<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] 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
$$
\lim_{x\rightarrow 0}f(x)=1\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{4}{2}\right)=1.
$$
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