ASYMPTOTY Beti: wyznacz asymptoty

	5X3+1
f(x)=
	X2−3X−4

D: x≠4 x≠−1 lim_x→4⁻=−∞ lim_x→4⁺=+∞ asymptota pionowa x=4 lim_x→−1⁻=−∞ lim_x→−1⁺=+∞ asymptota pionowa x=−1 asymptota pozioma

	5X3+1
limx→+∞=		= brak
	X2−3X−4

	5X3+1
limx→−∞=		= brak
	X2−3X−4

asymptota ukośna

	F(x)		5x3+1		1   X3(5+ )   X3
a=limx→∞		=		=		=5
	x		x3−3x2−4x		3   4   X3(1− − )   X   X2

	5X3+1
b=lim{x→∞)=		−5x= 15
	X2−3X−4

y=5x+15 asymptota ukośna