Answer
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^3}}} = 0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^3}}} \cr
& {\text{Evaluate the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^3}}} = \frac{{{{\left( {\ln \infty } \right)}^2}}}{{{{\left( \infty \right)}^3}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{{\left( {\ln x} \right)}^2}} \right]}}{{\frac{d}{{dx}}\left[ {{x^3}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2\left( {\ln x} \right)}}{{3{x^2}}}\left( {\frac{1}{x}} \right) \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2\left( {\ln x} \right)}}{{3{x^3}}} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {2\left( {\ln x} \right)} \right]}}{{\frac{d}{{dx}}\left[ {3{x^3}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{2}{{9{x^2}}}\left( {\frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{2}{{9{x^3}}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{2}{{9{x^3}}} = \frac{2}{{9{{\left( \infty \right)}^3}}} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^3}}} = 0 \cr} $$
