$\Rightarrow (1+x)^{\Large{\frac{1}{x}}}=e^{\Large{\frac{1}{x}}{ln(1+x)}}=e^{\Large\frac{1}{x}(x-x^{2}/2+x^{3}/3.....)}$
$\Rightarrow e^{\Large({1-\frac{x}{2}+\frac{x^2}{3}- .....})}$
$\therefore \lim_{x\rightarrow0} \{e^{\Large(1-x/2+x^{2}/3.....)}-e\}/x$
Will produce $\frac{0}{0}$ form.
Now apply L'hospital rule
$\Rightarrow - \color{Red}{\Large \frac{e}{2}}$