$$\mathop {\lim }\limits_{n \to \infty } \frac{{{n^p}{{\sin }^2}\left( {n!} \right)}}{{n + 1}},\,0 < p < 1$$      is equal to :

Solution :
Given limit can be written as
$$\eqalign{ & \mathop {\lim }\limits_{n \to 0} \frac{{{{\sin }^2}\left( {n!} \right)}}{{{n^{1 - p}}\left( {1 + \frac{1}{n}} \right)}}\,\,\left( {0 < p < 1} \right) \cr & = \frac{{{\text{some real number in }}\left[ {0,\,1} \right]}}{\infty } \cr & = 0\,\,\,\left( {\because \,1 - p > 0} \right) \cr} $$