<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] We need to use the limit $$\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}\tag{1}$$ which can be easily established either via L'Hospital's Rule or Taylor's theorem. Now using $(1)$ we have $$\frac{\sin 3x - 3x}{x^{3}}\to - \frac{9}{2}\tag{2}$$ and hence $$\frac{\sin 3x}{x^{3}}+\frac{a}{x^{2}}+b\to 0\tag{3}$$ is equivalent to (via subtracting equation $(2)$ from equation $(3)$) $$\frac{3+a}{x^{2}}+b\to \frac{9}{2}\tag{4}$$ or $$\frac{3+a+bx^{2}}{x^{2}}\to \frac{9}{2}\tag{5}$$ It is now clear that $3+a+bx^{2}\to 0$ so that $a=-3$. And then $b=9/2$ follows from the  penultimate equation. Note that everything apart from limit $(1)$ is based on algebra of limits. 
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