Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$ 
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$

The answer stated is $\displaystyle {1 \over 6}$. 

What I've tried: 
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x) -1+1- \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{1- \cos x}{x^4} - \frac {1-\cos (\sin x)}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{2 \sin^2(\frac {x}{2})}{x^4} - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
$$=\displaystyle \lim_{x\to0} \left(\frac{\sin(\frac {x}{2})}{x} \right)^2. \left( \dfrac{1}{2x^2} \right) - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
I'm not sure how I can evaluate the limit by proceeding this way. All help will be appreciated. 
P.S. I'd prefer not using L'Hôpital's rule, it can get really messy.
EDIT: I should have mentioned that I would prefer if the solution does not use taylor series approximations (or any approximations) for that matter.
 A: By using trigonometry identity and Taylor series,\begin{align} &\lim_{x\to0} \frac{\cos (\sin x)-\cos(x)}{x^4}\\
&=-2\lim_{x\to 0}\frac{\sin \left( \frac{\sin x - x}2\right) \sin\left(\frac{\sin x + x}2 \right)}{x^4}\\
&= -2\lim_{x \to 0 }\frac{\sin \left( \frac{-x^3}{2(6)}\right)\sin \left( \frac{x+x}{2}\right)}{x^4}\\
&= -2 \lim_{x \to 0}\frac{-\frac{x^3}{6(2)}\cdot x}{x^4}\\
&= \frac16\end{align}
A: Note that from the expansions $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ and $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have
$$\begin{align}
\cos(\sin(x))&=\cos\left(x-\frac{x^3}{6}+O(x^5)\right)\\\\
&=1-\frac12\left(x-\frac{x^3}{6}+O(x^5)\right)^2+\frac{1}{24}\left(x-\frac{x^3}{6}+O(x^5)\right)^4+O(x^6)\\\\
&=1-\frac12x^2+\frac{5}{24} x^4+O(x^6)
\end{align}$$
Hence, 
$$\frac{\cos(\sin(x))-\cos(x)}{x^4}=\frac16+O(x^2)$$
from which we find the coveted limit.
A: For small $x$, $\frac{\cos(\sin x)-\cos x}{\sin x-x}\sim\cos^\prime x\sim -x$, while $\frac{\sin x-x}{x^4}\sim-\frac{1}{6x}$, so the limit is $\frac16$.
