| 4x3 | 4 | 2 | ||||
= lim | = lim | = | ||||
| 5x4 | 5x | 5 |
| 0 | ||
Jest funkcja dostajesz [ | ] wiec mozesz | |
| 0 |
| x4−16 | (x−2)(x3+2x2+4x+8) | ||
= | = | ||
| x5−32 | (x−2)(x4+2x3+4x2+8x+16) |
| x3+2x2+4x+8 | ||
= | ||
| x4+2x3+4x2+8x+16 |
| x4−16 | 32 | 2 | ||||
więc lim x→2 | wynosi | = | ||||
| x5−32 | 80 | 5 |
| 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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