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If lim(x→0) [cos(x) + a³ * sin(b⁶ * x)]^(1/x) = e⁵¹², find the value of a * b².

1. -512
2. 512
3. 8
4. 8 sqrt(8)

Correct answer: 8

Solution

Let f(x) = cos x + a³ sin(b⁶ x) - 1. For small x: f(x) ≈ -x²/2 + a³ b⁶ x. So f(x)/x → a³ b⁶. Thus the limit = e^(a³ b⁶) = e⁵¹², giving a³ b⁶ = 512. Therefore a*b² = (a³ b⁶)^(1/3) = 512^(1/3) = 8.

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