Answer
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^m}}}{{{e^{nx}}}} = 0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^m}}}{{{e^{nx}}}} \cr
& {\text{Evaluate the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^m}}}{{{e^{nx}}}} = \frac{{{{\left( \infty \right)}^m}}}{{{e^\infty }}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^m}}}{{{e^{nx}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{x^m}} \right]}}{{\frac{d}{{dx}}\left[ {{e^{nx}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{m{x^{m - 1}}}}{{n{e^{nx}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{m{x^{m - 1}}}}{{n{e^{nx}}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {m{x^{m - 1}}} \right]}}{{\frac{d}{{dx}}\left[ {n{e^{nx}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{m\left( {m - 1} \right){x^{m - 2}}}}{{{n^2}{e^{nx}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {m\left( {m - 1} \right){x^{m - 2}}} \right]}}{{\frac{d}{{dx}}\left[ {{n^2}{e^{nx}}} \right]}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{m\left( {m - 1} \right)\left( {m - 2} \right){x^{m - 3}}}}{{{n^3}{e^{nx}}}} \cr
& {\text{If we continue indefinitely we obtain}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{m!}}{{{n^3}{e^{nx}}}} \cr
& {\text{Recall that }}m\left( {m - 1} \right)\left( {m - 2} \right)\left( {m - 3} \right) \cdots = m! \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{m!}}{{{n^3}{e^{nx}}}} = \frac{{m!}}{\infty } = 0 \cr} $$
