The value of lim x → 0 2((1-cos x √(cos 2 x) ∛(cos 3 x) … … . ¹⁰√(cos 10 x))/x²) is __________.

Solution

Maths sir · 4 steps

Near zero, each (cos kx)^(1/k) is about 1 - kx²/2, so the product turns into a sum.

1. Approximate each factor. For small x, cos kx ≈ 1 - k²x²/2. Taking the k-th root gives (cos kx)^(1/k) ≈ 1 - 1/k·k²x²/2 = 1 - kx²/2.
2. Multiply the factors. When you multiply ten factors of the form 1 - (small), keep only the x² terms. Products of two or more of them are of order x⁴ and vanish in the limit, so the x² terms just add up.

   Πₖ₌₁¹⁰(cos kx)^(1/k) ≈ 1 - x²/2(1+2+…+10) = 1 - 55x²/2

3. Take the limit. The numerator is 1 - (1 - 55x²/2) = 55x²/2. Dividing by x² gives 55/2, and the factor 2 outside makes it 2 × 55/2.
4. So. The limit equals 55.

The common slip: 385. Ignoring the roots and using cos kx ≈ 1 - k²x²/2 for every factor gives Σ k² = 385, not 55. The k-th root cuts k² down to k.

The idea to keep. For 1 - Π fₖ(x) over x² near 0, expand each factor as 1 - aₖ x². The limit is just Σ aₖ, because cross terms are of order x⁴.

Answer from the NTA final answer key. Solution by PrepWiser.