<p>Find \(\displaystyle\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2}\, x}\).</p>
Step-by-Step Solution
Key Concept: Recognize that √(1 - cos 2x) = √(2sin²x) = √2|sin x|, then determine the sign of sin x as x approaches 0 from both sides to resolve the absolute value.
<p><strong>Step 1:</strong> Use the identity 1 - cos 2x = 2sin²x.</p><p>√(1 - cos 2x) = √(2sin²x) = √2|sin x|</p><p><strong>Step 2:</strong> Substitute into the limit:</p><p>$$\lim_{x \to 0} \frac{\sqrt{2}|\sin x|}{\sqrt{2}x} = \lim_{x \to 0} \frac{|\sin x|}{x}$$</p><p><strong>Step 3:</strong> Evaluate one-sided limits:</p><p>For x → 0⁺: |sin x| = sin x, so $$\lim_{x \to 0^+} \frac{\sin x}{x} = 1$$</p><p>For x → 0⁻: |sin x| = -sin x, so $$\lim_{x \to 0^-} \frac{-\sin x}{x} = -1$$</p><p><strong>Step 4:</strong> Since left and right limits are different, the limit does not exist (or is undefined).</p><p>∴ Answer: D (Limit does not exist)</p>
Correct Answer: D