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.

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 $\frac16$.
