Find out constants$~a,b,c,d~$such that$~\lim_{x\to0}\frac{\sin^{}\left(3x\right)-\left(ax^{2}+bx+c\right)}{x^{3}}=d~$is satisfied

$$\left(a,b,c,d:=\text{constants}\right)~~\wedge~~\left(d\neq0\right)$$

I want to find out the formula(s)or value(s)of the above constants which satisfy the following equation.

$$\lim_{x\to0}\frac{\sin^{}\left(3x\right)-\left(ax^{2}+bx+c\right)}{x^{3}}=d\tag{1}$$

$$\sin^{}\left(3x\right)^{\left(n\right)}=3^{n}\sin^{}\left(x+\frac{n\pi}{2}\right)$$

And by using taylor expansion expression, the following is said.

$$\sin^{}\left(3x\right)=\sum_{i=0}^{\infty}\frac{\sin^{}\left(3x\right)^{\left(i\right)}|_{x=0}x^{i}}{i!}$$

$$\sin^{}\left(3x\right)^{\left(i\right)}|_{x=0}=3^{i}\sin^{}\left(\frac{i\pi}{2}\right)$$

$$\sin\left(3x\right)=\sum_{i=0}^{\infty}\frac{x^{i}}{i!}\cdot 3^{i}\sin\left(\frac{i\pi}{2}\right)$$

I think I made no mistake so far, however I can't get how$~o(x^{3})~$has been appeared.

$$\sin\left(3x\right)=3x-\frac{9}{2}x^{3}+\underbrace{\sum_{i=4}^{\infty}\frac{x^{i}}{i!}\cdot 3^{i}\sin\left(\frac{i\pi}{2}\right)}_{\text{min degree is}~5}$$

$$=3x-\frac{9}{2}x^{3}+\mathcal{O}\left(x^{5}\right)~~\leftarrow~~\text{Newest my progress so far}$$

$$\lim_{x\to0}\frac{\sin^{}\left(3x\right)-\left(ax^{2}+bx+c\right)}{x^{3}}=d\tag{1}$$

$$=\lim_{x\to 0}\frac{3x-\frac{9}{2}x^{3}+\mathcal{O}\left(x^{5}\right)-ax^{2}-bx-c}{x^{3}}~~\leftarrow~~\text{What can I do for next?}$$

Solution 1:

We can use Taylor - McLaurin expansion, for $\sin(3x)$: $$\sin(3x)=3x-\frac{9}{2}x^3+o(x^3)$$ The limit becomes: $$\lim_{x\to0}\frac{\sin\left(3x\right)-\left(ax^{2}+bx+c\right)}{x^{3}}=d\implies \lim_{x\to0}\frac{3x-\frac{9}{2}x^3+o(x^3)-\left(ax^{2}+bx+c\right)}{x^{3}}=d$$ Note that if, for example, $b\neq 3 \land a=c=0$, you are going to have $+\infty$. Can you justify the other cases?

Consequently, the only case acceptable is: $$b=3 \,\,\land \,\, a=c=0\implies \lim_{x\to 0}\frac{\frac{9}{2}x^3+o(x^3)}{x^3}=\frac{9}{2}$$