Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = {\infty ^{1/\infty }} = {\infty ^0} \cr
& {\text{This limit has the form }}{\infty ^0}.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {x^{1/x}} = {e^{\frac{1}{x}\ln x}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \sqrt x } \right)^{1/x}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln x}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln x = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{x} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = \frac{1}{\infty } = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^0} = 1 \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = 1 \cr} $$
