<s>[INST] $\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ 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?
 [/INST] \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, $\lim_{\scriptscriptstyle{x\to0}} f(x) = 1$.
</s>