A limit

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac {\sin (3x)}{x^3} + \frac a{x^2} + b \right) \\ & = \lim_{x \to 0} \left({\color{#3D99F6} \frac {\sin (3x) + ax}{x^3}} + b \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \lim_{x \to 0} \left({\color{#3D99F6} \frac {3\cos (3x) + a}{3x^2}} + b \right) & \small \color{#3D99F6} \text{For limit to exist }3\cos 0 + a = 0 \implies a = -3 \\ & = \lim_{x \to 0} \left({\color{#3D99F6} \frac {-9\sin (3x)}{6x}} + b \right) & \small \color{#3D99F6} \text{Apply L'Hôpital's rule again} \\ & = \lim_{x \to 0} \left({\color{#3D99F6} \frac {-27\cos (3x)}{6}} + b \right) & \small \color{#3D99F6} \text{and again.} \\ & = - \frac 92 + {\color{#3D99F6}b} = 0 & \small \color{#3D99F6} \implies b = \frac 92 \end{aligned}$

Therefore, $a+b = - 3 + \dfrac 92 = \boxed{1.5}$ .

First recall the Maclaurin series $\sin(u) = u - \dfrac{u^{3}}{3!} + O(x^{5})$ , so in the case $u = 3x$ we have

$\dfrac{\sin(3x)}{x^{3}} + \dfrac{a}{x^{2}} + b = \dfrac{1}{x^{3}}\left(3x - \dfrac{27x^{3}}{6} + O(x^{5})\right) + \dfrac{a}{x^{2}} + b = \dfrac{3}{x^{2}} - \dfrac{9}{2} + O(x^{2}) + \dfrac{a}{x^{2}} + b = \dfrac{3 + a}{x^{2}} + b - \dfrac{9}{2} + O(x^{2})$ ,

which will only approach $0$ as $x \to 0$ if $a = -3$ and $b = \dfrac{9}{2}$ , giving us $a + b = -3 + \dfrac{9}{2} = \dfrac{3}{2} = \boxed{1.5}$ .

The given expression is of the form 0/0. Applying L'Hospital's rule we get the same limit in the form of a ratio : 3cos(3x) +a +3b(x^2) and 3(x^2). This must be of the form 0/0, implying a=-3. Applying the rule twice more, we get b=4.5 Therefore a+b= 4.5 - 3 = 1.5