| 1 | ||
lim x−> 0+ (3x + 5x) do potęgi | ||
| x |
| 31/n + 51/n | ||
limn→∞ ( | )n ? | |
| 2 |
| 31/n+51/n | ||
=limn→∞(1+( | −1))n | |
| 2 |
| 31/n+51/n | ||
a= | −1→0 | |
| 2 |
| 3t+5t−2 | ||
limn→∞ n*a = limt→0+ | = | |
| 2t |
| 1 | etln3−1 | etln5−1 | |||
*limt→0+ ln3* | +ln5* | = | |||
| 2 | tln3 | tln5 |
| ln3+ln5 | ||
= | ||
| 2 |
| 31/n+51/n | ||
więc limn→∞(1+( | −1))n = eln15/2 = √15 | |
| 2 |
| a1/n+b1/n+c1/n | ||
jestem ciekaw czy ( | )n→3√a*b*c | |
| 3 |
| a11/n+a21/n+...+ak1/n | ||
( | )n→(a1*...*ak)1/k | |
| k |
| a11/n+...+ak1/n | ||
limn→∞ (1+( | −1))n | |
| k |
| a11/n+...+ak1/n | ||
teraz niech p= | −1→0 | |
| k |
| 1 | a1t−1 | akt−1 | ||||
limn→∞ n*p = | limt→0+ | +...+ | = | |||
| k | t | t |
| ln(a1*...*ak) | ||
= | ||
| k |
| a11/n+...+ak1/n | ||
limn→∞ (1+( | −1))n = (a1*...*ak)1/k | |
| k |
