| 2n+5 | ||
ciąg (an) określo jest wzorem an= | . Zbadaj znak różnicy an+1−an, gdzie n jest | |
| n+2 |
| 5 | 2 | ||
− | = 7 | ||
| x+2 | x+3 |
| 5 | 2 | ||
− | −7= 0 | ||
| x+2 | x+3 |
| 5x+15−2x−4−[7(x+2)(x+3)] | |
= 0 | |
| (x+2)(x+3) |
| 5x+15−2x−4−[(7x+14)(x+3)] | |
= 0 | |
| (x+2)(x+3) |
| 5x+15−2x−4−[7x2+21+14x+42] | |
= 0 | |
| (x+2)(x+3) |
| −7x2−11x−52 | |
= 0 | |
| (x+2)(x+3) |
| sin2n+1x | ||
cosx * cos2x* cos4x*...*cos2nx= | ||
| 2n+1sinx |
|
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Odwzorowanie liniowe A: R2 → R2 dane jest wzorem A | = | . Znaleźć macierz | ||||||||||||||
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tego odwzorowania w bazie [ | , | ] | ||||||||||||||
| lim | x+1 | xx2+1x2 | |||
= | = | = | |||
| x→∞ | x2+4x+1 | x2x2+4xx2+1x2 |
| 1x+1x2 | 0 | |||
= | =0 | |||
| 1+4x+1x2 | 1 |

| 1 | ||
Wykaz z definicji: lim( | )=0 n→∞ | |
| 2n−7 |
| 1 | ||
| | −0|<ε | |
| 2n−7 |
| 1 | ||
| | |<ε | |
| 2n−7 |
| 1 | |
<ε | |
| |2n−7| |
| 1 | ||
|2n−7|> | ||
| ε |
| (n+1)10+(n+2)10+...+(n+100)10 | |
| n10+1010 |