<p>Evaluate: \(\displaystyle\lim_{x \to 0} \dfrac{x\cot(4x)}{\sin^2 x \cot^2(2x)}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the expression using cot = cos/sin, then apply the standard limit sin(x)/x → 1 systematically to each trigonometric term. The key is recognizing that you need to manipulate the expression into forms containing sin(kx)/(kx) patterns.
<p><strong>Step 1:</strong> Rewrite using cot θ = cos θ/sin θ</p><p>$$\lim_{x \to 0} \frac{x\cot(4x)}{\sin^2 x \cot^2(2x)} = \lim_{x \to 0} \frac{x \cdot \frac{\cos(4x)}{\sin(4x)}}{\sin^2 x \cdot \frac{\cos^2(2x)}{\sin^2(2x)}}$$</p><p><strong>Step 2:</strong> Simplify the complex fraction</p><p>$$= \lim_{x \to 0} \frac{x \cos(4x) \sin^2(2x)}{\sin(4x) \sin^2 x \cos^2(2x)}$$</p><p><strong>Step 3:</strong> Rearrange to isolate standard limit forms</p><p>$$= \lim_{x \to 0} \frac{x}{\sin(4x)} \cdot \frac{\sin^2(2x)}{\sin^2 x} \cdot \frac{\cos(4x)}{\cos^2(2x)}$$</p><p><strong>Step 4:</strong> Apply standard limits: $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and continuity of cosine</p><p>$$= \frac{1}{4} \cdot \frac{\sin^2(2x)}{\sin^2 x} \cdot \frac{\cos(4x)}{\cos^2(2x)}$$</p><p><strong>Step 5:</strong> Evaluate $\frac{\sin^2(2x)}{\sin^2 x}$ using $\sin(2x) = 2\sin x \cos x$</p><p>$$\frac{\sin^2(2x)}{\sin^2 x} = \frac{4\sin^2 x \cos^2 x}{\sin^2 x} = 4\cos^2 x \to 4$$</p><p><strong>Step 6:</strong> Combine results</p><p>$$= \frac{1}{4} \cdot 4 \cdot \frac{1}{1} = 1$$</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: A