| sin2x−x2 cos2x | ||
Oblicz limx−>0 | ||
| x2 sin2x |
| sin2x−x2cos2x | (1+x2)sin(2x)−2xcos2x | |||
limx→0 | = H = limx→0 | = H | ||
| x4 | 4x3 |
| 4xsin(2x)+2(1+x2)cos(2x)−2cos2x | ||
= limx→0 | = H = | |
| 12x2 |
| (2−4x2)sin(2x)+12xcos(2x) | 1 | 1 | 2 | |||||
= limx→0 | = | + | = | |||||
| 24x | 6 | 2 | 3 |
| sin2x−x2cos2x | sin2x−x2cos2x | ||
zamiast | |||
| x4 | x2sin2x |
| sin2x−x2cos2x | sin2x−x2cos2x | x2 | 2 | |||
= | → | |||||
| x2sin2x | x4 | sin2x | 3 |
| sinx | ||
ponieważ limx→0 | = 1 | |
| x |