For continuity at $x=0$, the limit $\lim_{x\to 0}\frac{\sin 3x+\alpha\sin x-\beta\cos 3x}{x^3}$ must exist and be finite, and then equal to $f(0)$.

Expand the numerator near $x=0$ using Taylor series: $\sin 3x = 3x-\frac{(3x)^3}{6}+\cdots = 3x-\frac{9}{2}x^3+\cdots$ $\sin x = x-\frac{x^3}{6}+\cdots$ $\cos 3x = 1-\frac{(3x)^2}{2}+\cdots = 1-\frac{9}{2}x^2+\cdots$

So, $\sin 3x+\alpha\sin x-\beta\cos 3x$ $= \left(3x-\frac{9}{2}x^3\right)+\alpha\left(x-\frac{x^3}{6}\right)-\beta\left(1-\frac{9}{2}x^2\right)+\cdots$ $= -\beta +(3+\alpha)x+\frac{9\beta}{2}x^2+\left(-\frac{9}{2}-\frac{\alpha}{6}\right)x^3+\cdots$

Since the denominator is $x^3$, for the limit to be finite, the constant, $x$, and $x^2$ terms in the numerator must vanish.

Therefore,

• Constant term: $-\beta=0 \Rightarrow \beta=0$
• Coefficient of $x$: $3+\alpha=0 \Rightarrow \alpha=-3$
• Coefficient of $x^2$: automatically becomes $0$ because $\beta=0$

Substitute $\alpha=-3$ and $\beta=0$: $\sin 3x-3\sin x$

Then $f(0)=\lim_{x\to 0}\frac{\sin 3x-3\sin x}{x^3}$

Use the expansions:

So the correct option is: $\boxed{\text{C: }-4}$