Limits, Continuity & Differentiability
Limits
Grade 12

Question:

<p>\(\lim_{x \to 0} \dfrac{x \cot(4x)}{\sin^2 x \cot^2(2x)}\) is equal to __________.</p>

Step-by-Step Solution

Key Concept: Rewrite the expression using cot = cos/sin and recognize that you need to isolate standard limits like (sin x)/x → 1 and (x/sin x) → 1 as x → 0. Strategically group terms to form these standard limits.
<p><strong>Step 1:</strong> Rewrite using cot θ = cos θ/sin θ:</p><p>$$\lim_{x \to 0} \frac{x \cdot \frac{\cos(4x)}{\sin(4x)}}{\sin^2 x \cdot \frac{\cos^2(2x)}{\sin^2(2x)}}$$</p><p><strong>Step 2:</strong> Simplify to:</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 as a product of standard limits:</p><p>$$\lim_{x \to 0} \left(\frac{x}{\sin(4x)}\right) \cdot \left(\frac{\sin^2(2x)}{\sin^2 x}\right) \cdot \left(\frac{\cos(4x)}{\cos^2(2x)}\right)$$</p><p><strong>Step 4:</strong> Evaluate each factor:</p><p>• $\frac{x}{\sin(4x)} = \frac{1}{4} \cdot \frac{4x}{\sin(4x)} \to \frac{1}{4}$</p><p>• $\frac{\sin^2(2x)}{\sin^2 x} = \left(\frac{\sin(2x)}{x}\right)^2 \cdot \left(\frac{x}{\sin x}\right)^2 = 4 \cdot 1 = 4$</p><p>• $\frac{\cos(4x)}{\cos^2(2x)} \to \frac{1}{1} = 1$</p><p><strong>Step 5:</strong> Multiply: $\frac{1}{4} \times 4 \times 1 = 1$</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1

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