Limits, Continuity & Differentiability
L'Hôpital's Rule / Limits involving integrals
Grade 12
Question:
<p>The value of \(\lim_{x \to 0^+} \dfrac{\displaystyle\int_0^{\arctan x} \sin t^2\, dt}{x\cos x - x}\) is equal to:</p>
<p>(a) \(\dfrac{1}{3}\)</p>
<p>(b) \(\dfrac{-1}{3}\)</p>
<p>(c) \(\dfrac{2}{3}\)</p>
<p>(d) \(\dfrac{-2}{3}\)</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule after recognizing the 0/0 indeterminate form, then apply the fundamental theorem of calculus to differentiate the integral in the numerator.
<p><strong>Step 1:</strong> Check the form as x→0⁺. Numerator: ∫₀^(arctan 0) sin(t²)dt = 0. Denominator: 0·(cos 0 - 1) = 0·0 = 0. This is 0/0 indeterminate form.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule. Differentiate numerator using Fundamental Theorem of Calculus with chain rule:<br/>d/dx[∫₀^(arctan x) sin(t²)dt] = sin((arctan x)²)·d/dx(arctan x) = sin((arctan x)²)·1/(1+x²)</p><p><strong>Step 3:</strong> Differentiate denominator:<br/>d/dx[x cos x - x] = cos x - x sin x - 1</p><p><strong>Step 4:</strong> New limit: lim(x→0⁺) [sin((arctan x)²)·1/(1+x²)] / [cos x - x sin x - 1]</p><p><strong>Step 5:</strong> As x→0⁺: arctan x → 0, so sin((arctan x)²) → 0. Also cos 0 - 0 - 1 = -1. Numerator → 0·1/1 = 0, denominator → -1. This gives 0/(-1) = 0.</p><p><strong>Step 6:</strong> Apply L'Hôpital's rule again or use Taylor expansion: sin((arctan x)²) ≈ (arctan x)² ≈ x² for small x. So numerator ≈ x²/(1+x²). For denominator, cos x - x sin x - 1 ≈ -x²/2 - x² = -3x²/2. Thus limit ≈ (x²) / (-3x²/2) = -2/3.</p><p>∴ Answer: B</p>
Correct Answer: B