Something to do with e?

$\lim_{x\rightarrow{0}} \dfrac{(1+x)^{\frac{1}{x}}-e}{x} =A \implies \lim_{x \rightarrow{0}} \dfrac{\log(1+x)}{x}= \log(Ax+e)$ Maclaurin series for $\log(1+x)= x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots$ $\lim_{x\rightarrow {0}} \dfrac{x-\frac{x^2}{2}+\cdots}{x}=\log(e)+\log(\frac{Ax}{e}+1)$ Using $\lim_{x\rightarrow{0}}\log(1+x)=x$ We get $1-\dfrac{x}{2} = 1+\dfrac{Ax}{e} \implies A=-\dfrac{e}{2}$ Using Maclaurin series of $e^{\frac{\log(1+x)}{x}}$ $e^{\frac{\log(1+x)}{x}}=\lim_{x\rightarrow{0}}(1+x)^{\frac{1}{x}}= e^{1-\frac{x}{2}+\frac{x^2}{3}+\cdots}$ $\lim_{x\rightarrow{0}} (1+x)^{\frac{1}{x}}= e-\dfrac{ex}{2}+\dfrac{11ex^2}{24}+\cdots$ $\lim_{x\rightarrow{0}} {\dfrac{(1+x)^{\frac{1}{x}}-e}{x}}=-\dfrac{e}{2}+\dfrac{11ex}{24}$ $\lim_{x\rightarrow{0}} \dfrac{{\dfrac{(1+x)^{\frac{1}{x}}-e}{x}-A}{}}{x}=\dfrac{11ex}{24x}= \dfrac{11e}{24}$