1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{e^{2x} - 1}{x}$$. 2. **Recall the formula and rules:** The limit $$\lim_{x \to 0} \frac{e^{kx} - 1}{x} = k$$ for any constant $k$, because the derivative of $e^{kx}$ at $x=0$ is $k e^{0} = k$. 3. **Apply the limit:** Here, $k=2$, so the limit is expected to be 2. 4. **Show intermediate work using series expansion:** $$e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \cdots = 1 + 2x + 2x^2 + \cdots$$ Substitute into the limit expression: $$\frac{e^{2x} - 1}{x} = \frac{1 + 2x + 2x^2 + \cdots - 1}{x} = \frac{2x + 2x^2 + \cdots}{x}$$ 5. **Simplify by canceling $x$:** $$= \frac{\cancel{2x} + 2x^2 + \cdots}{\cancel{x}} = 2 + 2x + \cdots$$ 6. **Evaluate the limit as $x \to 0$:** $$\lim_{x \to 0} (2 + 2x + \cdots) = 2$$ **Final answer:** $$\boxed{2}$$