Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}3x}}{x} \cr
& {\text{Rewrite }}{\sin ^2}3x{\text{ as }}\left( {\sin 3x} \right)\left( {\sin 3x} \right) \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sin 3x} \right)\left( {\sin 3x} \right)}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin 3x}}{x}} \right)\left( {\sin 3x} \right) \cr
& {\text{Multiplying and dividing by }}3 \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin 3x}}{{3x}}} \right)\left( {3\sin 3x} \right) \cr
& {\text{Use the product law of limits}}{\text{, the limit of a product is the }} \cr
& {\text{product of the limits}}{\text{, then}} \cr
& = \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}} \right)\mathop {\lim }\limits_{x \to 0} \left( {3\sin 3x} \right) \cr
& {\text{Letting }}\theta = 3x,{\text{ then }}\theta \to 0{\text{ as }}x \to 0,{\text{ so by }}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1 \cr
& = 3\overbrace {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}^{\theta = 3x,{\text{ }}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1}\left[ {\mathop {\lim }\limits_{x \to 0} \left( {3\sin 3x} \right)} \right] \cr
& = 3\left( 1 \right)\left[ {3\left( {\sin 3\left( 0 \right)} \right)} \right] \cr
& = 3\left( 1 \right)\left( 0 \right) \cr
& = 0 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}3x}}{x} = 0 \cr} $$
