Answer
$r=-3, s=\dfrac{9}{2}$
Work Step by Step
Write the Taylor series for $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
Now, $\lim\limits_{x \to 0} (\dfrac{\sin 3x}{x^3}+\dfrac{r}{x^2}+s)=0 \\ \implies \lim\limits_{x \to 0} [\dfrac{x(3-\dfrac{(3x)^3}{3!}+....)}{x^3}+\dfrac{r}{x^2}+s]=0 \\ \implies \lim\limits_{x \to 0} [\dfrac{3}{x^2}-\dfrac{9}{2}+\dfrac{81 x^2}{40}+......\dfrac{r}{x^2}+s]=0 \\ \implies \dfrac{3}{x^2}+\dfrac{r}{x^2}=0; s-\dfrac{9}{2}=0$
So, $r=-3 \space and \space s=\dfrac{9}{2}$
