Answer
$ r=-3, s=\frac{9}{2}$
Work Step by Step
The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
Now, $\lim\limits_{x \to 0} (\dfrac{\sin 3x}{x^3}+\dfrac{r}{x^2}+s)=0$
So, $\lim\limits_{x \to 0} [\dfrac{x(3-\dfrac{(3x)^3}{3!}+....)}{x^3}+\dfrac{r}{x^2}+s]=0$
or, $=\lim\limits_{x \to 0} [\dfrac{3}{x^2}-\dfrac{9}{2}+\dfrac{81 x^2}{40}+......\dfrac{r}{x^2}+s]=0$
Now, $\dfrac{3}{x^2}+\dfrac{r}{x^2}=0 \implies r=-3$
and $ s-\dfrac{9}{2}=0 \implies s=\frac{9}{2}$
So, $ r=-3, s=\frac{9}{2}$
