Answer
$e^2$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0}(e^x+x)^{1/x}$
Re-write the given function as: $y=(e^x+x)^{1/x} \implies \ln y=\dfrac{\ln (e^x+x)}{x}$
But $\lim\limits_{x \to 0}\dfrac{\ln (e^x+x)}{x}=\dfrac{0}{0}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0} \ln y=\lim\limits_{x \to 0} \dfrac{1}{(e^x+x)}(e^x+1)\\= \dfrac{e^0+1}{e^0+0}\\=2$
So, $\lim\limits_{x \to 0} \ln y=2 \implies \lim\limits_{x \to 0} y=e^2$
