Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{x} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}\infty {\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } = {\left( {\ln \infty } \right)^2} = \infty \cr
& \mathop {\lim }\limits_{x \to \infty } x = \infty = \infty \cr
& {\text{Since the limits of both numerator and denominator are }}\infty {\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{D_x}\left( {{{\left( {\ln x} \right)}^2}} \right)}}{{{D_x}\left( x \right)}} \cr
& {\text{use chain rule for }}{D_x}\left( {{{\left( {\ln x} \right)}^2}} \right){\text{ and the power rule for }}{D_x}\left( {{x^{ - 2}}} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2\left( {\ln x} \right)\left( {1/x} \right)}}{1} \cr
& {\text{simplifying}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2\ln x}}{x} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{2\ln x}}{x} = \frac{{2\ln \infty }}{\infty } = \frac{\infty }{\infty } \cr
& {\text{Since the limits of both numerator and denominator are }}\infty {\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{D_x}\left( {2\ln x} \right)}}{{{D_x}\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2\left( {1/x} \right)}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{2}{x} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{2}{x} = \frac{2}{\infty } = 0 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\ln x} \right)}^2}}}{x} = 0 \cr} $$
