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?

Use $\sin2x=2\sin x\cos x$ and $\sin3x=3\sin x-4\sin^3x$, so your function can be written
$$
\frac{\sin x}{x}\frac{3-4\sin^2x+2A\cos x+B}{x^4}
$$
In order for the limit to be finite, you need that the numerator in the second fraction has limit $0$, so
$$
3+2A+B=0
$$
so $B=-2A-3$. Thus you get, removing $\frac{\sin x}{x}$, that does not contribute to the limit,
$$
\frac{-4\sin^2x+2A\cos x-2A}{x^4}
$$
Set $x=2y$, so the function becomes
$$
\frac{-2\sin^22y+A\cos2y-A}{8y^4}=\frac{-8\sin^2y\cos^2y-2A\sin^2y}{8y^4}=
-\frac{\sin^2y}{y^2}\frac{4\cos^2y+A}{4y^2}
$$
and we need $A=-4$, so the function is
$$
\frac{\sin^4y}{y^4}
$$
which has limit $1$.
Thus $B=5$ and the value to give to the function to ensure continuity is $1$.

Just a confirmation with Taylor expansion:
\begin{align}
\sin 3x+A\sin 2x+B\sin x
&=  3x-\frac{(3x)^3}{6}+\frac{(3x)^5}{120}\\
&\qquad  +2Ax-A\frac{(2x)^3}{6}+A\frac{(2x)^5}{120}\\
&\qquad  +Bx-B\frac{x^3}{6}+B\frac{x^5}{120}+o(x^5)
\end{align}
Thus we need
$$
\begin{cases}
3+2A+B=0 \\
-27-8A-B=0
\end{cases}
$$
so $A=-4$ and $B=5$. The coefficient of $x^5$ is
$$
\frac{1}{120}(3^5+2^5A+B)=1
$$
