Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {x^2}{\left( {\ln x} \right)^2} \cr
& {\text{use the property }}{x^n} = \frac{1}{{{x^{ - n}}}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\left( {\ln x} \right)}^2}}}{{{x^{ - 2}}}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} = {\left( {\ln 0} \right)^2} = \infty \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{ - 2}} = \frac{1}{{{x^2}}} = \frac{1}{{{{\left( 0 \right)}^2}}} = \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 {0^ + }} \frac{{{D_x}\left( {{{\left( {\ln x} \right)}^2}} \right)}}{{{D_x}\left( {{x^{ - 2}}} \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 {0^ + }} \frac{{2\left( {\ln x} \right)\left( {1/x} \right)}}{{ - 2{x^{ - 3}}}} \cr
& {\text{simplifying}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left( {\ln x} \right){x^{ - 1}}}}{{ - {x^{ - 3}}}} \cr
& = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{{x^{ - 2}}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{{x^{ - 2}}}} = - \frac{{\ln 0}}{{{0^{ - 2}}}} = - \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 {0^ + }} \frac{{{D_x}\left( {\ln x} \right)}}{{{D_x}\left( {{x^{ - 2}}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1/x}}{{ - 2{x^{ - 3}}}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^{ - 2}}}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{x \to {0^ + }} {x^2} \cr
& {\text{Find the limit }} \cr
& = - \frac{1}{2}{\left( 0 \right)^2} \cr
& = 0 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^2}{\left( {\ln x} \right)^2} = 0 \cr} $$