$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left(   \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$ 
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$
now ?

The MacLaurin expansion of $\sin3x$ up to order $3$ is
$$\sin3x = 3x - \frac{9x^3}{2} + \mathcal O(x^5)$$
Plugging in the limit we obtain
$$\lim_{x \to 0} \frac{(3 + a)x + (b - \frac92)x^3}{x^3} =0$$
If the numerator is not zero then the limit is infinite, because the denominator goes to $0$ as order $3$. Therefore it has to be
$$\left\{\begin{align*}
a &= -3\\
b &= \frac92
\end{align*}\right.$$
