Lim x→0(cos (sin x)-cos x)/(x^4 equals

Expand to order $x^4$. We have $\sin x=x-\dfrac{x^3}{6}+\cdots$, so $\sin^2x=x^2-\dfrac{x^4}{3}+\cdots$ and $\sin^4x=x^4+\cdots$. Then $\cos(\sin x)=1-\dfrac{\sin^2x}{2}+\dfrac{\sin^4x}{24}=1-\dfrac{x^2}{2}+\dfrac{x^4}{6}+\dfrac{x^4}{24}+\cdots$ and $\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots$. Subtracting, the $x^2$ and the $\tfrac{x^4}{24}$ parts cancel, leaving $\dfrac{x^4}{6}$. Hence the limit is $\dfrac16$. Dropping the $-\tfrac{x^4}{3}$ correction in $\sin^2x$ would wrongly give $0$. This is why the question cannot be done by naively replacing $\cos(\sin x)$ with $\cos x$: the two functions agree up to $x^3$ and differ only at the fourth order, so a first-order approximation would wrongly suggest the limit is zero. Series expansion is the efficient method.