oblicz granice ciągu
kacha: ln(3n+6n+2n)n+arctg*n!n2
3 mar 12:19
Godzio:
Niech Twój ciąg nazywa się a
n wtedy:
| ln(6n) | | | | ln(6n + 6n + 6n) | | | |
| − |
| ≤ an ≤ |
| + |
| |
| n | | n2 | | n | | n2 | |
ln(6n + 6n + 6n) = ln(3 * 6n) = ln3 + ln6n = ln3 + n * ln6
| | π | | ln3 | | π | |
ln6 − |
| ≤ an ≤ |
| + ln6 + |
| |
| | 2n2 | | n | | 2n2 | |
L → ln6, P → ln6 ⇒ a
n → ln6
3 mar 13:10