lim x → 0⁺ (tan (5(x)^(1/3)) log ₑ(1+3 x²))/(tan ⁻¹ 3 √x)²(e^(5(x)^(4/3))-1) is equal to

Solution

Maths sir · 3 steps

Swap each of the four factors for its small-angle form, then compare the powers of x.

1. Standard limits. As x → 0⁺, every inner quantity goes to 0. For a small u, each of tan u, tan⁻¹ u, ln(1+u) and eᵘ-1 behaves like u itself, because each ratio such as (tan u)/u tends to 1.

   Factor	Small quantity u	Behaves like
   tan(5x^(1/3))	5x^(1/3)	5x^(1/3)
   ln(1+3x²)	3x²	3x²
   (tan⁻¹ 3√x)²	3√x, then squared	(3√x)² = 9x
   e^(5x^(4/3))-1	5x^(4/3)	5x^(4/3)

2. Powers of x. Numerator: 5x^(1/3) × 3x² = 15x^(7/3). Denominator: 9x × 5x^(4/3) = 45x^(7/3). The powers match, so x^(7/3) cancels.

   lim(x → 0⁺) (15 x^(7/3))/(45 x^(7/3)) = 15/45 = 1/3

3. So. The limit is 1/3, which is option (c).

The idea to keep. For a product of small-angle factors, replace each tan u, tan⁻¹ u, ln(1+u) and eᵘ-1 by u. Then check that the powers of x cancel and read off the constant.

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