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