<s>[INST] $\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ 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 ?
 [/INST] Why not to use Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin(3x)=3 x-\frac{9 x^3}{2}+\frac{81 x^5}{40}+O\left(x^6\right)$$
$$\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b =\frac{a+3}{x^2}+\left(b-\frac{9}{2}\right)+\frac{81 x^2}{40}+O\left(x^3\right)$$ Then $???$
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