$\underset{h→0}{\lim}\frac{\sin(a+3h)-3\sin(a+2h)+3\sin(a+h)-\sin a}{h^3}$ is:

$-\cos a$

Applying L'Hopital's rule repeatedly

$\underset{h→0}{\lim}\frac{(\sin(a+3h)-3\sin a+2h)+3\sin(a+h)-\sin a}{h^3}$

$=\underset{h→0}{\lim}\frac{3\cos(a+3h)-6\cos(a+2h)+3\cos(a+h)}{3h^2}$

$=\underset{h→0}{\lim}\frac{-9\sin(a+3h)+12\sin(a+2h)-3\sin(a+h)}{6h}$

$=\underset{h→0}{\lim}\frac{-27\cos(a+3h)+24\cos(a+2h)-3\cos(a+h)}{6}$

$=-\cos a$