Answer
$$\frac{1}{{{e^3}}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^{ - x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^{ - x}} = {1^{ - \infty }} \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left( {1 + \frac{3}{x}} \right)^{ - x}} = {e^{ - x\ln \left( {1 + \frac{3}{x}} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^{ - x}} = \mathop {\lim }\limits_{x \to + \infty } {e^{ - x\ln \left( {1 + \frac{3}{x}} \right)}} = {e^{ - \mathop {\lim }\limits_{x \to + \infty } x\ln \left( {1 + \frac{3}{x}} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = - \mathop {\lim }\limits_{x \to + \infty } x\ln \left( {1 + \frac{3}{x}} \right) = - \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln \left( {1 + \frac{3}{x}} \right)}}{{1/x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = - \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln \left( {1 + \frac{3}{x}} \right)}}{{1/x}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{ - \frac{3}{{{x^2}}}}}{{1 + \frac{3}{x}}}}}{{ - \frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\frac{3}{{{x^2}}}}}{{\frac{{x + 3}}{x}}}}}{{\frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3x}}{{{x^2}\left( {x + 3} \right)}}}}{{\frac{1}{{{x^2}}}}} \cr
& = - \mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^3}}}{{{x^2}\left( {x + 3} \right)}} = - \mathop {\lim }\limits_{x \to + \infty } \frac{{3x}}{{x + 3}} \cr
& - \mathop {\lim }\limits_{x \to + \infty } \frac{{3x}}{{x + 3}} = - \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{1 + 3/x}} = - 3 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^{ - x}} = \mathop {\lim }\limits_{x \to + \infty } {e^{ - x\ln \left( {1 + \frac{3}{x}} \right)}} = {e^{ - 3}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{3}{x}} \right)^{ - x}} = \frac{1}{{{e^3}}} \cr} $$
