If lim _(t → 0)(∫₀¹(3 x+5)ᵗ d x)^(1/t)=α/(5 e)(8/5)^(2/3), then α is equal to ________________.

Solution

Maths sir · 5 steps

This is a 1 to the power infinity limit, so the answer is e raised to the integral of ln(3x+5).

1. Spot the form. Call the integral I(t). As t → 0, (3x+5)ᵗ → 1, so I(t) → ∫₀¹ 1 dx = 1, while the power 1/t blows up. This is a 1^∞ form, so the limit is L = e^(lim (I(t)-1)/t).
2. Bring the limit inside. Write (I(t)-1)/t = ∫₀¹ ((3x+5)ᵗ - 1)/t dx. Since (aᵗ-1)/t → ln a, the integrand becomes ln(3x+5). Put u = 3x+5, so dx = du/3 and u runs from 5 to 8.

   ln L = ∫₀¹ ln(3x+5) dx = 1/3[u ln u - u]₅⁸ = 8/3 ln 8 - 5/3 ln 5 - 1

3. Undo the log. L = 8^(8/3) × 5^(-5/3) × e⁻¹. Since 8^(8/3) = 2⁸ = 256, this is L = 256/(5^(5/3) e).
4. Match the form. The given right side is α/5e × 8^(2/3)/5^(2/3) = 4α/(5^(5/3) e), because 8^(2/3) = 4.
5. So. 4α = 256, which gives α = 64.

The common slip: 256. Forgetting that (8/5)^(2/3) on the right also carries a factor 8^(2/3) = 4 gives 256, not 64.

The idea to keep. For a 1^∞ limit fᵍ, the answer is eˡⁱᵐᵍ⁽ᶠ⁻¹⁾. And (aᵗ-1)/t → ln a turns a power inside an integral into a logarithm.

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