Answer
$$\frac{1}{{{e^6}}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{ - 3/x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{ - 3/x}} = {\left( {1 + 2\left( 0 \right)} \right)^{ - 3/0}} = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }.{\text{ }} \cr
& {\left( {1 + 2x} \right)^{ - 3/x}} = {e^{ - \frac{3}{x}\ln \left( {1 + 2x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{ - 3/x}} = \mathop {\lim }\limits_{x \to 0} {e^{ - \frac{3}{x}\ln \left( {1 + 2x} \right)}} = {e^{ - \mathop {\lim }\limits_{x \to 0} \frac{3}{x}\ln \left( {1 + 2x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = - \mathop {\lim }\limits_{x \to 0} \frac{3}{x}\ln \left( {1 + 2x} \right) = - \frac{{3\ln \left( {1 + 2\left( 0 \right)} \right)}}{x} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = - 3\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {1 + 2x} \right)} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} = - 3\mathop {\lim }\limits_{x \to {0^ + }} \frac{2}{{1 + 2x}} \cr
& L = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\ln x}} = - 3\left( {\frac{2}{{1 + 2\left( 0 \right)}}} \right) = - 6 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{ - 3/x}} = \mathop {\lim }\limits_{x \to 0} {e^{ - \frac{3}{x}\ln \left( {1 + 2x} \right)}} = {e^{ - \mathop {\lim }\limits_{x \to 0} \frac{3}{x}\ln \left( {1 + 2x} \right)}} = {e^{ - 6}} \cr
& \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2x} \right)^{ - 3/x}} = \frac{1}{{{e^6}}} \cr} $$
