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Find the value of lim(x -> 0) (1 - cos⁵(x) * cos(2x) * cos³(x)) / x².

1. 6
2. 22
3. 11
4. 16

Correct answer: 11

Solution

Let P = cos⁵(x)*cos(2x)*cos³(x) = cos⁸(x)*cos(2x). Taking log: ln P = 8*ln(cos x) + ln(cos 2x). Near x=0: ln(cos kx) ~ -(kx)²/2. So ln P ~ -8*(x²/2) - (2x)²/2 = -4x² - 2x² = -6x². Wait, that gives P ~ e^(-6x²) ~ 1 - 6x², so (1-P)/x² ~ 6. But let me re-read: cos⁵(x)*cos(2x)*cos³(x) = cos⁵(x)*cos³(x)*cos(2x). This is ambiguous notation. If it means cos⁵(x) * cos²(x) * cos³(x) = cos¹⁰(x)... or cos(5x)*cos(2x)*cos(3x)?

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