<p>Let <em>f<sub>n</sub>(x)</em> = <em>lim</em><sub>t→x</sub> <br> \(f_n(x) = \lim_{t \to x} \frac{\sin^{-1}(nt)}{2t} = \frac{\sin^{-1}(nx)}{2x}\). Find the value of \(\lim_{x \to 0} \left[\frac{\sin^2 2x}{2x}\right] + \left[\frac{\sin^{-1} 4x}{2x}\right]\).</p>
Step-by-Step Solution
Key Concept: Recognize that as x→0, sin⁻¹(4x)/(2x) → 2 (using sin⁻¹(u)/u → 1 as u→0), while sin²(2x)/(2x) → 0. The floor function [·] applied to each term separately determines the final answer.
<p><strong>Step 1:</strong> Analyze the first term as x→0:</p><p>lim(x→0) sin²(2x)/(2x) = lim(x→0) sin(2x)·sin(2x)/(2x)</p><p>Since sin(2x)~2x as x→0: = lim(x→0) (2x)·sin(2x)/(2x) = lim(x→0) sin(2x) = 0</p><p>Therefore: [sin²(2x)/(2x)] = [0] = <strong>0</strong></p><p><strong>Step 2:</strong> Analyze the second term as x→0:</p><p>lim(x→0) sin⁻¹(4x)/(2x). Let u = 4x, then as x→0, u→0</p><p>= lim(u→0) sin⁻¹(u)/(u/2) = 2·lim(u→0) sin⁻¹(u)/u = 2·(1) = <strong>2</strong></p><p>Therefore: [sin⁻¹(4x)/(2x)] = [2] = <strong>2</strong></p><p><strong>Step 3:</strong> Combine results:</p><p>[sin²(2x)/(2x)] + [sin⁻¹(4x)/(2x)] = 0 + 2 = <strong>2</strong></p><p>∴ Answer: C</p>
Correct Answer: C