Let f(x)=∫₀ˣ(t+sin (1-eᵗ)) d t, x ∈ ℝ. Then, lim x → 0 f(x)/x³ is equal to

Solution

Maths sir · 4 steps

The integral's derivative is just the integrand, so apply L'Hopital until the 0/0 goes away.

1. Form 0/0. At x = 0 the integral runs from 0 to 0, so f(0) = 0, and x³ = 0 too. By the fundamental theorem of calculus, f'(x) = x + sin(1 - eˣ).
2. First two rounds. The first round gives (x + sin(1-eˣ))/3x², which is still 0/0 since sin 0 = 0. The second round gives (1 - eˣ cos(1-eˣ))/6x, and at x = 0 the top is 1 - 1 = 0, so it is still 0/0.
3. Third round. Differentiate eˣ cos(1-eˣ) by the product rule. The chain rule gives d/dx cos(1-eˣ) = eˣ sin(1-eˣ).

   lim(x→ 0)(-eˣ cos(1-eˣ) - e²ˣ sin(1-eˣ))/6 = (-1 - 0)/6 = -1/6

4. So. The limit is -1/6, option (b). As a check, sin(1-eᵗ) ≈ -t - t²/2, so the integrand is about -t²/2 and f(x) ≈ -x³/6.

The idea to keep. For a limit with ∫₀ˣ g(t) dt on top, L'Hopital turns the integral into g(x). Keep differentiating while it stays 0/0, and watch the signs in every chain rule.

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