−3		cos(3x)−1
limx→0		= limx→0		= limx→0		=
	x3		3x2		x2

	−2sin2(3x/2)		9	−sin2(3x/2)
limx→0		= limx→0			=
	x2		2	(3x/2)2

	9
= −
	2

Mariusz: Można bez de l'Hospitala Rozwińmy sinusa sin(3x)=sin(2x)cos(x)+cos(2x)sin(x) sin(3x)=(sin(x)cos(x)+cos(x)sin(x))cos(x)+(cos(x)cos(x)−sin(x)sin(x))sin(x) sin(3x)=2sin(x)cos(x)²+(cos(x)²−sin(x)²)sin(x) sin(3x)=2sin(x)(1−sin²(x))+(1−2sin(x)²)sin(x) sin(3x)=3sin(x)−4sin³(x)

	3sin(x)−4sin3(x)−3x
limx→0		=
	x3

	sin(x)		sin(x)−x
−4*limx→0(		)3+3*limx→0
	x		x3

	sin(3x)−3x
limx→0		=−4+3g
	x3

Niech 3x=t t→0 gdy x→0

	sin(t)−t		sin(t)−t
limt→0		=27*limt→0
	(t/3)3		t3

−4+3g=27g −4=24g

	1
g=−
	6

	sin(3x)−3x		3		9
limx→0		=−4−		=−
	x3		6		2

lol: Z rozwinięcia sinusa w szereg potęgowy. sin x = x −x³/3! + o(x³) Zatem lim_x−>0 (sin(3x)−3x)/x³ = lim_x−>0 (3x−27/6x³+o(x³)−3x)/x³ = = lim_x−>0 −27/6 + o(x³)/x³ = −27/6 = −9/2