I didn't solve it pls help me
Problem:
Limit x->0+ [(x+e^x)^(1/x)]
Tanks for your interest
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.
$\displaystyle{\lim_{x\to 0}}\left((x+e^x)^{1/x}\right) =e^2$