$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ without series expansion or L Hospital rule 
If $$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},$$ $x\neq 0$, is continuous at $x=0$, then find $A,B$ and $f(0)$. Do not use series expansion or L Hospital's rule.

As $f(x)$ is continuous at $x=0$,therefore its limit at $x=0$ should equal to its value.
Note that this question is to be solved without series expansion or L Hospital's rule,
I tried to find the limit $\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$
$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3\sin x-4\sin^3x+2A\sin x\cos x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}\times\frac{\sin x}{x}$
$=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}$
As the denominator is zero,so numerator has to be zero,in order the limit to be finite.
So, $3+2A+B=0. (1)$
I tried but I could not get the second equation between $A$ and $B$. I am stuck here. How do I continue?
 A: Using trigonometric identities we have
\begin{align}
3-4\sin^2 x+2A\cos x+B&=4(1-\sin^2 x)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\
&=4\cos^2\left(2\cdot\frac{x}{2}\right)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\
&=4\left(1-2\sin^2\frac{x}{2}\right)^2+2A\left(1-2\sin^2 \frac{x}{2}\right)+B-1\\
&=16\sin^4\frac{x}{2}-16\sin^2\frac{x}{2}-4A\sin^2\frac{x}{2}+2A+B+3\\
&=16\sin^4\frac{x}{2}-4(A+4)\sin^2\frac{x}{2}+2A+B+3\\
\end{align}
In order to make the limit finite we must have
$$A+4=0\quad\text{and}\quad 2A+B+3=0\qquad\iff\qquad \color{blue}{A=-4}\quad\text{and}\quad \color{blue}{B=5}$$
By taking those values we get
\begin{align}
\lim_{x\to 0}\frac{\sin 3x\color{blue}{-4}\sin 2x+\color{blue}{5}\sin x}{x^5}&=\left(\lim_{x\to 0}\frac{16\sin^4\frac{x}{2}}{x^4}\right)\left(\lim_{x\to 0}\frac{\sin x}{x}\right)\\
&=\left(\lim_{x\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^4\left(1\right)\\
&=\color{blue}{1}
\end{align}
Since $f$ is continuous at $0$ it follows $f(0)=1$.
A: Your first step is correct, you need $3+2A+B=0$ for the function to converge.  Let's substitute $B=-3-2A$ into the original function. Them you have
$$
f(x)=\frac{\sin(3x)+A\sin(2x)+(-3-2A)\sin(x)}{x^5}
$$
Using the angle sum formulas:
$$
\sin(3x)=\sin(x)\cos(2x)+\cos(x)\sin(2x)=\sin(x)(\cos^2(x)-\sin^2(x))+2\sin(x)\cos(x)^2.
$$
Now, using the standard trig identity, $\sin^2(x)=1-\cos^2(x)$.  Therefore, 
$$
\sin(3x)=4\sin(x)\cos(x)^2-\sin(x)
$$
Also using the angle sum formulas:
$$
\sin(2x)=2\sin(x)\cos(x)
$$
Substituting these into the original function, we have
$$
f(x)=\frac{4\sin(x)\cos^2(x)-\sin(x)+2A\sin(x)\cos(x)-3\sin(x)-2A\sin(x)}{x^5}.
$$
By combining the terms with $A$ and those without, we have
$$
f(x)=\frac{4\sin(x)(\cos^2(x)-1)+2A\sin(x)(\cos(x)-1)}{x^5}=\frac{\sin(x)}{x}\cdot\frac{4(\cos^2(x)-1)+2A(\cos(x)-1)}{x^4}
$$
Factoring a difference of squares, the numerator becomes
$$
f(x)=\frac{\sin(x)}{x}\cdot\frac{4(\cos(x)-1)(\cos(x)+1)+2A(\cos(x)-1)}{x^4}$$
Factoring again, we have
$$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)+1)+2A}{x^2}
$$
Observe that 
$$
\lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}=\lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}\cdot\frac{\cos(x)+1}{\cos(x)+1}=\lim_{x\rightarrow 0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}
$$
Then, using the standard trig identities, we get that this limit equals
$$
\lim_{x\rightarrow 0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}=\lim_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)^2\frac{-1}{\cos(x)+1},
$$
which we can see has a limit of $-\frac{1}{2}$.
Continuing from here, you also need the numerator to vanish for the limit to exist because the denominator vanishes.  In other words, $\lim_{x\rightarrow 0}4(\cos(x)+1)+2A=8+2A$ must be $0$, so $A=-4$.  After substitution, the entire formula becomes 
$$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)-1)}{x^2}
$$
Substituting the values that we've calculated the limit as
$$
\lim_{x\rightarrow 0}f(x)=1\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{4}{2}\right)=1.
$$
A: \begin{align*}
f(x) &= \frac{\sin(3x) + A\sin(2x) + B\sin{x}}{x^{5}} \\
&= \frac{(-4\sin^{3}x + 3\sin{x}) + 2A\sin{x}\cos{x} + B\sin{x}}{x^{5}} \\
&= \frac{\sin{x}}{x} \cdot \frac{-4\sin^{2}x + 3 + 2A\cos{x} + B}{x^{4}} \\
&= \frac{\sin{x}}{x} \cdot \frac{-4\sin^{2}x + 3 + 2A\cos{x} + B}{x^{4}} \\
&= \frac{\sin{x}}{x} \cdot \frac{4(1 - \sin^{2}x) - 1 + 2A\cos{x} + B}{x^{4}} \\
&= \frac{\sin{x}}{x} \cdot \frac{-4(1 - \cos^{2}x) + 3 + 2A\cos{x} + B}{x^{4}} .
\end{align*}
The given function is defined at $0$ if, and only if, $3 + 2A + B = 0$. So, $B = -3 - 2A$, and
\begin{align*}
f(x) &= \frac{\sin{x}}{x} \cdot \frac{-4(1 - \cos^{2}x) - 2A(1 - \cos{x})}{x^{4}} \\
&= \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{-4 - 2A - 4\cos{x}}{x^{2}} .
\end{align*}
Again, the given function is defined at $0$ if, and only if, $-4 - 2A -4 = 0$. So, $A = -4$, $B = 5$, and
\begin{align*}
f(x) &= \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{4(1 - \cos{x})}{x^{2}} \\
&= 4 \cdot \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{1 - \cos{x}}{x^{2}} .
\end{align*}
So, $\lim_{\scriptscriptstyle{x\to0}} f(x) = 1$.
