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