<p>\[\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}\] is equal to: (where \([\,]\) denotes greatest integer function)</p>
Step-by-Step Solution
Key Concept: For small x near 0, use tan(πsin²x) ≈ πsin²x ≈ πx² and analyze |x| - sin(x[x]) carefully by noting [x] = -1 for x ∈ [-1,0) and [x] = 0 for x ∈ [0,1), making the second term vanish as x→0.
<p><strong>Step 1:</strong> Analyze the numerator as x→0. Use Taylor expansion: tan(πsin²x) ≈ πsin²x ≈ πx² as x→0.</p><p><strong>Step 2:</strong> For the second term, analyze left and right limits separately:</p><p>• For x→0⁺: [x] = 0, so (|x| - sin(0))² = x²</p><p>• For x→0⁻: [x] = -1, so (|x| - sin(-x))² = (|x| + sin(x))² → x² (since sin(x) ≈ x for small x, contributions of order x³ vanish)</p><p><strong>Step 3:</strong> Both sides give the same behavior. The numerator ≈ πx² + x² = (π+1)x²</p><p><strong>Step 4:</strong> Therefore: $$\lim_{x \to 0} \frac{(\pi+1)x^2}{x^2} = \pi + 1$$</p><p>∴ Answer: B</p>
Correct Answer: B