1. **Problem:** Evaluate $$\lim_{x \to 0} \frac{e^{2x} - e^{-2x}}{\sin x}$$ 2. **Formula and rules:** - Recall the Taylor expansions near 0: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ - $$\sin x = x - \frac{x^3}{6} + \cdots$$ - For limits of the form $$\frac{f(x)}{g(x)}$$ as $$x \to 0$$, use expansions or L'Hôpital's Rule. 3. **Intermediate work:** - Expand numerator: $$e^{2x} = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots$$ $$e^{-2x} = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \cdots$$ - So, $$e^{2x} - e^{-2x} = (1 + 2x + 2x^2 + \frac{4x^3}{3}) - (1 - 2x + 2x^2 - \frac{4x^3}{3}) = 4x + \frac{8x^3}{3} + \cdots$$ - Denominator: $$\sin x = x - \frac{x^3}{6} + \cdots$$ 4. **Evaluate limit:** $$\lim_{x \to 0} \frac{4x + \frac{8x^3}{3}}{x - \frac{x^3}{6}} = \lim_{x \to 0} \frac{4x(1 + \frac{2x^2}{3})}{x(1 - \frac{x^2}{6})} = \lim_{x \to 0} 4 \cdot \frac{1 + \frac{2x^2}{3}}{1 - \frac{x^2}{6}}$$ As $$x \to 0$$, the terms with $$x^2$$ vanish, so the limit is: $$4 \times \frac{1}{1} = 4$$ **Final answer:** $$4$$