<p>\(\lim_{x \to 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that √[(1-cos 2x)/2] simplifies to |sin x| using the half-angle identity, transforming the limit into the standard form lim(sin x/x). The absolute value requires careful evaluation as x→0 from both sides.
<p><strong>Step 1:</strong> Use the identity <strong>1 - cos 2x = 2sin²x</strong></p><p>$$\sqrt{\frac{1}{2}(1-\cos 2x)} = \sqrt{\frac{1}{2} \cdot 2\sin^2 x} = \sqrt{\sin^2 x} = |\sin x|$$</p><p><strong>Step 2:</strong> Rewrite the limit</p><p>$$\lim_{x \to 0} \frac{|\sin x|}{x}$$</p><p><strong>Step 3:</strong> Evaluate left and right limits</p><p>$$\lim_{x \to 0^+} \frac{\sin x}{x} = 1 \quad (\text{since } \sin x > 0 \text{ for } x > 0)$$</p><p>$$\lim_{x \to 0^-} \frac{-\sin x}{x} = -1 \quad (\text{since } \sin x < 0 \text{ for } x < 0)$$</p><p><strong>Step 4:</strong> Since left limit ≠ right limit, the limit does not exist.</p><p>∴ Answer: <strong>D (Limit does not exist)</strong></p>
Correct Answer: D