<p>Given <br>\(\lim_{x \to 0} \dfrac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}\)</p><p>Find the value of the limit.</p>
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying by the conjugate to eliminate the square root, then apply standard limit formulas like lim(x→0) sin(x)/x = 1 and Taylor expansions for trigonometric functions.
<p><strong>Step 1: Identify the indeterminate form</strong><br>As x → 0: sin²x → 0 and √2 - √(1+cos x) → √2 - √2 = 0, giving 0/0 form.</p><p><strong>Step 2: Rationalize the denominator</strong><br>Multiply numerator and denominator by the conjugate √2 + √(1+cos x):<br>$$\lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}} \cdot \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$$</p><p><strong>Step 3: Simplify the denominator using difference of squares</strong><br>$$= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)}$$<br>$$= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$$</p><p><strong>Step 4: Apply Taylor expansion for cos x</strong><br>Use cos x = 1 - x²/2 + x⁴/24 - ...<br>Then: sin²x = 1 - cos(2x))/2 = 1 - (1 - 2x² + ...)/2 = x² - x⁴/12 + ...<br>And: 1 - cos x = x²/2 - x⁴/24 + ...</p><p><strong>Step 5: Find the limit of the product</strong><br>$$\lim_{x \to 0} \frac{\sin^2 x}{1 - \cos x} = \lim_{x \to 0} \frac{x^2 - x^4/12 + ...}{x^2/2 - x^4/24 + ...} = \lim_{x \to 0} \frac{x^2(1 - x^2/12 + ...)}{x^2(1/2 - x^2/24 + ...)} = \frac{1}{1/2} = 2$$</p><p><strong>Step 6: Evaluate the conjugate term</strong><br>$$\lim_{x \to 0} (\sqrt{2} + \sqrt{1 + \cos x}) = \sqrt{2} + \sqrt{1 + 1} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$</p><p><strong>Step 7: Combine the results</strong><br>$$\lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x} = 2 \times 2\sqrt{2} = 4\sqrt{2} = 4 \times 1.4142... = 5.6568...$$</p><p><strong>∴ Answer: 5.6568</strong></p>
Correct Answer: 5.6568