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?