Answer
$${\text{Limit is not in the form }}\frac{0}{0}{\text{ or }}\frac{\infty }{\infty }$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{{e^x}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{2{e^{2x}}}}{{{e^x}}}{\text{ }}\left( {{\text{Step 1}}} \right) \cr
& = \mathop {\lim }\limits_{x \to 0} 2{e^x}{\text{ }}\left( {{\text{Step 2}}} \right) \cr
& = 2{\text{ }}\left( {{\text{Step 3}}} \right) \cr
& {\text{From the first step it is not necessary to use l'hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{{e^x}}} = \frac{{{e^{2\left( 0 \right)}} - 1}}{{{e^0}}} = \frac{{1 - 1}}{1} = 0 \cr
& {\text{The result is not in the form }}\frac{0}{0},\frac{\infty }{\infty },{\text{ so the rule}} \cr
& {\text{does not apply. }} \cr} $$
