| 4n−3 | ||
lim n−>∞ | ||
| √9n4−2n3+7n+5−3n2+2n |
| 2 | 7 | |||
√n4(9− | + | ...) no i wychodzi mi że ten pierwiastek jest równy 3n2 | ||
| n | n3 |
| 4n−3 | |
| 3n2−3n2+2n |
| 4n−3 | 4n−3 | a+b | (4n−3)*(a+b) | |||||
i masz | = | * | = | |||||
| a−b | a−b | a+b | a2−b2 |
| 1 | 1 | |||
√n2+1 + n = n* (√1+ | ) + n*1= n ( ( √1+ | ) +1) | ||
| n2 | n2 |