| 3√x−1 | ||
lim | ||
| √x−1 |
| 2√x+10−2√2x+11 | ||
lim | ||
| 3√x+1 |
| 1−cos(5x) | ||
lim | ||
| x2 |
| 1−cos(ax) | ||
3) zastosować wzór: lim | = 0 albo 'ze szpitala' jeżeli już był (ale wątpię) | |
| ax |
| 3√x−1 | (3√x−1)*(3√x2+3√x+1) | ||
= | = | ||
| √x−1 | (√x−1)(3√x2+3√x+1) |
| x−1 | (√x−1)*(√x+1) | |||
= | = | |||
| (√x−1)*(3√x2+3√x+1) | (√x−1)*(3√x2+3√x+1) |
| √x+1 | 2 | |||
limx→1 | = | |||
| (3√x2+3√x+1) | 3 |
| (√x+10−√2x+11)*(√x+10+√2x+11) | |
= | |
| (3√x+1)*(√x+10+√2x+11) |
| x+10−2x−11 | ||
= | = | |
| (3√x+1)*(√x+10+√2x+11) |
| −(x+1) | ||
= | = | |
| (3√x+1)*(√x+10+√2x+11) |
| −(3√x+1)*(3√x2−3√x+1) | ||
= | = | |
| (3√x+1)*(√x+10+√2x+11) |
| −(3√x2−3√x+1) | ||
= | ||
| (√x+10+√2x+11) |
| −(3√x2−3√x+1) | 1 | |||
limx→−1 | =− | |||
| (√x+10+√2x+11) | 2 |