$\displaystyle\lim_{x \to \infty} \frac{1}{\sin^2 x} - \frac{1}{\sin h^2 x}$ is equal to :

Question

( A ) $-\frac{2}{3}$
( B ) $0$
( C ) $\frac{2}{3}$
( D ) $\frac{1}{3}$