| x2−x−6 | ||
lim | ||
| x2+4x+4 |
| 0+ | ||
Gdy podstawiam 2+ to wychodzi mi | ||
| 0 |
| (x+2)(x−3) | x−3 | |||
U = | = | |||
| (x+2)2 | x+2 |
| 1 | ||
lim U = lim(x−3)*lim | = (−2−3)*(+∞) = −5*(+∞) = −∞ | |
| x+2 |
| (x − 3)(x + 2) | x − 3 | ||
= | |||
| (x + 2)2 | x + 2 |
| x − 3 | −5 | |||
limx→ −2+ | → | → −∞ | ||
| x + 2 | 0+ |
. Nie zauważyłem wzoru skróconego mnożenia...