<p>The value of \(\displaystyle\lim_{x \to \frac{\pi}{2}} \dfrac{4(x-\pi)\cos^2 x}{\pi(\pi - 2x)\tan\!\left(x - \dfrac{\pi}{2}\right)}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that tan(x - π/2) = -cot(x) and use the substitution u = x - π/2 to convert this into a standard limit form involving sin and cos near zero.
<p><strong>Step 1:</strong> Simplify the trigonometric expression. Note that tan(x - π/2) = -cot(x) = -cos(x)/sin(x).</p><p><strong>Step 2:</strong> Rewrite (π - 2x) = -2(x - π/2). Substitute u = x - π/2, so x = u + π/2, and as x → π/2, we have u → 0.</p><p><strong>Step 3:</strong> The limit becomes:<br>$$\lim_{u \to 0} \frac{4u\cos^2(u + \pi/2)}{\pi(-2u) \cdot (-\cot(u + \pi/2))}$$</p><p><strong>Step 4:</strong> Since cos(u + π/2) = -sin(u) and cot(u + π/2) = -tan(u), we get:<br>$$\lim_{u \to 0} \frac{4u\sin^2(u)}{-2\pi u \cdot (-\tan(u))} = \lim_{u \to 0} \frac{4u\sin^2(u)}{2\pi u\tan(u)}$$</p><p><strong>Step 5:</strong> Simplify:<br>$$\lim_{u \to 0} \frac{4\sin^2(u)}{2\pi\tan(u)} = \lim_{u \to 0} \frac{4\sin^2(u) \cdot \cos(u)}{2\pi\sin(u)} = \lim_{u \to 0} \frac{2\sin(u)\cos(u)}{\pi}$$</p><p><strong>Step 6:</strong> Using sin(u) → 0 and cos(u) → 1:<br>$$= \frac{2 \cdot 0 \cdot 1}{\pi} = 0$$</p><p>∴ Answer: A (which equals <strong>0</strong>)</p>
Correct Answer: A