<p>Evaluate: <span>\(\lim_{x \to \pi} \dfrac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}\)</span>. If this limit equals <span>\(k\)</span>, find <span>\(k\)</span>.</p>
Step-by-Step Solution
Key Concept: Use Taylor expansion around x = π by substituting u = π - x, then expand cos(π - u) = -cos(u) ≈ -1 + u²/2 near u = 0 to convert the indeterminate form into a rational expression.
<p><strong>Step 1:</strong> Recognize the indeterminate form. At x = π: numerator = √(2 - 1) - 1 = 0 and denominator = 0, giving 0/0 form.</p><p><strong>Step 2:</strong> Substitute u = π - x, so x = π - u and as x → π, u → 0. The limit becomes:<br>$$\lim_{u \to 0} \frac{\sqrt{2 + \cos(π - u)} - 1}{u^2}$$</p><p><strong>Step 3:</strong> Use cos(π - u) = -cos(u). Expand cos(u) ≈ 1 - u²/2 + O(u⁴):<br>$$\cos(π - u) = -\cos(u) ≈ -(1 - \frac{u^2}{2}) = -1 + \frac{u^2}{2}$$</p><p><strong>Step 4:</strong> Substitute into the numerator:<br>$$\sqrt{2 + (-1 + \frac{u^2}{2})} - 1 = \sqrt{1 + \frac{u^2}{2}} - 1$$</p><p><strong>Step 5:</strong> Apply binomial expansion √(1 + x) ≈ 1 + x/2 for small x:<br>$$\sqrt{1 + \frac{u^2}{2}} ≈ 1 + \frac{1}{2} \cdot \frac{u^2}{2} = 1 + \frac{u^2}{4}$$</p><p><strong>Step 6:</strong> Evaluate the limit:<br>$$\lim_{u \to 0} \frac{(1 + \frac{u^2}{4}) - 1}{u^2} = \lim_{u \to 0} \frac{\frac{u^2}{4}}{u^2} = \frac{1}{4}$$</p><p><strong>∴ Answer: k = 1/4 or 0.25</strong></p>
Correct Answer: D