Calculus

This is of indeterminate form $1^\infty$. $\ln L = \lim_{x \to 0} \frac{\ln(1 + 2x)}{x}$ Applying L'Hopital's rule: $\ln L = \lim_{x \to 0} \frac{\frac{2}{1 + 2x}}{1} = 2$ $L = e^2 \approx 7.389$