Limits, Continuity & Differentiability
Continuity
Grade None
Question:
<p>If the function <span>\[ f(x) = \begin{cases} \dfrac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases} \]</span> is continuous at <span>\( x = \pi \)</span>, then <span>\( k \)</span> equals</p>
<p>0</p>
<p>\( \dfrac{1}{2} \)</p>
<p>2</p>
<p>\( \dfrac{1}{4} \)</p>
Step-by-Step Solution
Key Concept: For continuity at x = π, we need lim(x→π) f(x) = f(π) = k. Use Taylor expansion of √(2 + cos x) around x = π where cos π = -1, so 2 + cos x ≈ 1 + (x-π)²/2 near π.
<p><strong>Step 1:</strong> For continuity at x = π, we need: k = lim(x→π) [√(2 + cos x) - 1]/(π - x)²</p><p><strong>Step 2:</strong> Near x = π, use Taylor expansion: cos x = cos π + cos'(π)(x-π) + (cos''(π)/2)(x-π)² + ... = -1 + 0·(x-π) - (1/2)(x-π)² + O((x-π)⁴)</p><p><strong>Step 3:</strong> Therefore: 2 + cos x = 1 - (1/2)(x-π)² + O((x-π)⁴)</p><p><strong>Step 4:</strong> Apply √(1 + u) ≈ 1 + u/2 - u²/8 + ... with u = -(x-π)²/2:</p><p>√(2 + cos x) = √[1 - (x-π)²/2] ≈ 1 - (1/4)(x-π)² + O((x-π)⁴)</p><p><strong>Step 5:</strong> Substitute into the limit:</p><p>k = lim(x→π) [1 - (1/4)(x-π)² - 1]/(x-π)² = lim(x→π) [-(1/4)(x-π)²]/(x-π)² = -1/4</p><p>∴ Answer: k = <strong>-1/4</strong> (Option D)</p>
Correct Answer: D