If $\underset{x\to 0}{\lim }\frac{\sin \left(ax\right)-\sin x-x}{x^3}$exists and finite then a =

Use expansion

$l=\underset{x\to 0}{lim} \frac{\sin ax-\sin x-x}{x^3}$

$=\underset{x\to 0}{lim} \frac{\left(ax-\frac{a^3{x}^3}{3!}+\frac{a^5{x}^5}{5!}\dots ..\right)-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-.....\right)-x}{x^3}$

$=\underset{x\to 0}{lim} \frac{(a-2)x+\left(\frac{1}{6}-\frac{a^3}{6}\right)x^3+\dots \dots }{x^3}$

Limit exists 

$(\mathrm{a}-2)=0$

$a=2$