| √1+x−1 | ||
a) lim x−>0 | ||
| x |
| 0 | ||
podstawiajac za x wartosc 0 to mamy | zatem symbol nieoznaczony. | |
| 0 |
| √1+x−1 | √1+x+1 | 1+x−1 | ||||
lim | * | = lim | = | |||
| x | √1+x+1 | x(√1+x+1) |
| x | 1 | |||
lim | = lim | = ... | ||
| x(√1+x+1) | √1+x+1 |
| 1 | 1 | |||
... = | = | |||
| √1+1 | 2 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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