oblicz granice lim x-->0
abmaker: hej, pomoze ktos z rozwiazaniem granicy?
| | 1 | | x−2 | |
lim x−−>0 |
| − |
| |
| | x2 | | x3−x | |
4 lis 18:59
abmaker: chociaz ktos mnie nakieruje?:(
4 lis 19:26
jc: | | (x2−1)−x(x−2) | |
= |
| = U{ |
| | x2(x2−1) | |
4 lis 19:40
4 lis 19:42
abmaker: czemu w mianowniku x2(x2−1)?
4 lis 19:44
jc: | 1 | | 2−x | | 1 | | 1 | | 2−x | |
| − |
| = |
| ( |
| − |
| ) |
| x2 | | x−x3 | | x | | x | | 1−x2 | |
| | 1 | | (1−x2) − x(2−x) | | 1 | | 1−2x | |
= |
| * |
| = |
| * |
| |
| | x | | x(1−x2) | | x | | x(1−x2) | |
4 lis 20:05
Mila:
| | x3−x−x2*(x−2) | |
lim x→0 |
| = |
| | x2*(x3−x) | |
| | 1−2x | | 1 | |
=lim x→0 |
| =[ |
| =∞ |
| | x2*(1−x2) | | 0+] | |
4 lis 20:06
abmaker: Mila, czemu w drugiej linijce w mianowniku zostalo x
2(x
2−1)?
(((
4 lis 20:20
abmaker: a dobra, juz wiem
4 lis 20:28
Mila:
4 lis 21:19