Question

If $\underset{x\to 0}{lim}(x^{-3}\sin 3x+a{x}^{-2}+b)=0$ , then $a+2b$ is equal to​​​​​​​

Answer 1

Let, $L=\underset{x\to 0}{lim}(x^{-3}\sin 3x+a{x}^{-2}+b)$
$=\underset{x\to 0}{lim}\frac{\sin 3x+ax+b{x}^3}{x^3}=\frac{0}{0}$ form
$=\underset{x\to 0}{lim}\frac{3\cos 3x+a+3b{x}^2}{3x^2}\text{(Using L' Hospital rule)}$
For existence of limit, $(3+a)=0\Rightarrow a=-3$
Therefore,
$\underset{x\to 0}{lim}\frac{3\cos 3x-3+3b{x}^2}{3x^2}$
$=\underset{x\to 0}{lim}\frac{-9\sin 3x+6bx}{6x}$
$=\underset{x\to 0}{lim}\frac{-27\cos 3x+6b}{6}\text{( Using L' Hospital rule)}$
But given,
$\underset{x\to 0}{lim}\frac{-27\cos 3x+6b}{6}=0$
$\Rightarrow -27+6b=0\Rightarrow b=\frac{9}{2}$
$\therefore a+2b=-3+2\times \frac{9}{2}=6$

Alternate:
$\underset{x\to 0}{lim}\frac{\sin 3x+ax+b{x}^3}{x^3}$

$\underset{x\to 0}{lim}\frac{3x-\frac{(3x{)}^3}{3!}+\frac{(3x{)}^5}{5!}\cdots +ax+b{x}^3}{x^3}$
For limit to exist,
$3+a=0\Rightarrow a=-3$
$\Rightarrow -\frac{27}{6}+b=0\text{(Given: L=0)}$
$\Rightarrow b=\frac{9}{2}$
$\therefore a+2b=-3+2\times \frac{9}{2}=6$