Calculus

Using the Maclaurin series expansion of $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + O(u^4)$ with $u = 2x$: $e^{2x} = 1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + O(x^4) = 1 + 2x + 2x^2 + \frac{8x^3}{6} + O(x^4)$ Subtracting $(1 + 2x + 2x^2)$: $e^{2x} - 1 - 2x - 2x^2 = \frac{4}{3}x^3 + O(x^4)$ Dividing by $x^3$ and taking the limit as $x \to 0$: $\lim_{x \to 0} \frac{\frac{4}{3}x^3 + O(x^4)}{x^3} = \frac{4}{3} \approx 1.33$