<p>\(\lim_{x \to 0} \dfrac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}\) equals __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying by the conjugate, then use the standard limit sin(x)/x → 1 and Taylor series for cos(x) near x = 0.
<p><strong>Step 1:</strong> Rationalize the denominator by multiplying by the conjugate:</p><p>$$\lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}} \cdot \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$$</p><p><strong>Step 2:</strong> Simplify the denominator using (a-b)(a+b) = a² - b²:</p><p>$$= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)} = \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$$</p><p><strong>Step 3:</strong> Use the identity 1 - cos(x) = 2sin²(x/2), so:</p><p>$$= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{2\sin^2(x/2)}$$</p><p><strong>Step 4:</strong> Since sin(x) = 2sin(x/2)cos(x/2), we have sin²(x) = 4sin²(x/2)cos²(x/2):</p><p>$$= \lim_{x \to 0} \frac{4\sin^2(x/2)\cos^2(x/2)(\sqrt{2} + \sqrt{1 + \cos x})}{2\sin^2(x/2)} = \lim_{x \to 0} 2\cos^2(x/2)(\sqrt{2} + \sqrt{1 + \cos x})$$</p><p><strong>Step 5:</strong> Substitute x → 0: cos(0) = 1, cos(0) = 1:</p><p>$$= 2(1)^2(\sqrt{2} + \sqrt{1 + 1}) = 2(\sqrt{2} + \sqrt{2}) = 2 \cdot 2\sqrt{2} = 4\sqrt{2}$$</p><p>$$4\sqrt{2} = 4 \times 1.41421... = 5.6569$$</p><p><strong>∴ Answer: 5.6569</strong></p>
Correct Answer: 5