Question

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

then the values of $a$ and $b$ are?

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

Answer 1

Answer: (A)

$a=-3$ & $b=\frac{9}{2}$

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

Applying L'Hospital's Rule,

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

As the limit doesn't exist, so

$3\cos 3x+a+3b{x}^2=0$

$x\to 0$ so, $(3+a)=0=>a=-3$

Again,

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

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

As the limit is 0, then

$(-27+6b)=0$ so,

$b=\frac{27}{6}=\frac{9}{2}$