Granice Ola: Oblicz granice ciągów

	(12)n−2n
1) an =
	2−n

2) b_n = (3+n³)^1_n

Janek191:

	1   ( )n − 2n   2		1
an =		= (		)n*2n − 2n*2n = 1 − (2n)2
	2−2		2

więc lim a_n = −∞ n→∞

Janek191: b_n =ⁿ√3 + n³ a_n = ⁿ√n³ =(ⁿ√n)³ c_n = ⁿ√ 2 *n³ =n{2}*(ⁿ√n)³ Mamy a_n ≤ b_n ≤ c_n oraz lim a_n = 1 i lim c_n = 1*1 = 1 n→∞ n→∞ więc na podstawie tw. o trzech ciagach lim b_n = 1 n→∞

Janek191: c_n = ⁿ√2 *n³ = ⁿ√2*( ⁿ√n)³