Question

Let $f\left(x\right)=\frac{\sin \left(3x\right)+A\sin \left(5x\right)+B\sin \left(x\right)}{x^4{\tan }^{-1}x},x\ne 0$ and $f\left(0\right)=C.$ If $f$ is continuous at $x=0$, then the value of $\frac{AB+C}{A}$ is

Answer 1

Using expansion of $\sin \left(3x\right),\sin \left(5x\right),\sin \left(x\right)$,we get
$\underset{x\to 0}{lim}\frac{[\frac{3x}{1!}-\frac{(3x{)}^3}{3!}+\frac{(3x{)}^5}{5!}\cdots ]+A[\frac{5x}{1!}-\frac{(5x{)}^3}{3!}+\frac{(5x{)}^5}{5!}\cdots ]+B[\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\cdots ]}{x^5}$
For limit to exist,
coefficient of $x=0$ and coefficient of $x^3=0$
$3+5A+B=0$ and $\frac{9}{2}+\frac{125A}{6}+\frac{B}{6}=0$
Solving, we get $A=-\frac{1}{5};B=-2$
$\therefore f\left(0\right)=C=$ coefficient of $x^5=-\frac{16}{5}$
Hence, $\frac{AB+C}{A}=14$