granica jeszcze raz
diana: Granica : dąży do zera
lim cosx − cos2xsin2x
3 lut 23:01
diana: w liczniku jest cosx − cos2x
3 lut 23:03
Basia:
L(x) = cosx − cos2x → cos0 − cos0 = 1−1 =0
M(x) = sin
2x → sin
20 = 0
2 = 0
| | cosx−cos2x | | L(x) | |
G=limx→0 |
| = limx→0 |
| = |
| | sin2x | | M(x) | |
L'(x) = −sinx + 2sin2x = −sinx +2*2sinx*cosx = sinx(4cosx−1)
M'(x) = 2sinxcosx
| L'(x) | | 4cosx−1 | | 4cos0−1 | | 4*1−1 | | 3 | |
| = |
| → |
| = |
| = |
| |
| M'(x) | | 2cosx | | 2cos0 | | 2*1 | | 2 | |
3 lut 23:14
BlackDragon: Nie powinno być:
| L'(x) | | −cosx + 4sinx*cosx | | −1+4sinx | | −1 | |
| = |
| = |
| = |
| = infinite |
| M'(x) | | 2sinx*cosx | | 2sinx | | 0 | |
4 lut 17:16