| 2n−1 | ||
Lim | =2 | |
| n |
a w jaki sposob mozna inaczej to udowodnic?
| 2n − 1 | ||
lim | = 2 ⇔ | |
| n |
| 2n − 1 | ||
∀ε > 0 ∃ n0 : ∀ n > n0 | | − 2| < ε | |
| n |
| 2n − 1 | ||
| | − 2| < ε | |
| n |
| 2n − 1 | 2n | |||
| | − | | < ε | ||
| n | n |
| −1 | ||
| | | < ε | |
| n |
| 1 | |
< ε | |
| n |
| 1 | |
< n | |
| ε |
| 1 | ||
Niech n0 = [ | ] + 1, | |
| ε |
| 2n − 1 | ||
wtedy: ∀ n > n0 | | − 2| < ε | |
| n |
| 2n − 1 | ||
Zatem lim | = 2 | |
| n |
