1. We need to evaluate $L=\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}.$

2. Since both numerator terms approach $1$ as $x\to 0$, this is a $0/0$ form. We use series expansions near $x=0$.

3. First, expand $\sin x$: $\sin x = x-\frac{x^3}{6}+O(x^5).$ So, $\sin x - x = -\frac{x^3}{6}+O(x^5).$

4. Let $u=\sin x, \quad v=x.$ Then $\cos u - \cos v = -\sin v\,(u-v)+O((u-v)^2),$ but using direct expansion is cleaner here.

5. Expand $\cos t$ around $t=0$: $\cos t = 1-\frac{t^2}{2}+\frac{t^4}{24}+O(t^6).$ Hence, $\cos(\sin x)=1-\frac{(\sin x)^2}{2}+\frac{(\sin x)^4}{24}+O(x^6),$ $\cos x = 1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6).$ Therefore, $\cos(\sin x)-\cos x = -\frac{(\sin x)^2-x^2}{2}+\frac{(\sin x)^4-x^4}{24}+O(x^6).$

6. Now expand $(\sin x)^2$ and $(\sin x)^4$. Using $\sin x = x-\frac{x^3}{6}+O(x^5),$ we get $(\sin x)^2 = \left(x-\frac{x^3}{6}+O(x^5)\right)^2 = x^2-\frac{x^4}{3}+O(x^6).$ So, $(\sin x)^2-x^2 = -\frac{x^4}{3}+O(x^6).$

Also, $(\sin x)^4 = \left(x^2-\frac{x^4}{3}+O(x^6)\right)^2 = x^4+O(x^6),$ so $(\sin x)^4-x^4=O(x^6).$

7. Substitute into the numerator: $\cos(\sin x)-\cos x = -\frac{1}{2}\left(-\frac{x^4}{3}+O(x^6)\right)+\frac{1}{24}O(x^6)+O(x^6).$ Thus, $\cos(\sin x)-\cos x = \frac{x^4}{6}+O(x^6).$

8. Divide by $x^4$: $\frac{\cos(\sin x)-\cos x}{x^4} = \frac{1}{6}+O(x^2).$ Hence, $L=\frac{1}{6}.$

9. Checking options:

• A: $\frac13$
• B: $\frac14$
• C: $\frac16$ ✅
• D: $\frac1{12}$

So the correct option is C.