If the function $\text{⨍ (x)=}\frac{\cos \text{(sinx)-cosx}}{{\mathrm{x}}^4}$ is continuous at each point in its domain and 

$⨍ (0)=\frac{1}{k}$ then $\mathrm{k}$ is _________

$\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos (\sin \mathrm{x})-\cos \mathrm{x}}{{\mathrm{x}}^4}$ is continuous at eachpoint

$f\left(0\right)=\frac{1}{k}\begin{array}{} \end{array}$

$\Rightarrow \frac{1}{k}=f\left(0\right)\begin{array}{} \\ \\ \end{array}$

$\mathrm{ }=\underset{\mathrm{x}\to 0}{\lim } \frac{\cos (\sin \mathrm{x})-\cos \mathrm{x}}{\mathrm{x}}$

$\mathrm{ }=\underset{\mathrm{x}\to 0}{\lim } \frac{-2\sin \frac{\sin \mathrm{x}+\mathrm{x}}{2}\sin \frac{\sin \mathrm{x}-\mathrm{x}}{2}}{{\mathrm{x}}^4}$

$=-2\underset{\mathrm{x}\to 0}{\lim } \frac{\left.\sin \frac{\sin \mathrm{x}+\mathrm{x}}{2}\right)}{\frac{\sin \mathrm{x}+\mathrm{x}}{2}}\cdot \frac{\left.\sin \frac{\sin \mathrm{x}}{2}\right)}{\frac{\sin \mathrm{x}-\mathrm{x}}{2}}\cdot \frac{\sin \mathrm{x}+\mathrm{x}\cdot \frac{\sin -\mathrm{x}}{2}}{{\mathrm{x}}^4}$$=-2\underset{x\to 0}{\lim } \frac{1}{2}\cdot \left(\frac{\sin x+x}{x}\right)\underset{x\to 0}{\lim } \frac{1}{2}\cdot \frac{\sin x-x}{x^3}$

$=-2\cdot \frac{1}{2}\left(2\right)-\frac{1}{2}\underset{x\to 0}{\lim } \frac{\sin x-x}{x^3}$

$= + \underset{\mathrm{x}\to 0}{\lim } \frac{1 - \cos \mathrm{x}}{3{\mathrm{x}}^2}$

$=\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}$