<p>Evaluate:
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x(1 - \sin x)}{-8\left(x - \dfrac{\pi}{2}\right)^3}\]</p>
Step-by-Step Solution
Key Concept: Since this limit has the indeterminate form 0/0 at x = π/2, we use Taylor series expansions around x = π/2. Substitute h = x - π/2 and express sin x and cot x in terms of h to find the leading behavior.
<p><strong>Step 1: Recognize the indeterminate form</strong><br>As x → π/2: sin x → 1, so (1 - sin x) → 0; cot x → 0; and the denominator → 0. This is a 0/0 indeterminate form.</p><p><strong>Step 2: Substitute h = x - π/2</strong><br>Let h = x - π/2, so x = π/2 + h and h → 0 as x → π/2.<br>The limit becomes:<br>$$\lim_{h \to 0} \frac{\cot(\pi/2 + h)(1 - \sin(\pi/2 + h))}{-8h^3}$$</p><p><strong>Step 3: Apply trigonometric identities</strong><br>• sin(π/2 + h) = cos h<br>• cot(π/2 + h) = -tan h<br><br>Therefore:<br>$$\lim_{h \to 0} \frac{-\tan h(1 - \cos h)}{-8h^3} = \lim_{h \to 0} \frac{\tan h(1 - \cos h)}{8h^3}$$</p><p><strong>Step 4: Use Taylor series expansions</strong><br>• tan h = h + h³/3 + O(h⁵)<br>• cos h = 1 - h²/2 + h⁴/24 + O(h⁶)<br>• 1 - cos h = h²/2 - h⁴/24 + O(h⁶)</p><p><strong>Step 5: Multiply the series</strong><br>$$\tan h \cdot (1 - \cos h) = \left(h + \frac{h^3}{3} + O(h^5)\right) \left(\frac{h^2}{2} - \frac{h^4}{24} + O(h^6)\right)$$<br><br>The leading terms up to h⁵:<br>$$= h \cdot \frac{h^2}{2} + \frac{h^3}{3} \cdot \frac{h^2}{2} + O(h^6) = \frac{h^3}{2} + \frac{h^5}{6} + O(h^6)$$</p><p><strong>Step 6: Evaluate the limit</strong><br>$$\lim_{h \to 0} \frac{\tan h(1 - \cos h)}{8h^3} = \lim_{h \to 0} \frac{\frac{h^3}{2} + \frac{h^5}{6} + O(h^6)}{8h^3}$$<br><br>$$= \lim_{h \to 0} \left(\frac{1}{16} + \frac{h^2}{48} + O(h^3)\right) = \frac{1}{16}$$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A