What is the proper way to handle the limit with little-$o$?
Note that if $\lim_{n\to\infty}f(n)=0$ and $\lim_{n\to\infty}g(n)=\infty$ then $$ \lim_{n\to\infty}(1+f(n))^{g(n)}=e^{\lim_{n\to\infty}f(n)g(n)}. $$ In your case, you have $f(n)=-\frac xn+o\left(\frac{2x}n\right)$ and $g(n)=n$ which of course satisfies the above condition. Since $$ {\lim_{n\to\infty}f(n)g(n)}={\lim_{n\to\infty}}\left\{-\frac xn+o\left(\frac{2x}n\right)\right\}n=-x+0=-x $$ then your limit equals to $e^{-x}$.