find the limit of the natural log of the expression and take the exponential of that limit.

$\ln\left((x+e^x)^{1/x}\right) = \dfrac 1 x \ln(x+e^x)$

Now apply L'Hopital's rule

$f(x)=\ln(x+e^x)$

$g(x)=x$

$f'(x)=\dfrac{1+e^x}{x+e^x}$

$g'(x)=1$

$\displaystyle{\lim_{x \to 0}}\dfrac{f'(x)}{g'(x)}=\dfrac{1+1}{0+1}=2$

Now take the exponential of this to obtain the limit of the original expression.

$\displaystyle{\lim_{x\to 0}}\left((x+e^x)^{1/x}\right) =e^2$