| x + 1 | ||
lim x→∞ ( | )x2 | |
| 2x + 1 |
| x | ||
lim x→+∞ ( | )x | |
| x + 1 |
| x+1 | 1 | |||
pierwsza to zero, bo | → | , a x2→∞ | ||
| 2x+1 | 2 |
| x | x+1−1 | −1 | ||||
druga: ( | )x=( | )x=(1+ | )x+1 *(x/x+1)→e−1 | |||
| x+1 | x+1 | x+1 |
| x | a | |||
bo | →1 i korzystamy z faktu (1+ | )n→ea gdy n→∞ | ||
| x+1 | n |
| x | 1 | 1 | |||||||||||||
x →∞, ( | )x = | → | |||||||||||||
| x+1 |
| e |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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