The value of $\underset{x\to 0}{\lim }\left\{\frac{\sin x-x+\frac{x^3}{6}}{x^5}\right\}\text{ is }\frac{1}{k}$$\text{ then }k\text{ is }$

$=\underset{x\to 0}{\lim }\left\{\frac{\sin x-x+\frac{x^3}{6}}{x^5}\right\}$

$=\underset{x\to 0}{\lim }\frac{\cos x-1+\frac{3x^2}{6}}{5x^4}$     $\left( by L.H. rule\right)$

$=\underset{x\to 0}{\lim }\frac{-\sin x+\frac{6x}{6}}{20x^3}$

$=\underset{x\to 0}{\lim }\frac{-\cos x+1}{60x^2}$

$=\underset{x\to 0}{\lim }\frac{\sin x}{120x}=\underset{x\to 0}{\lim }\frac{\cos x}{120}=\frac{1}{120}$