Limits, Continuity & Differentiability Question - None

Question

Evaluate: $\lim_{x \to 0^+} \dfrac{\displaystyle\int_0^{\tan^{-1} x} (\sin t^2)\, dt}{\dfrac{-x^3}{2}}$

Accepted Answer

Step 1: Check the form. As x→0⁺, both numerator ∫₀^(tan⁻¹x) sin(t²)dt → 0 and denominator -x³/2 → 0, giving 0/0 form.

Step 2: Apply L'Hôpital's rule. Differentiate numerator using Leibniz integral rule: d/dx[∫₀^(tan⁻¹x) sin(t²)dt] = sin((tan⁻¹x)²) · d/dx(tan⁻¹x) = sin((tan⁻¹x)²) · 1/(1+x²)

Differentiate denominator: d/dx(-x³/2) = -3x²/2

Step 3: New limit is: lim(x→0⁺) [sin((tan⁻¹x)²)·1/(1+x²)] / (-3x²/2). Still 0/0 form.

Step 4: Apply L'Hôpital's rule again. For near x=0: tan⁻¹x ≈ x - x³/3 + ..., so (tan⁻¹x)² ≈ x² - 2x⁴/3 + ... ≈ x²

Thus sin((tan⁻¹x)²) ≈ sin(x²) ≈ x² (using sin(u) ≈ u for small u)

Step 5: The limit becomes: lim(x→0⁺) [x² · 1/(1+x²)] / (-3x²/2) = lim(x→0⁺) [x²/(1+x²)] / (-3x²/2) = lim(x→0⁺) [1/(1+x²)] / (-3/2) = 1/(-3/2) = -2/3

∴ Answer: B

Question:

Step-by-Step Solution

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