obliczanie granicy ciągu
Karolina : Jak obliczyć tę granicę?
lim (3n−13n+1)n+4
n−>∞
18 gru 16:02
Adamm: | | 3n−1 | | 2 | |
( |
| )n+4=(1− |
| )n+4→e−2/3 |
| | 3n+1 | | 3n+1 | |
18 gru 16:04
Karolina : A mógłbyś to troszkę rozpisać?
18 gru 16:05
KKrzysiek: (1 + U{−2}{3n+1))
18 gru 19:45
18 gru 19:45
Adamm: | | 2 | |
(1− |
| )(3n+1)(n+4)/(3n+1) |
| | 3n+1 | |
| | 2 | | 1 | |
mamy (1− |
| )(3n+1)→e−2, (n+4)/(3n+1)→ |
| |
| | 3n+1 | | 3 | |
| | 2 | |
(1− |
| )(3n+1)(n+4)/(3n+1)→e−2/3 |
| | 3n+1 | |
18 gru 19:55