| 8n2+2 | 2 | |||
3√ | całość jest pod pierwiastkiem 3−go stopnia granica wychodzi | ? | ||
| 125n2−1 | 5 |
| 3√8+0 | ||
3√n2 przed nawias lim = | = 2/5 | |
| 3√125−0 |
| 8n2+2 | 23(n2−2−1) | |||
3√ | = 3√ | = | ||
| 125n2−1 | 53(n2−5−3) |
| 2 | n2−2−1 | |||
= | 3√ | , | ||
| 5 | n2−5−3 |
| 2 | 1 | |||
3√8+ | a w mianowniku 3√5− | |||
| n2 | n2 |
| 2 | |
dąży do 0 | |
| n2 |
| 1 | |
dąży do 0 | |
| n2 |
