| 3x2+2x−1 | ||
limx→∞ ( | )2n2 =
| |
| 3x2−x+3 |
| 3x−4 | ||
limx→∞ (1+ | )2n2 =
| |
| 3x2−x+3 |
| 3x−4 | ||
limx→∞ ((1+ | )3x2−x+33x−4 ) 6n2−8n23x2−x+3 =
| |
| 3x2−x+3 |
| 6n3 − 8n2 | −8 | |||
limx→∞ ey gdzie y = limx→∞ y= ( | ) = | |||
| 3x2−x+3 | 3 |
| 6x3−8x2 | ||
= limx→∞ ey, gdzie y = | ||
| 3x2−x+3 |
| 6x−8 | ||
= e[∞], ponieważ y = limx→∞ | = ∞ | |
| 3 |
dzieki