matematykaszkolna.pl
zad granica ciagu:
 4n−2 
lim(
)3
 2n−1 
25 cze 11:28
granica ciagu:
(4n−2)(16n2−16n+4) 8{4n2−4n+1} 
=
=8
(2n−1)(4n2−4n+1 4n2−4n+1 
25 cze 11:29
kjh: 0g
25 cze 18:36