oblicz granice funkcji
mat: limx−>1 (1/(1−x)−1/(1−x3 ))
7 lis 00:07
irena_1:
| | 1 | | 1+x+x2−1 | | x2+x | |
( |
| −U{1−x3})= |
| = |
| |
| | 1−x | | 1−x3 | | 1−x3 | |
| | x2+x | | 2 | |
limx→1− |
| =[ |
| ]=−∞ |
| | 1−x3 | | 0− | |
| | x2+x | | 2 | |
limx→1+ |
| =[ |
| ]=∞ |
| | 1−x3 | | 0+ | |
7 lis 08:16
irena_1: W pierwszej linijce powinno być
| 1 | | 1 | | 1+x+x2−1 | | x2+x | |
| − |
| = |
| = |
| |
| 1−x | | 1−x3 | | 1−x3 | | 1−x3 | |
Korzystałam z:
1−x
3=(1−x)(1+x+x
2)
7 lis 08:17