| n+7 | ||
Oblicz granicę ciągu ( | )3n+1 | |
| n+9 |
| 2 | 3n+1 | |||
= lim [(1 − | )n]K , gdzie: K = | i lim K = 3 | ||
| n+9 | n |
| 3n+1 | ||
pomyłka ... K = | i lim K = 3 | |
| n + 9 |
| 2 | ||
= lim[(1 − | )n+9]K | |
| n+9 |
A czy w tym przykładzie mogę tak zrobić:
| n+1 | n(1+1n) | |||
lim ( | )n=( | n=1n=1 | ||
| 2n+5 | n(2+5n) |
| 1 | ||
Nie, bo lim n→∞(1+ | )n=e | |
| n |
| 1 | n+1 | |||
= lim( | )n*( | )n = 0 | ||
| 2 | n+5/2 |