Question

If $\underset{x\to 0}{\lim }\left(x^{-3}\sin 3x+a{x}^{-2}+b\right)$ exists and is equal to zero, then the value of $a+2b$, is $\dots$

Answer 1

Answer: 6

$\underset{x\to 0}{\lim }\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b$
$=\underset{x\to 0}{\lim }\frac{\sin 3x+ax+b{x}^3}{x^3}$
$=\underset{x\to 0}{\lim }\frac{3\frac{\sin 3x}{3x^3}+a+b{x}^2}{x^2}$
for existence of limit $3+a=0$
$\Rightarrow a=-3$
$l=\underset{x\to 0}{\lim }\frac{\sin 3x-3x+b{x}^3}{x^3}$
$=27\cdot \frac{\sin t-t}{t^3}+b=0$
$=-\frac{27}{6}+b=0$
$\Rightarrow b=\frac{9}{2}$
OR [Usign L’Hospital’s rule]
Hence, $a+2b=-3+2\times \frac{9}{2}=6$