Granica
A: Granica funkcji:
lim x→0 [cos(sinx)](1/x2)
18 lis 17:57
grzest:
lim
x→0 [cos(sinx)]
1/x2 = lim
x→0 e
1/x2ln[cos(sin x)].
Dalej rozpatruję tylko wykładnik liczby e:
| | ln[cos(sin x)] | | −sin(sin x)cos x | |
limx→0 |
| = H = limx→0 |
| = H = |
| | x2 | | 2xcos(sin x) | |
| | −cos(sin x)cos2 x | | 1 | |
limx→0 |
| =− |
| . |
| | 2cos(sin x)−2xsin(sin x)cos x | | 2 | |
Granica
| | 1 | |
limx→0[cos(sinx)]1/x2=e−1/2= |
| . |
| | √e | |
18 lis 19:27