| (x+1)3 | x3−1 | |||
lim [ | − | ] | ||
| x2−1 | x2+1 |
| (x+1)3 | x3−1 | |||
lim x→+∞( | − | ) = | ||
| x2−1 | x2+1 |
| (x+1)2 | (x−1)(x2+x+1) | |||
= lim x→+∞( | − | ) = | ||
| x−1 | x2+1 |
| (x+1)2(x2+1) − (x−1)2(x2+x+1) | ||
= lim x→+∞ | = | |
| (x−1)(x2+1) |
| (x+1)2(x2+1) − (x−1)2(x2+1) − (x−1)2x | ||
= lim x→+∞ | = | |
| (x−1)(x2+1) |
| (x2+1) [(x+1)2 − (x−1)2] − (x−1)2x | ||
= lim x→+∞ | = | |
| (x−1)(x2+1) |
| (x2+1) (x+1−x+1) (x+1+x−1) − (x−1)2x | ||
= lim x→+∞ | = | |
| (x−1)(x2+1) |
| (x2+1)*4x − (x−1)2x | ||
= lim x→+∞ | = | |
| (x−1)(x2+1) |
| 4x | (x−1)x | |||
= lim x→+∞( | − | ) = | ||
| x−1 | x2+1 |
| 4x | (1−1x)x2 | |||
= lim x→+∞( | − | ) = | ||
| x(1−1x) | (1+1x2)x2 |
| 4 | 1−1x | |||
= lim x→+∞( | − | ) = | ||
| 1−1x | 1+1x2 |
| 4 | 1−0 | |||
= | − | = 4−1 = 3 . ... ![]() | ||
| 1−0 | 1+0 |
. Mam tylko pytanie. Co się stało w liczniku z kroku 2. na 3.?
(x+1)2(x2+1) − (x−1)2(x2+x+1)
(x+1)2(x2+1) − (x−1)2(x2+1) − (x−1)2x