oblicz granice korzystając ze wzorów
Tomek-granice: 1)
lim (1+cosx)(4/cosx) (1+cosx jest podniesione do potęgi 4/cosx)
x→π/2
2)
lim (1−cos2x)/5x2
x→0
3)
lim sin3x/tg4x
x→0
4)
lim (4−π)/(sin(π/4)x)
x→8
3 gru 14:25
piotr: 1)
=e
ln(1+cosx)4/cosx
| 4ln(1+cosx) | | 4(ln(1+cosx))' | |
limx→π/2 |
| = limx→π/2 |
| = |
| cosx | | (cosx)' | |
| −4sinx/(1+cosx) | |
=limx→π/2 |
| = 4 |
| −sinx | |
ostatecznie:
lim
x→π/2(1+cosx)
4/cosx = e
4
3 gru 14:40
Jerzy:
| sin3x | | tg4x | | 3x | | 3 | |
3) = limx→0 |
| * |
| * |
| = |
| |
| 3x | | 4x | | 4x | | 4 | |
3 gru 14:43
Jerzy:
| 2sin2x | | 4cos2x | | 4 | |
2) = limx→0 |
| = limx→0 |
| = |
| |
| 5x | | 5 | | 5 | |
3 gru 14:45
piotr: | 1−cos2x | | (1−cos2x)' | |
limx→0 |
| = limx→0 |
| = |
| 5x2 | | (5x2)' | |
| sin2x | | 2 | | sin2x | | 2 | |
=limx→0 |
| = |
| limx→0 |
| = |
| |
| 5x | | 5 | | 2x | | 5 | |
3 gru 14:46
Jerzy:
@piotr .... popraw
3 gru 14:51
Jerzy:
(−cos2x)' = 2sin(2x) oraz (2sin(2x))' = 4cos(2x)
3 gru 14:53
piotr: Jerzy, a ile to jest (5x
2)'
3 gru 15:22