Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
The limit is indeed $\frac{1}{2}$. Due to Taylor's formula with integral remainder,
$$ \sum_{k=0}^{n}\frac{1}{k!} = e-\int_{0}^{1}\frac{(1-t)^n}{n!}\,e^t\,dt=e\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)=\exp\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)\tag{1} $$ while $$ \left(1+\frac{1}{n}\right)^{n+x}=\exp\left[(n+x)\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right]=\exp\left[1+\frac{x-\frac{1}{2}}{n}+o\left(\frac{1}{n}\right)\right]\tag{2} $$ so by equating the RHSs of $(1)$ and $(2)$ we get $\lim_{n\to +\infty}x_n=\frac{1}{2}$ as expected.
Answer. $x_n\to \dfrac{1}{2}$
Explanation. Taylor series remainder $$ \mathrm{e}=1+\frac{1}{1!}+\cdots+\frac{1}{n!}+\frac{\mathrm{e}^{\xi_n}}{(n+1)!} $$ for some $\xi_n\in(0,1)$. Since $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!},$$ Then $$ n+x_n=\frac{\log\left(\mathrm{e}-\dfrac{\mathrm{e}^{\xi_n}}{(n+1)!}\right)}{\log\left(1+\frac{1}{n}\right)}=\frac{1+\log\left(1-\dfrac{\mathrm{e}^{\xi_n-1}}{(n+1)!}\right)}{\dfrac{1}{n}-\dfrac{1}{2n^2}+{\mathcal O}(n^{-3})}=n\cdot\frac{1+{\mathcal O}\left(\frac{1}{(n+1)!}\right)}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})}\\=n+\frac{\dfrac{1}{2}+{\mathcal O}(n^{-1})}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})} $$ and hence $$ x_n\to \frac{1}{2} $$
If $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$ then we have $$\begin{align}x_n&=\frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\ln\left( 1+\frac{1}{n}\right)}-n\\&\sim \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\left(\frac{1}{n}-\frac{1}{2n^2}\right)}-n \\&= n\left( \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{1-\frac{1}{2n}}-1\right)\\&\sim n\left( \ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)-1+\color{blue}{\frac{1}{2n}}\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\right)&\to \color{red}{\frac12} \end{align}$$
Given that $$\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\to 1$$