| f(xo + h) − f(xo) | ||
Wzór to limh−>∞ = | ||
| h |
| sin (3x + 3h)−sin (3x) | sin 3x cos 3h + cos 3x sin 3h − sin 3x | |||
lim | = lim | = | ||
| h | h |
| sin 3x (cos 3h − 1) | sin 3h | |||
= lim [ ( | ) + cos 3x | ] = | ||
| h | h |
| cos 3h − 1 | sin 3h | |||
= sin (3x) lim ( | ) + lim 3cos 3x | = | ||
| h | 3h |
| cos2 3h − 1 | ||
= sin (3x) lim ( | ) + 3cos (3x) *1 = | |
| h(cos 3h + 1) |
| sin 3h | sin 3h | |||
= sin (3x) lim ( | * | ) + 3cos (3x) = | ||
| h | cos 3h + 1 |
| 0 | ||
= sin 3x * 1 * | + 3cos(3x) = 3cos(3x) | |
| 2 |
| sin 3h | ||
ostatnia linijka .... lim | = 3, a nie jak napisałem 1 (co i tak nie zmienia wyniku) | |
| h |