<p>Evaluate: \[\lim_{x \to \frac{\pi}{2}} \frac{4(x-\pi)\sin^2\left(\frac{\pi}{2}-x\right)}{-2\pi\left(\frac{\pi}{2}-x\right)^2 \cdot \frac{\tan\left(x-\frac{\pi}{2}\right)}{\left(x-\frac{\pi}{2}\right)}}\]</p>
Step-by-Step Solution
Key Concept: Substitute u = x - π/2 to convert the limit to u → 0, then use the standard limits: sin(u)/u → 1 and tan(u)/u → 1 as u → 0. Recognize that sin²(π/2 - x) = cos²(x) = sin²(u) when u = x - π/2.
<p><strong>Step 1:</strong> Let u = x - π/2, so as x → π/2, u → 0. Then π/2 - x = -u.</p><p><strong>Step 2:</strong> Rewrite sin²(π/2 - x) = sin²(-u) = sin²(u) and substitute into the expression:</p><p>$$\lim_{u \to 0} \frac{4u \cdot \sin^2(u)}{-2\pi \cdot u^2 \cdot \frac{\tan(u)}{u}}$$</p><p><strong>Step 3:</strong> Simplify the denominator: -2πu² · (tan(u)/u) = -2πu · tan(u)</p><p>$$\lim_{u \to 0} \frac{4u \cdot \sin^2(u)}{-2\pi u \cdot \tan(u)}$$</p><p><strong>Step 4:</strong> Cancel u from numerator and denominator (valid as u → 0):</p><p>$$\lim_{u \to 0} \frac{4\sin^2(u)}{-2\pi \tan(u)} = \lim_{u \to 0} \frac{4\sin^2(u)}{-2\pi \cdot \frac{\sin(u)}{\cos(u)}}$$</p><p><strong>Step 5:</strong> Simplify:</p><p>$$\lim_{u \to 0} \frac{4\sin^2(u) \cdot \cos(u)}{-2\pi \sin(u)} = \lim_{u \to 0} \frac{4\sin(u)\cos(u)}{-2\pi}$$</p><p><strong>Step 6:</strong> As u → 0: sin(u) → 0 and cos(u) → 1</p><p>$$= \frac{4 \cdot 0 \cdot 1}{-2\pi} = 0$$</p><p>∴ Answer: <strong>0 (Option A)</strong></p>
Correct Answer: A