If for some real number $a,\underset{x\to 0}{lim} \frac{\sin 2x+a\sin x}{x^3}$

exists, then the limit is equal to

$\underset{x\to 0}{lim} \frac{\sin 2x+a\sin x}{x^3}=\underset{x\to 0}{lim} \frac{2\cos 2x+a\cos x}{3x^2}\left(\frac{0}{0}\text{ form }\right)$

The last limits will exist if $a=-2$ In this case the
last limit is equal to

$\underset{x\to 0}{lim} \frac{-4\sin 2x+2\sin x}{6x}=\underset{x\to 0}{lim} \left[-\frac{4}{3}\frac{\sin 2x}{2x}+\frac{1}{3}\frac{\sin x}{x}\right]$

$=-\frac{4}{3}+\frac{1}{3}=-1$