Question

If $f\left(x\right)$ =$\frac{sin3x+Asin2x+Bsinx}{x^5}(x\ne 0)$ is continuous at $x=0$, then find $A$ +$B$.

Answer 1

$f\left(x\right)=\underset{x\to 0}{lim}\frac{sin3x+Asin2x+Bsinx}{x^5}$

$=\underset{x\to 0}{lim}\frac{sinx}{x}\left(\frac{3-4si{n}^{2}x+2Acosx+B}{x^4}\right)$

$=\underset{x\to 0}{lim}\frac{1+2cos2x+2Acosx+B}{x^4}$

$=\underset{x\to 0}{lim}\frac{1}{x^4}(1+2(1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}+.....)+2A(1-\frac{x^2}{2!}+\frac{x^4}{4!})+B)$

$=\underset{x\to 0}{lim}\frac{1}{x^4}(3+2A+B+x^2(-4-A)+x^4(\frac{4}{3}+\frac{A}{12})+.....)$

$\Rightarrow$ $2A+B+3=0$ and $-4-A=0$
$\Rightarrow$ $A=-4,B=5$

$A+B=-4+5=1$