$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ 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 ?

With Taylor expansion
$$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\sin \left(3x\right)+ax+bx^3}{x^3}\right)
& = \lim _{x\to \:0}\left(\frac{\left(3x-\frac{9}{2}x^3+\frac{81}{40}x^5+o\left(x^5\right)\right)+ax+bx^3}{x^3}\right)
\\ & = \lim _{x\to \:0}\left(\frac{81x^2}{40}\right) = 0
\end{aligned}
$$
Then $a$ and $b$ must have the following values
$$a = -3, b = 9/2$$ 
So:
$$\lim _{x\to \:0}\left(\frac{\sin \left(3x\right)-3x+\frac{9}{2}x^3}{x^3}\right) = 0$$
