oblicz granice
szymi: limx→0 (cos3x)1/x
tj. cos3x do potęgi 1/x
16 cze 17:48
pigor: .., np. tak :
lim
x→0 (cos3x)
1x= [1
∞]= lim
x→0 (1+cos3x−1)
1x=
= lim
x→0 (1+cos3x−1)
1cos3x−1 * 1x * (cos3x−1) =
= e
lim x→0 1x * (cos3x−1) , gdzie
| | cos3x−1 | | 0−1 | | −1 | |
lim x→0 |
| = [ |
| ] = [ |
| ]= − ∞ , zatem |
| | x | | 0 | | 0 | |
| | 1 | | 1 | |
e−∞= |
| = |
| = 0 . ...  |
| | e∞ | | ∞ | |
16 cze 18:06
Mila: albo skorzystaj z logarytmu
cosx>0 w otoczeniu 0
f(x)=(cos(3x))
1x
| | 1 | | ln(cos(3x)) | |
limx→0ln(f(x))=limx→0 |
| ln(cos(3x))=limx→0 |
| =H |
| | x | | x | |
| | −3sin(3x) | |
=limx→0 |
| =0 |
| | cos(3x) | |
Lim{x→0} f(x)=e
0=1
16 cze 21:17