Limits, Continuity & Differentiability
L'Hôpital's Rule
Grade 12
Question:
<p>The value of \(\lim_{x \to \frac{\pi}{4}} \frac{\int_2^{\csc^2 x} g(t)dt}{x^2 - \frac{\pi^2}{16}}\) is:</p>
<p>(a) \(\frac{2}{\pi}g(2)\)</p>
<p>(b) \(-\frac{4}{\pi}g(2)\)</p>
<p>(c) \(-\frac{16}{\pi}g(2)\)</p>
<p>(d) \(-4g(2)\)</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule for $\frac{0}{0}$ indeterminate form, combined with Leibniz rule for differentiating integrals with variable limits.
<p><strong>Step 1:</strong> Apply L'Hôpital's rule. The numerator and denominator both approach 0 as $x \to \frac{\pi}{4}$.</p><p><strong>Step 2:</strong> Differentiate numerator: $\frac{d}{dx}\int_2^{\csc^2 x} g(t)dt = g(\csc^2 x) \cdot (-2\csc x \cot x)$.</p><p><strong>Step 3:</strong> Differentiate denominator: $\frac{d}{dx}\left(x^2 - \frac{\pi^2}{16}\right) = 2x$.</p><p><strong>Step 4:</strong> At $x = \frac{\pi}{4}$: $\csc\frac{\pi}{4} = \sqrt{2}$, $\cot\frac{\pi}{4} = 1$. Numerator becomes $g(2)(-2\sqrt{2})(1) = -2\sqrt{2}g(2)$. Denominator: $2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$. Result: $-\frac{16}{\pi}g(2)$.</p>
Correct Answer: c