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?
 A: You have to use L'Hopitals 3 times we have $$\begin{align} \lim_{x\to 0}\frac{9x(\text{cos}(4x)-1}{\text{sin}(8x)-8x}&=\lim_{x\to 0}\frac{(9 (\text{cos}(4 x)-1)-36 x \text{sin}(4 x))}{(8 \text{cos}(8 x)-8)}\\
&=\lim_{x\to 0}\frac{-1}{64}\frac{(-72 \text{sin}(4 x)-144 x \text{cos}(4 x))}{\text{sin}(8x)}\\&=\lim_{x\to 0}\frac{-1}{512}\frac{(576 x \text{sin}(4 x)-432 \text{cos}(4 x))}{\text{cos}(8x)}\\&=\frac{432}{512}\\&=\frac{27}{32}.\end{align}$$
A: 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, $$\lim_{x \to 0} \frac{\sin2x}{\sin4x}=\lim_{x \to 0} \frac{2\cos2x}{4\cos4x}=\frac24=\frac12$$
$$(2) \lim_{x \to 0}\frac{x\sin4x}{\cos8x-1}=\lim_{x \to 0}\frac{x\sin4x}{-2\sin^24x}$$
$$=-\lim_{x \to 0}\frac{x}{2\sin4x}\text{  as } x\to0,\sin4x\to0\implies \sin4x\ne0$$
$$=-\lim_{x \to 0}\frac1{2\cdot4\cos4x}=-\frac18$$
