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