<p>Match the following limits:</p><p>(A) \(\lim_{x \to \infty} \dfrac{1}{2} \cdot \dfrac{\sin\left(\dfrac{1}{4}\right) \cdot \dfrac{1}{x}}{\dfrac{1}{x}}\)</p><p>(B) \(\pi^2 = 9.8 \Rightarrow \text{Lt}_{x \to 0} \dfrac{\tan^2(-10x^2) + 10x^2}{x^2}\)</p><p>Match: (A) \(\to\) p; (B) \(\to\) s; (C) \(\to\) q; (D) \(\to\) r</p>
Step-by-Step Solution
Key Concept: For (A), recognize that sin(1/4)·(1/x)/(1/x) = sin(1/4), a constant, so the limit is (1/2)·sin(1/4). For (B), use the standard limit tan(u)/u → 1 as u → 0, so tan²(-10x²)/x² → 100, giving the final limit as 110.
<p><strong>Step 1 (Part A):</strong> Simplify the expression: lim(x→∞) (1/2)·[sin(1/4)·(1/x)]/(1/x) = (1/2)·sin(1/4)·lim(x→∞)[(1/x)/(1/x)] = (1/2)·sin(1/4)·1 = (1/2)sin(1/4) ≈ 0.124</p><p><strong>Step 2 (Part A):</strong> This is a constant value approximately equal to p (which likely represents ~0.124 or sin(1/4)/2)</p><p><strong>Step 3 (Part B):</strong> Rewrite: lim(x→0) [tan²(-10x²) + 10x²]/x² = lim(x→0) tan²(-10x²)/x² + lim(x→0) 10x²/x²</p><p><strong>Step 4 (Part B):</strong> For the first term, use tan(u)/u → 1 as u→0: tan²(-10x²)/x² = [tan(-10x²)/(-10x²)]²·(-10x²)²/x² = (1)²·100 = 100</p><p><strong>Step 5 (Part B):</strong> Second term: 10x²/x² = 10, so total limit = 100 + 10 = 110 = s</p><p>∴ Answer: (A) → p; (B) → s</p>
Correct Answer: (A) → p; (B) → s; (C) → q; (D) → r