| 1 | ||
a) f(x) = (1 − | )n2 tutaj jest nawias do potęgi n2 | |
| n3 |
| 1 | ||
b) f(x) = (1 + | )3ne nawias do potęgi 3ne | |
| 3n−2 |
| 2−n−1 | ||
c) f(x) = | sin(n2−2) | |
| 3n |
| −1 | n2 | |||
a) = lim [(1 + | )n3]K = eK , gdzie: K = | i lim K = 0 | ||
| n3 | n3 |
| 1 | 3ne | |||
b) = lim [(1 + | ]K , gdzie: K = | i lim K = e | ||
| (3n−1)3n−2 | 3n−2 |
Natalka pewnie już jest na wykładach
| 1 | ||
b) poprawa zapisu: = lim [ (1 + | )3n−2]K | |
| 3n−2 |
... wróci, to poczyta