lim _(x → 0) (sin ²( π cos ⁴x ))/x⁴ is equal to :

Solution

Maths sir · 4 steps

The angle tends to π, not 0, so first write sin θ = sin(π - θ) to get a small angle.

1. Make angle small. As x → 0, π cos⁴ x → π, so (sin t)/t → 1 cannot be used yet. Since sin θ = sin(π - θ), rewrite the sine as sin(π(1-cos⁴ x)), whose angle does tend to 0.
2. Factor the angle. 1-cos⁴ x = (1-cos² x)(1+cos² x) = sin² x (1+cos² x). Near 0 this behaves like x² × 2 = 2x².

   lim(x→ 0)(sin(π sin² x (1+cos² x))/(π sin² x (1+cos² x)))² ((π sin² x (1+cos² x))/x²)²

3. Take each limit. The first bracket tends to 1² = 1. In the second, (sin² x)/x² → 1 and 1+cos² x → 2, so it tends to (2π)² = 4π².
4. So. The limit is 1 × 4π² = 4π², which is option (c).

The idea to keep. Before using (sin t)/t → 1, check that the angle really goes to 0. If it goes to π, rewrite it with sin θ = sin(π-θ) first.

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