Limits, Continuity & Differentiability
Limits of Trigonometric Functions
Grade 12
<p>\(\lim_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{(\pi - 2x)^3}\) equals</p>
Step-by-Step Solution
Key Concept: This is a 0/0 indeterminate form requiring Taylor series expansion around x = π/2. Substitute u = π/2 - x to simplify, then expand cot(u) and cos(u) as power series in u.
<p><strong>Step 1:</strong> Verify the indeterminate form. At x = π/2: cot(π/2) = 0, cos(π/2) = 0, denominator = 0. This is 0/0 form.</p><p><strong>Step 2:</strong> Substitute u = π/2 - x, so x = π/2 - u. As x → π/2, u → 0. Note that π - 2x = π - 2(π/2 - u) = 2u, so (π - 2x)³ = 8u³.</p><p><strong>Step 3:</strong> Express in terms of u:</p><ul><li>cot(π/2 - u) = tan(u) = u + u³/3 + 2u⁵/15 + ...</li><li>cos(π/2 - u) = sin(u) = u - u³/6 + u⁵/120 + ...</li></ul><p><strong>Step 4:</strong> Compute the numerator:<br/>cot(π/2 - u) - cos(π/2 - u) = (u + u³/3 + ...) - (u - u³/6 + ...)<br/>= u³/3 + u³/6 + ... = u³(1/3 + 1/6) + ... = u³(1/2) + O(u⁵)</p><p><strong>Step 5:</strong> Calculate the limit:<br/>$$\lim_{u \to 0} \dfrac{u³/2 + O(u⁵)}{8u³} = \dfrac{1/2}{8} = \dfrac{1}{16}$$</p><p>∴ Answer: <strong>1/16</strong> (Option D)</p>
Correct Answer: D