<p>\(\lim_{x \to 2} \left( \dfrac{\sqrt{1 - \cos\{2(x-2)\}}}{x-2} \right)\)</p>
<p>equals \(\sqrt{2}\)</p>
<p>equals \(-\sqrt{2}\)</p>
<p>equals \(\dfrac{1}{\sqrt{2}}\)</p>
<p>does not exist</p>
Step-by-Step Solution
Key Concept: Recognize that {2(x-2)} denotes the fractional part function, which behaves like 2(x-2) near x=2 where 0 < 2(x-2) < 1. Use the half-angle identity: √(1-cos(θ)) = √2|sin(θ/2)| to simplify the numerator.
<p><strong>Step 1:</strong> For x near 2, let u = x - 2 where u → 0⁺. The fractional part {2u} = 2u when 0 < 2u < 1, i.e., 0 < u < 1/2.</p><p><strong>Step 2:</strong> Use identity √(1 - cos(θ)) = √2|sin(θ/2)|. Here θ = 2u, so:</p><p>√(1 - cos(2u)) = √2|sin(u)|</p><p><strong>Step 3:</strong> Since u → 0⁺, we have sin(u) > 0, so |sin(u)| = sin(u):</p><p>$$\lim_{u \to 0^+} \frac{\sqrt{2}\sin(u)}{u} = \sqrt{2} \cdot \lim_{u \to 0^+} \frac{\sin(u)}{u} = \sqrt{2} \cdot 1 = \sqrt{2}$$</p><p><strong>Step 4:</strong> The limit exists only from the right (u → 0⁺) due to the fractional part function restriction.</p><p>∴ Answer: D (which should be √2)</p>
Correct Answer: D