If lim _(x → 0) { 1/x⁸( 1 - cos x²/2 - cos x²/4 + cos x²/2 cos x²/4 ) } = 2⁻ᵏ
then the value of k is _______ .

Solution

Maths sir · 3 steps

Group the four terms into two brackets, and each bracket becomes a standard 1 - cos limit.

1. Factor the bracket. Write a = x²/2 and b = x²/4. Then 1 - cos a - cos b + cos a cos b = (1 - cos a)(1 - cos b), which you can check by expanding.
2. Use the standard limit. For small θ, (1 - cos θ)/θ² → 1/2. So 1 - cos a behaves like a²/2 = x⁴/8, and 1 - cos b behaves like b²/2 = x⁴/32.

   lim(x → 0) (1-cos a)(1-cos b)/x⁸ = 1/8 × 1/32 = 1/256 = 2⁻⁸

3. So. The limit is 2⁻⁸, so k = 8.

The common slip: 6. Using 1 - cos θ ≈ θ² and dropping the 1/2 in both factors gives 1/4 × 1/16 = 2⁻⁶, so k = 6, not 8.

The idea to keep. When a 0/0 limit has four terms like 1 - p - q + pq, try factoring into (1-p)(1-q) first. Then each factor is a standard limit you already know.

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