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. 
 A: Notice, 
$$\lim_{x\to 0}\frac{1-\cos(4x)}{\sin^2(7x)}$$
$$\lim_{x\to 0}\frac{1-\cos^2(4x)}{\sin^2(7x)(1+\cos (4x))}$$
$$=\lim_{x\to 0}\frac{\sin^2(4x)}{\sin^2(7x)(1+\cos (4x))}$$
$$=\lim_{x\to 0}\frac{\sin^2(4x)}{\sin^2(7x)}\cdot \lim_{x\to 0} \frac{1}{1+\cos (4x)}$$
$$=\lim_{x\to 0}\left(\frac{\sin(4x)}{\sin(7x)}\right)^2\cdot \frac{1}{1+1}$$
$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{4}{7}\frac{\frac{\sin(4x)}{4x}}{\frac{\sin(7x)}{7x}}\right)^2$$
$$=\frac{1}{2}\frac{16}{49}\lim_{x\to 0}\left(\frac{\frac{\sin(4x)}{(4x)}}{\frac{\sin(7x)}{(7x)}}\right)^2$$
$$=\frac{8}{49}\left(\frac{1}{1}\right)^2$$
$$=\color{red}{\frac{8}{49}}$$
A: Hint:
$$\lim_{t\to 0} \frac{\sin^2(4t)}{\sin^2(7t)}=\lim_{t\to 0}\frac{\left(\frac{\sin 4t}{t}\right)^2}{\left(\frac{\sin 7t}{t}\right)^2}=\frac{4^2}{7^2}\times\frac{\left(\lim_{t\to 0}\frac{\sin 4t}{4t}\right)^2}{\left(\lim_{t\to 0}\frac{\sin 7t}{7t}\right)^2}=\frac{16}{49}\times \frac{1^2}{1^2}$$
Also
\begin{align}
\lim_{t\to 0}\frac{1}{1+\cos(4t)}&=\frac{1}{1+1}=\frac{1}{2}
\end{align}
A: You could also use L'Hopital's rule, 
$$\lim_{x\to 0}\frac{1-\cos(4x)}{\sin^2(7x)}$$
$$=\lim_{x\to 0}\frac{4\sin(4x)}{7\sin(14x)}$$
$$=\lim_{x\to 0}\frac{16\cos(4x)}{7\times14\cos(7x)}$$
$$=\frac{16\cos(0)}{7\times14\cos(0)} = \frac{8}{49}$$
A: HINT:
Using $\cos2A=1-2\sin^2A,$
$$\dfrac{1-\cos4x}{\sin^27x}=2\cdot2^2\cdot\left(\dfrac{\sin2x}{2x}\right)^2\cdot\dfrac1{7^2\cdot\left(\dfrac{\sin7x}{7x}\right)^2}$$
