Official solution
Consider \[\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]} \over {{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi - \pi {{\sin }^2}x} \right)} \over {{x^2}}}\]
\[\left[ \, \right.\] As \[\sin \left( {\pi - \theta } \right) = \sin \theta \] \[\left. \, \right]\]
\[ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}} \times {{\pi {{\sin }^2}x} \over {{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 0} 1 \times \pi {\left( {{{\sin x} \over x}} \right)^2} = \pi \]