If $\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos \frac{x}{2})}{2^m{x}^n}$  is equal to the left hand derivative of $f\left(x\right)=e^{-\left|x\right|}$  at  $x=0,$ then find the value of  $(4n+m).$

 $$\begin{gathered} f\left(x\right)=\left[\begin{array}{cc}{e}^{-x},& ifx\ge 0\\ e^x,& ifx<0\end{array}\right] \\ f\text{'}\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{e^{-h}-1}{-h}=1 \end{gathered}$$
Hence  $\underset{x\to 0}{\lim }\frac{[1-\cos (1-\cos \frac{x}{2})]}{{(1-\cos \frac{x}{2})}^2}\frac{{(1-\cos \frac{x}{2})}^2}{2^m.x^n}=1$
$$\begin{gathered} [Using\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{{\theta }^2}=\frac{1}{2}] \\ \underset{x\to 0}{\lim }\frac{{(1-\cos \frac{x}{2})}^2}{2^m{x}^n}=2;\underset{x\to 0}{\lim }\frac{4{\sin }^4\frac{x}{2}}{2^m{x}^n}=2 \end{gathered}$$
;  
 for limit to exist  $n=4$
$\frac{2}{2^{8+m}}=1\Rightarrow 2^{8+m}=2$
$\therefore m=-7\Rightarrow n=4$ and  $m=-7$
Hence  $n-10m=4+70=74$