lim x → 0(((1-cos ²(3 x)/(cos ³(4 x)))((sin ³(4 x))/(log ₑ(2 x+1))⁵)) is equal to _____________.

Solution

Maths sir · 4 steps

Turn every trig and log factor into its small-x power of x, and the x⁵ on top cancels the x⁵ below.

1. Tidy the numerator. First write 1-cos²(3x) = sin²(3x). The limit is now (sin²(3x) sin³(4x))/(cos³(4x) (ln(1+2x))⁵), which is of the form 0/0.
2. Standard limits. As x → 0, sin(3x) ≈ 3x, sin(4x) ≈ 4x and ln(1+2x) ≈ 2x. Also cos(4x) → 1, so cos³(4x) just becomes 1.

   lim(x → 0)((3x)² (4x)³)/(1 × (2x)⁵) = (9 × 64 x⁵)/(32 x⁵)

3. Cancel powers. The powers of x match: x² × x³ = x⁵ above and x⁵ below, so they cancel. What remains is 576/32 = 18.
4. So. The limit equals 18, which is option (b).

The idea to keep. For a 0/0 product, replace sin ax by ax, ln(1+ax) by ax and 1-cos ax by a²x²/2. Then check that the powers of x balance before you read off the number.

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