Answer
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} = 0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} \cr
& {\text{Evaluate the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} = \frac{{{{\left( \infty \right)}^2}}}{{{e^{5\left( \infty \right)}}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{x^2}} \right]}}{{\frac{d}{{dx}}\left[ {{e^{5x}}} \right]}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2x}}{{5{e^{5x}}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2x}}{{5{e^{5x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {2x} \right]}}{{\frac{d}{{dx}}\left[ {5{e^{5x}}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{2}{{25{e^{5x}}}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{2}{{25{e^{5x}}}} = \frac{2}{{25{e^{\left( \infty \right)}}}} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^{5x}}}} = 0 \cr} $$
