Answer
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{e^{2x}}}} = 0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{e^{2x}}}} \cr
& {\text{Evaluate the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{e^{2x}}}} = \frac{{{{\left( \infty \right)}^3}}}{{{e^{2\left( \infty \right)}}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{e^{2x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{x^3}} \right]}}{{\frac{d}{{dx}}\left[ {{e^{2x}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{2{e^{2x}}}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{2{e^{2x}}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{2{e^{2x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {3{x^2}} \right]}}{{\frac{d}{{dx}}\left[ {2{e^{2x}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{4{e^{2x}}}} = \frac{\infty }{\infty } \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {6x} \right]}}{{\frac{d}{{dx}}\left[ {4{e^{2x}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{6}{{8{e^{2x}}}} = \frac{6}{\infty } = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{e^{2x}}}} = 0 \cr} $$
