Answer
$${e^4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {{e^x} + x} \right)^{2/x}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {{e^x} + x} \right)^{2/x}} = {\left( {{e^0} + 0} \right)^{2/{0^ + }}} = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left( {{e^x} + x} \right)^{2/x}} = {e^{\frac{2}{x}\ln \left( {{e^x} + x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {{e^x} + x} \right)^{2/x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\frac{2}{x}\ln \left( {{e^x} + x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \frac{2}{x}\ln \left( {{e^x} + x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{2}{x}\ln \left( {{e^x} + x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2\ln \left( {{e^0} + 0} \right)}}{0} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = 2\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {{e^x} + x} \right)} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{{{e^x} + 1}}{{{e^x} + x}}}}{1} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^x} + 1}}{{{e^x} + x}} \cr
& = 2\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^x} + 1}}{{{e^x} + x}} = 2\left( {\frac{{{e^0} + 1}}{{{e^0} + 0}}} \right) = 2\left( {\frac{2}{1}} \right) = 4 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {e^{\frac{2}{x}\ln \left( {{e^x} + x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \frac{2}{x}\ln \left( {{e^x} + x} \right)}} = {e^4} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left( {{e^x} + x} \right)^{2/x}} = {e^4} \cr} $$
