| √x−8 | |
| 3√x−4 |
| √1+2x−3 | |
| √x−2 |
| √x−8 | 1/2 x−1/2 | 3 | ||||
limx→64 | = limx→64 | = | 641/6 = 3 | |||
| 3√x−4 | 1/3 x−2/3 | 2 |
| t3−8 | 3t2 | |||
limt→2 | = limt→2 | = 3 | ||
| t2−2 | 2t |
| √1+2x−3 | (√1+2x−3)' | |||
limx→4 | = limx→4 | = | ||
| √x−2 | (√x−2)' |
| 1/√1+2x | 1/√1+2*4 | 8 | ||||
limx→4 | = | = | ||||
| 1/(2√x) | 1/(2√4) | 3 |
| √64−8 | 0 | |||
1. Jak podstawimy 64 za iks to otrzymujemy: | = | |||
| 3√64−4 | 0 |
| a3−b3 | ||
korzystamy ze wzoru a3−b3 = (a−b)(a2+ab+b2) czyli a−b= | ||
| a2+ab+b2 |
| √x+8 | ||
oraz mnozymy razy "sprzezenie" mianownika, czyli razy | ||
| √x+8 |
| √x−8 | 3√x2+43√x+16 | √x+8 | |||
* | * | = | |||
| 3√x−4 | 3√x2+43√x+16 | √x+8 |
| (x−64)(3√x2+43√x+16) | (3√x2+43√x+16) | |||
= | = | |||
| (x−64)(√x+8) | (√x+8) |
| (3√x2+43√x+16) | 3√642+43√64+16 | 16+16+16 | ||||
lim | = | = | = | |||
| (√x+8) | √64+8 | 8+8 |
| 48 | ||
= | =3 | |
| 16 |