Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\ln }^3}x}}{{\sqrt x }} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\ln }^3}x}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\ln }^3}\left( \infty \right)}}{{\sqrt \infty }} = \frac{\infty }{\infty } \cr
& {\text{Apply the LHopital's rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\ln }^3}x}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{{\ln }^3}x} \right]}}{{\frac{d}{{dx}}\left[ {\sqrt x } \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{3{{\ln }^2}x\left( {\frac{1}{x}} \right)}}{{\frac{1}{{2\sqrt x }}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{6\sqrt x {{\ln }^2}x}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{6{{\ln }^2}x}}{{\sqrt x }} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{6{{\ln }^2}x}}{{\sqrt x }} = \frac{{6{{\ln }^2}\infty }}{{\sqrt \infty }} = \frac{\infty }{\infty } \cr
& {\text{Apply the LHopital's rule}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{6{{\ln }^2}x}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{12\left( {\frac{{\ln x}}{x}} \right)}}{{\frac{1}{{2\sqrt x }}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{24\ln x}}{{\sqrt x }} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{24\ln x}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{24\left( {\frac{1}{x}} \right)}}{{\frac{1}{{2\sqrt x }}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{48}}{{\sqrt x }} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{48}}{{\sqrt x }} = \frac{{48}}{{\sqrt \infty }} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\ln }^3}x}}{{\sqrt x }} = 0 \cr} $$
