Calculating limit of function

To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4}$.
I differentiated it using L Hospital's rule. I got

$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.

Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.

Answer

Using Prosthaphaeresis Formulas,

$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$

So,

$$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$

We know, $\lim_{h\to0}\frac{\sin h}h=1$

$$\text{Apply L'Hospital's Rule on }\lim_{x\to0}\frac{x-\sin x}{x^3}$$

$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$