oblicz:
| √1+x−1 | ||
lim | ||
| x |
| (√1 + x −1)(√1 + x + 1) | x | 1 | |||
= | = | ||||
| x*(√1 + x + 1) | x*(√1 + x + 1) | √1 + x + 1 |
| 1 | ||
Dla x = 0 lim = | ||
| 2 |
| √1 + x − 1 |
| 1 | ||||||||||
f(x) = | = | = | ||||||||||
| x | x | √1+x +1 |
| 1 | 1 | 1 | ||||
lim | = | = | ||||
| √1 + x + 1 | √ 1 + 0 + 1 | 2 |
| 1+x−1√1+x+1 | ||
Mam pytanie jak Pan to obliczył | ||
| x |
| a2 − b2 | ||
a − b = | ||
| a + b |
| a2 − b2 | ||
a2 − b2 = ( a − b)*(a + b) ⇒ a − b = | ||
| a + b |
| x | ||
A w wyniku nie powinno być | ||
| √1+x+1 |
| x | 1 | |||||||||||||
... = | = | = | |||||||||||||
| x | x*(√1 + x + 1) | √1 + x + 1 |
| a | a | ||
: c = | |||
| b | b*c |