| n + 3 | ||
lim n→∞= ( | )4n | |
| n − 2 |
| n(1+3n | ||
..= lim n→∞( | )4n= | |
| n(1−2n) |
| (1+3n)4n | ||
= lim n→∞ | = | |
| (1−2n)4n |
| (1+3n) n3 *12 | ||
= lim n→∞ | = | |
| (1−2n) (−n2) *(−8) |
| e12 | ||
= | = e20 − szukana granica . .. ![]() | |
| e−8 |
| n−2 | 5 | 5 | ||||
czyli będzie tak lim ( | + | )4n = lim (1+ | )4n .... i dalej | |||
| n−2 | n−2 | n−2 |
| n−2+5 | 5 | |||
... = lim n→∞ ( | )4n= lim n→∞ (1+ | )4n= itd. ![]() | ||
| n−2 | n−2 |