| √x+x | ||
Lim x−−>0. | =1. Dlaczego? Wychodzi mi −1... | |
| √x−x |
| √x + x | ( √x + x)2 | |||||||||
f(x) = | = | = | ||||||||
| x*(1 − x) |
| x + 2x√x + x2 | 1 + 2√x + x | |||
= | = | |||
| x*(1 − x) | 1 − x |
| 1 + 0 + 0 | ||
lim f(x) = | = 1 | |
| 1 − 0 |
| a2 − b2 | ||
Do mianownika stosujemy wzór : a − b = |  | |
| a + b |
| √x(1+√x) | ||
Prościej...= lim | = 1 | |
| √x(1−√x) |