Evaluate the limit: $$\lim_{x \to 0} \frac{\tan(2x) - 2x}{x^3}$$ (Round off to two decimal places).

Answer Key:

2.67

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Series Expansion:** $\tan(\theta) = \theta + \frac{\theta^3}{3} + O(\theta^5)$. 2. **Stage 2: Substitute $\theta = 2x$:** $\tan(2x) = 2x + \frac{8x^3}{3} + O(x^5)$. 3. **Stage 3: Numerator:** $\tan(2x) - 2x = \frac{8}{3}x^3 + O(x^5)$. 4. **Stage 4: Limit Computation:** $\lim_{x \to 0} \frac{8/3 x^3}{x^3} = 8/3 \approx 2.67$. 5. **Stage 5: Tolerance:** 2.60 to 2.70.

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