<p>The value of \(\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{4(x-\pi)\cos^2 x}{\pi(\pi-2x)\tan\!\left(x-\dfrac{\pi}{2}\right)}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Substitute u = x - π/2 to transform the limit to u → 0, then use tan(u) ~ u and cos²(π/2 + u) = sin²(u) ~ u² as u → 0 to evaluate the indeterminate form.
<p><strong>Step 1:</strong> Let u = x - π/2, so x = u + π/2. As x → π/2, u → 0.</p><p><strong>Step 2:</strong> Rewrite the limit:</p><p>• Numerator: 4(x - π)cos²(x) = 4(u + π/2 - π)cos²(u + π/2) = 4(u - π/2)sin²(u)</p><p>• Denominator: π(π - 2x)tan(x - π/2) = π(π - 2(u + π/2))tan(u) = π(-2u)tan(u) = -2πu·tan(u)</p><p><strong>Step 3:</strong> The limit becomes:</p><p>$$\lim_{u\to 0} \frac{4(u - π/2)\sin^2(u)}{-2πu\tan(u)}$$</p><p><strong>Step 4:</strong> Use standard limits: sin(u) ~ u and tan(u) ~ u as u → 0:</p><p>$$\lim_{u\to 0} \frac{4(u - π/2)·u^2}{-2πu·u} = \lim_{u\to 0} \frac{4(u - π/2)}{-2π} = \frac{4(-π/2)}{-2π} = \frac{-2π}{-2π} = 1$$</p><p>∴ Answer: <strong>A (which is 1)</strong></p>
Correct Answer: A