Limits, Continuity & Differentiability
L'Hôpital's Rule / Limits involving integrals
Grade 12
Question:
<p>Let \(f: \mathbb{R} \to \mathbb{R}\) be a differentiable function having \(f(2) = 6\), \(f'(2) = \left(\dfrac{1}{48}\right)\). Then \(\lim_{x \to 2} \int_6^{f(x)} \dfrac{4t^3}{x-2}\, dt\) equals</p>
<p>24</p>
<p>36</p>
<p>12</p>
<p>18</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule on the limit by recognizing the integral creates a 0/0 indeterminate form. Differentiate numerator using Leibniz rule: d/dx∫₆^(f(x)) 4t³ dt = 4[f(x)]³·f'(x).
<p><strong>Step 1: Verify indeterminate form</strong></p><p>At x = 2: numerator = ∫₆^(f(2)) 4t³ dt = ∫₆⁶ 4t³ dt = 0 and denominator = 2 - 2 = 0. This is 0/0 form. ✓</p><p><strong>Step 2: Apply L'Hôpital's rule</strong></p><p>$$\lim_{x \to 2} \frac{\int_6^{f(x)} 4t^3\, dt}{x-2} = \lim_{x \to 2} \frac{\frac{d}{dx}\int_6^{f(x)} 4t^3\, dt}{\frac{d}{dx}(x-2)}$$</p><p><strong>Step 3: Differentiate using Leibniz rule (chain rule)</strong></p><p>By Leibniz integral rule:</p><p>$$\frac{d}{dx}\int_6^{f(x)} 4t^3\, dt = 4[f(x)]^3 \cdot f'(x)$$</p><p>And $\frac{d}{dx}(x-2) = 1$</p><p><strong>Step 4: Evaluate the limit</strong></p><p>$$\lim_{x \to 2} 4[f(x)]^3 \cdot f'(x) = 4[f(2)]^3 \cdot f'(2)$$</p><p>$$= 4 \cdot (6)^3 \cdot \frac{1}{48} = 4 \cdot 216 \cdot \frac{1}{48} = \frac{864}{48} = 18$$</p><p><strong>∴ Answer: D (or 18)</strong></p>
Correct Answer: D