Granica ciągu
Sławomir: Czy mógłby mi ktoś pomóc z tymi dwiema granicami?
limn→∞ (3n+17n−2)n
odp: 0
limn→∞ (2n−12n−3)3n−2
odp: e−6
8 lis 23:55
jc:
n ≥ 1
| | 3n+1 | | 4n | |
0 < |
| ≤ |
| =4/5 |
| | 7n−2 | | 5n | |
| | 3n+1 | |
0 < ( |
| )n ≤ (4/5)n |
| | 7n−2 | |
trzy ciągi, granica = 0
| | 2n−1 | | 2n−1 | | 2n−3 | |
= ( |
| )−2 ( |
| )3n : ( |
| )3n |
| | 2n−3 | | 2n | | 2n | |
| | 2n−3 | | 1 | | 3 | |
= ( |
| )2 (1− |
| )3n : (1− |
| )3n |
| | 2n−1 | | 2n | | 2n | |
→ e
−3/2 : e
−9/2 = e
−3/2+9/2 = e
3
9 lis 07:56