Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{\tan \left( {{x^2}} \right)}} \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{{{{\sin }^2}0}}{{\tan \left( {{0^2}} \right)}} \cr
& = \frac{0}{0} \cr
& {\text{Using l'Hopital's rule, we get:}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {{{\sin }^2}x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\tan \left( {{x^2}} \right)} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin x\cos x}}{{{{\sec }^2}{x^2}\left( {2x} \right)}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{2\sin 0\cos 0}}{{{{\sec }^2}{{\left( 0 \right)}^2}\left( {2\left( 0 \right)} \right)}} \cr
& = \frac{0}{0} \cr
& {\text{Using l'Hopital's rule again, we get:}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {2\sin x\cos x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {{{\sec }^2}{x^2}\left( {2x} \right)} \right)}} \cr
& {\text{Then}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {2\sin x\left( { - \sin x} \right) + 2\cos x\cos x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {2{{\sec }^2}{x^2} + 2x\left( {2\sec {x^2}} \right)\left( {\sec {x^2}} \right)\left( {\tan {x^2}} \right)\left( {2x} \right)} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2{{\cos }^2}x - 2{{\sin }^2}x} \right)}}{{\left( {2{{\sec }^2}{x^2} + 8{x^2}{{\sec }^2}{x^2}\tan {x^2}} \right)}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{\left( {2{{\cos }^2}0 - 2{{\sin }^2}0} \right)}}{{\left( {2{{\sec }^2}{0^2} + 8{{\left( 0 \right)}^2}{{\sec }^2}{{\left( 0 \right)}^2}\tan {{\left( 0 \right)}^2}} \right)}} \cr
& = \frac{2}{{2 + 0}} \cr
& = 1 \cr} $$
