<p>\(\lim_{x \to 0} \dfrac{x + 2\sin x}{\sqrt{x^2 + 2\sin x + 1} - \sqrt{\sin^2 x - x + 1}}\) is __________.</p>
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying by the conjugate, then use Taylor expansions (sin x ≈ x - x³/6) to identify the leading terms and simplify the limit.
<p><strong>Step 1:</strong> Multiply numerator and denominator by the conjugate:</p><p>$$\lim_{x \to 0} \frac{(x + 2\sin x)[\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1}]}{(x^2 + 2\sin x + 1) - (\sin^2 x - x + 1)}$$</p><p><strong>Step 2:</strong> Simplify the denominator:</p><p>$$(x^2 + 2\sin x + 1) - (\sin^2 x - x + 1) = x^2 + 2\sin x - \sin^2 x + x = x^2 + x + 2\sin x - \sin^2 x$$</p><p><strong>Step 3:</strong> Use Taylor expansion: $\sin x = x - \frac{x^3}{6} + O(x^5)$</p><p>Numerator factor: $x + 2\sin x = x + 2x + O(x^3) = 3x + O(x^3)$</p><p>Denominator: $x^2 + x + 2(x - \frac{x^3}{6}) - (x - \frac{x^3}{6})^2 = x^2 + x + 2x + O(x^3) = 3x + x^2 + O(x^3) = x(3 + x + O(x^2))$</p><p><strong>Step 4:</strong> For the square roots as $x \to 0$:</p><p>$$\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1} \to \sqrt{1} + \sqrt{1} = 2$$</p><p><strong>Step 5:</strong> Compute the limit:</p><p>$$\lim_{x \to 0} \frac{3x \cdot 2}{x(3 + x)} = \lim_{x \to 0} \frac{6x}{x(3 + x)} = \lim_{x \to 0} \frac{6}{3 + x} = \frac{6}{3} = 2$$</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2