I need to prove the following limit exists, and evaluate it

lim

x-->0 1-cos(3x)/x^4+3x^2

Let f(x) = 1-cos(3x) and g(x) = x^4+3x^2

The f and g are differentiatiable, and f(0)=g(0)=0

Thus f and g satisfy the conditions of L'Hopitals Rule at 0.

Now

f'(x)=sin3x and g'(x)= 4x^3+6x

Thus by L'Hopitals Rule, the required limit exists and equals

lim

x-->0 f'(x)/g'(x) = lim x-->0 Sin3x/4x^3+6x = ?

This is where I am stuck. I know if just the basic end bit but could anyone help?