Use L'Hopital's rule to evaluate $\lim_{x \to 0} \frac{9x(\cos4x-1)}{\sin8x-8x}$ $$\lim_{x \to 0} \frac{9x(\cos4x-1)}{\sin8x-8x}$$
I have done this problem a couple of times and could not get the correct answer. Here is the work I have done so far http://imgur.com/GDZjX26 . The correct answer was $\frac{27}{32}$, did I differentiate wrong somewhere?

Using L'Hospital Rule repeatedly, 
$$\lim_{x \to 0} \frac{9x(\cos4x-1)}{\sin8x-8x}$$
$$=\lim_{x \to 0} \frac{9(\cos4x-1)+9x(-4\sin4x)}{8\cos8x-8}$$
$$=\frac98\left( \lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}\right)-\frac92\left( \lim_{x \to 0}\frac{x\sin4x}{\cos8x-1} \right)$$
$$(1)\lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}=\lim_{x \to 0} \frac{-4\sin4x}{-8\sin8x}=\frac12\lim_{x \to 0} \frac{\sin4x}{\sin8x}=\frac12\lim_{x \to 0} \frac{4\cos4x}{8\cos8x}=\frac12\frac48$$
$$(2)\lim_{x \to 0}\frac{x\sin4x}{\cos8x-1}= \lim_{x \to 0}\frac{\sin4x+4x\cos4x}{-8\sin8x}$$
$$=-\frac18\lim_{x \to 0}\frac{\sin4x}{\sin8x}-\frac12 \lim_{x \to 0}\frac{x}{\sin8x}\cdot \lim_{x \to 0}\frac{\cos4x}1$$
$$=-\frac18\cdot\frac48 \text{(already found)}-\frac12 \lim_{x \to 0}\frac1{8\cos8x}\cdot1$$
$$=-\frac1{16}-\frac12\cdot\frac18=-\frac18 $$
Can you take it home form here?
Alternatively, 
using $\cos2y=1-2\sin^2y,$
$$(1)\lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}=\left(\lim_{x \to 0} \frac{\sin2x}{\sin4x}\right)^2$$
Now, $$\lim_{x \to 0} \frac{\sin2x}{\sin4x}=\lim_{x \to 0} \frac{2\cos2x}{4\cos4x}=\frac24=\frac12$$
$$(2) \lim_{x \to 0}\frac{x\sin4x}{\cos8x-1}=\lim_{x \to 0}\frac{x\sin4x}{-2\sin^24x}$$
$$=-\lim_{x \to 0}\frac{x}{2\sin4x}\text{  as } x\to0,\sin4x\to0\implies \sin4x\ne0$$
$$=-\lim_{x \to 0}\frac1{2\cdot4\cos4x}=-\frac18$$
