| a | 1 | 1 | ||||
Wyznacz parametr a, jeśli wiadomo, że: lim ( | − | )= | ||||
| 1−x | x2−1 | 4 |
| −a(x+1) − 1 | −ax −a −1 | −a | ||||
= lim | = lim | = [H] = | ||||
| x2−1 | x2−1 | 2x |
| −a | 1 | 1 | |||
= | ⇔ a = = | ||||
| 2x | 4 | 2 |
| 1 | ||
a = − | ||
| 2 |
| a | 1 | a | 1 | a | 1 | a(1+x)+1 | |||||||
− | = | + | = | + | = | ||||||||
| 1−x | x2−1 | 1−x | 1−x2 | 1−x | (1−x)(1+x) | (1−x)(1+x) |
| a+ax+1 | ax−a+1+2a | −a(1−x)+1+2a | ||||
= | = | = | ||||
| (1−x)(1+x) | (1−x)(1+x) | (1−x)(1+x) |
| −a | 1+2a | |||
= | + | |||
| (1+x) | (1−x)(1+x) |
| 1 | ||
i widać, ze jeżeli a=− | to 1+2a=0 | |
| 2 |
| −a | 1 | |||
oraz limx→1 | = | |||
| (1+x) | 4 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
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