Limits, Continuity & Differentiability
Limits of Trigonometric Functions
Grade None
Question:
<p>\(\lim_{x \to 0} \dfrac{\sqrt{1 - \cos 2x}}{\sqrt{2}\, x}\) is</p>
<p>\(\lambda\)</p>
<p>\(-1\)</p>
<p>zero</p>
<p>does not exist</p>
Step-by-Step Solution
Key Concept: Recognize that √(1 - cos 2x) = √2|sin x| using the identity 1 - cos 2x = 2sin²x. The critical step is handling the absolute value carefully as x approaches 0 from both sides.
<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<sub>x→0</sub> [√2|sin x|]/(√2·x) = lim<sub>x→0</sub> |sin x|/x</p><p><strong>Step 3:</strong> Check left and right limits due to absolute value:</p><p>• As x→0⁺: |sin x|/x = sin x/x → 1</p><p>• As x→0⁻: |sin x|/x = -sin x/x = -(sin x/x) → -1</p><p><strong>Step 4:</strong> Since left limit (-1) ≠ right limit (1), the limit does not exist.</p><p>∴ Answer: D (limit does not exist)</p>
Correct Answer: D