Question

If ${lim}_{x\to 0}(x^{-3}\sin 3x+a{x}^{-2}+b)$ exists and is equal to $0$, then

• A. $a=-3$ and $b=\frac{9}{2}$
• B. $a=3$ and $b=\frac{9}{2}$
• C. $a=-3$ and $b=-\frac{9}{2}$
• D. $a=3$ and $b=-\frac{9}{2}$

Answer 1

The correct option is A $a=-3$ and $b=\frac{9}{2}$

$\underset{x\to 0}{lim}\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b=\underset{x\to 0}{lim}\frac{\sin 3x+ax+b{x}^3}{x^3}$

Applying L'Hopital's rule

$=\underset{x\to 0}{lim}\frac{3\cos 3x+a+3b{x}^2}{3x^2}$

If the numerator of the last limit were non zero then the limit would be $+\infty$, so we must have,

$\underset{x\to 0}{lim}\frac{3\cos 3x+a+3b{x}^2}{3x^2}=0$

$3+a=0$

$a=-3$

Again applying L'Hopital's rule

$=\underset{x\to 0}{lim}\frac{-9\sin 3x+6bx}{6x}$

Again applying L'Hopital's rule

$=\underset{x\to 0}{lim}\frac{-27\cos 3x+6b}{6}$

$-\frac{9}{2}+b=0$

$b=\frac{9}{2}$

Thus, we have

$a=-3$

$b=\frac{9}{2}$