Answer
$$12$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\theta \to 0} \frac{{3{{\sin }^2}2\theta }}{{{\theta ^2}}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{3{{\sin }^2}2\left( 0 \right)}}{{{{\left( 0 \right)}^2}}} \cr
& = \frac{0}{0} \cr
& \cr
& {\text{applying l'Hopital's rule twice}} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{\frac{d}{{d\theta }}\left[ {3{{\sin }^2}2\theta } \right]}}{{\frac{d}{{d\theta }}\left[ {{\theta ^2}} \right]}} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{6\sin 2\theta \left( {\cos 2\theta } \right)\left( 2 \right)}}{{2\theta }} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{6\sin 2\theta \cos 2\theta }}{\theta } \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{\frac{d}{{d\theta }}\left[ {6\sin 2\theta \cos 2\theta } \right]}}{{\frac{d}{{d\theta }}\left[ \theta \right]}} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{6\sin 2\theta \frac{d}{{d\theta }}\left[ {\cos 2\theta } \right] + \cos 2\theta \frac{d}{{d\theta }}\left[ {6\sin 2\theta } \right]}}{1} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \left[ {6\sin 2\theta \left( { - 2\sin 2\theta } \right) + \cos 2\theta \left( {12\cos 2\theta } \right)} \right] \cr
& = \mathop {\lim }\limits_{\theta \to 0} \left( { - 12{{\sin }^2}2\theta + 12{{\cos }^2}2\theta } \right) \cr
& {\text{evaluating the limit}} \cr
& = - 12{\sin ^2}2\left( 0 \right) + 12{\cos ^2}2\left( 0 \right) \cr
& = 12 \cr} $$
