Question:

<p><strong>Step 1:</strong> Rewrite the numerator by factoring:</p><p>3θ - sin(3θ) = 3θ(1 - sin(3θ)/3θ)</p><p><strong>Step 2:</strong> Express the limit as a product:</p><p>$$\lim_{\theta \to 0} \frac{3\theta - \sin 3\theta}{\theta^3} = \lim_{\theta \to 0} \frac{3\theta(1 - \sin(3\theta)/3\theta)}{\theta^3}$$</p><p><strong>Step 3:</strong> Separate into two parts:</p><p>$$= \lim_{\theta \to 0} \frac{3}{\theta^2} \cdot \left(1 - \frac{\sin 3\theta}{3\theta}\right)$$</p><p><strong>Step 4:</strong> Apply the standard limit with substitution u = 3θ:</p><p>As θ → 0, we have u → 0, so sin(u)/u → 1, thus sin(3θ)/(3θ) → 1</p><p><strong>Step 5:</strong> Therefore: (1 - sin(3θ)/3θ) → 0. Rewrite using L'Hôpital's principle for the indeterminate form:</p><p>$$\lim_{\theta \to 0} \frac{3\theta - \sin 3\theta}{\theta^3} = \lim_{\theta \to 0} \frac{\sin 3\theta - 3\theta\cos 3\theta}{\theta^3} \cdot (-1) = \lim_{\theta \to 0} \frac{3(1 - \cos 3\theta)}{\theta^2} \cdot \frac{1}{\theta}$$</p><p><strong>Alternative (avoiding expansion):</strong> Use 1 - cos(3θ) ≈ (3θ)²/2 conceptually:</p><p>$$\lim_{\theta \to 0} \frac{3\theta - \sin 3\theta}{\theta^3} = \lim_{\theta \to 0} \frac{\frac{(3\theta)^3}{6}}{\theta^3} = \frac{27}{6} \cdot \frac{1}{\theta^3} \to \frac{1}{6}$$</p><p>∴ Answer: <strong>1/6</strong></p>