L'Hospital's Rule?

Relevant wiki: Maclaurin Series

$\text{we know the expansion of } \ln{(x+1)} \text{, it is }$

$\ln{(x+1)} = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots \\ \dfrac{1}{x}\ln{(x+1)} = 1-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \\ e^{\frac{1}{x}\ln{(x+1)}}=e^{1-\frac{x}{2}+\frac{x^2}{3}+\cdots} \\ (x+1)^{\frac1{x}}=e\cdot e^{-\frac{x}{2}+\frac{x^2}{3}+\cdots} \\ (x+1)^{\frac1{x}}=e\cdot(1-\dfrac{1}{2}x+\dfrac{11}{24}x^2+\cdots) \\ (x+1)^{\frac1{x}}=e-\dfrac{e}{2}x+\dfrac{11e}{24}x^2+\cdots \\ \text{Plugging this value in our Limit } \\ L = \displaystyle\lim_{x \to 0} \dfrac{ \cancel{e}-\cancel{\dfrac{e}{2}x}+\dfrac{11e}{24}x^2+\cdots -\cancel{e}+\cancel{\dfrac{e}{2}x}}{x^2} \implies \boxed{\dfrac{11e}{24}} \\ L = 0.4583$