Question

$\underset{x\to 0}{lim}(x^{-3}\sin 3x+a{x}^{-2}+b)$ 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 $b=\frac{9}{2}$

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

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

$\text{For existence,}$

$(3+a)=0\text{or}a=-3$

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

$\Rightarrow 27\underset{t\to 0}{lim}\frac{\sin t-t}{t^3}+b=0........(3x=t)$
The left hand side reduces to $\frac{0}{0}$ form by substituting the limit

Then, using L'Hospital's rule

$\Rightarrow 27\underset{t\to 0}{lim}\frac{\cos t-1}{3t^2}+b=0$

Again, applying L'Hospital's rule

$\Rightarrow 27\underset{t\to 0}{lim}\frac{-\sin t}{6t}+b=0$

$\Rightarrow -\frac{27}{6}+b=0\text{or}b=\frac{9}{2}$