If $\underset{x\to 0}{\mathrm{Lim}} \left(x^{-3}\sin 3x+a{x}^{-2}+b\right)$ exists and is equal to zero, then the value of a+ 2b = 

$$\begin{gathered} \underset{x\to 0}{\mathrm{Lim}} \left(x^{-3}\sin 3x+a{x}^{-2}+b\right) \\ =\underset{x\to 0}{\mathrm{Lim}} \left(\frac{\sin 3x+ax+b{x}^3}{x^3}\right) \\ =\underset{x\to 0}{\mathrm{Lim}} \left(\frac{3x-\frac{{\left(3x\right)}^3}{3!}+\frac{{\left(3x\right)}^5}{5!}-\frac{{\left(3x\right)}^7}{7!}+\cdots +ax+b{x}^3}{x^3}\right) \end{gathered}$$
$a+3=0\Rightarrow a=-3$
$b-\frac{9}{2}=0\Rightarrow b=\frac{9}{2}\Rightarrow 2b=9$
$\therefore a+2b=6$