granica karmen:

	ln(n2 + nsin(n2) + 1)
Oblicz limn →∞
	sin(n2) + ln(n+1)

getin: −1 ≤ sin(n²) ≤ 1

	ln(n2+n*sin(n2)+1)
niech xn =
	sin(n2)+ln(n+1)

	ln(n2+n*(−1)+1)		ln(n2−n+1)
an =		=
	1+ln(n+1)		ln(n+1)+1

	ln(n2+n*1+1)		ln(n2+n+1)
bn =		=
	−1+ln(n+1)		ln(n+1)−1

a_n ≤ x_n ≤ b_n dla każdego n∊N₊

	ln(n2−n+1)
limn→∞ an = limn→∞		=H
	ln(n+1)+1

	2n−1   n2−n+1		(2n−1)(n+1)
= limn→∞		= limn→∞		= 2
	1   n+1		n2−n+1

	ln(n2+n+1)
limn→∞ bn = limn→∞		=H
	ln(n+1)−1

	2n+1   n2+n+1		(2n+1)(n+1)
= limn→∞		= limn→∞		= 2
	1   n+1		n2+n+1

Ponieważ a_n ≤ x_n ≤ b_n oraz lim a_n = lim b_n = 2, to również lim x_n = 2 twierdzenie o trzech ciągach