How to evaluate the limit $\lim_{x\to 0} \frac{1-\cos(4x)}{\sin^2(7x)}$ I am lost in trying to figure out how to evaluate the $$\lim_{x\to 0} \frac{1-\cos(4x)}{\sin^2(7x)}.$$
So far, I have tried the following:
Multiply the numerator and denominator by the numerator's conjugate $1+\cos(4x)$, which gives $\frac{\sin^2(4x)}{(\sin^2(7x))(1+\cos(4x))}$. However, I am not sure what to do after this step. 
Could anyone please help point me in the right direction as to what I am supposed to do next?
All help is appreciated. 

HINT:
Using $\cos2A=1-2\sin^2A,$
$$\dfrac{1-\cos4x}{\sin^27x}=2\cdot2^2\cdot\left(\dfrac{\sin2x}{2x}\right)^2\cdot\dfrac1{7^2\cdot\left(\dfrac{\sin7x}{7x}\right)^2}$$
