If the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos (\sin \mathrm{x})-\cos \mathrm{x}}{{\mathrm{x}}^4}$ is continuous at each point in its domain & $\mathrm{f}\left(0\right)=\frac{1}{\mathrm{k}}$, then k = 

$\mathrm{f}\left(0\right)={\mathrm{Lt}}_{\mathrm{x}\to 0} \frac{\cos (\sin \mathrm{x})-\cos \mathrm{x}}{{\mathrm{x}}^4}$

$\begin{array}{rl} & \mathrm{f}\left(0\right)={\mathrm{Lt}}_{\mathrm{x}\to 0}\frac{-2\sin \left[\frac{\sin \mathrm{x}+\mathrm{x}}{2}\right]\sin \left[\frac{\sin \mathrm{x}-\mathrm{x}}{2}\right]}{{\mathrm{x}}^4}\\ & ={\mathrm{Lt}}_{\mathrm{x}\to 0} \frac{-2\cdot \left[\frac{\sin \mathrm{x}+\mathrm{x}}{2}\right]\sin \left[\frac{\sin \mathrm{x}-\mathrm{x}}{2}\right]}{{\mathrm{x}}^4}={\mathrm{Lt}}_{\mathrm{x}\to 0}\frac{{\mathrm{x}}^2-{\sin }^2 \mathrm{x}}{{\mathrm{x}}^4}={\mathrm{Lt}}_{\mathrm{x}\to 0}\frac{\frac{-2\mathrm{x}}{2}[\sin \mathrm{x}-\mathrm{x}]}{{\mathrm{x}}^3}\text{ as }\mathrm{x}\to 0\sin \mathrm{x}\approx \mathrm{x}\\ & ={\mathrm{Lt}}_{\mathrm{x}\to 0}\frac{-\cos \mathrm{x}+1}{3{\mathrm{x}}^2}=\frac{+1}{6}\Rightarrow \mathrm{k}=+6\end{array}$