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 |