Do not use series expansion or L' Hospital's rule: $f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ In the following function
$$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},(x\neq 0)$$
is continuous at $x= 0$. Find $A$ and $B$. Also find $f(0)$.
I first thought of using L hospital. But my sir told me to 
'Do not use series expansion or L' Hospital's rule.�'
Now how can I proceed?

Hint. If one admits that
$$
\lim_{u \to 0}\frac{\sin u}u=1 \tag1
$$ then using 
$$
\sin(2x)=2\sin x \cos x,\quad \sin(3x)=3\sin x-4\sin^3x,
$$ one may observe that, for $x \neq0$,
$$
\begin{align}
\frac{\sin 3x + A \sin 2x + B\sin x}{x^5}
&=\frac{\sin x}{x}\cdot\frac{4\left(\cos x+\frac{A}4 \right)^2+\left(B-1-\frac{A^2}4\right)}{x^4}
\\&=\frac{\sin x}{x}\cdot\left[\left(\frac{\sin^2(\frac{x}2)}{(\frac{x}2)^2}-\frac{A+4}{2x^2} \right)^2+\frac{B-1-\frac{A^2}4}{x^4}\right] \tag2
\end{align}
$$ implying for the existence of the limit as $x \to 0$ that
$$
A+4=0,\quad B-1-\frac{A^2}4=0
$$ that is

$$
A=-4,\quad B=5.
$$ 

From $(2)$ one sees that in this case $f(0)=1$.
