Question

If $\underset{x\to 0}{lim}\frac{\sin 3x+a\sin 2x}{x^3}=\lambda$ and $\lambda$ is finite non zero real number, then $\lambda =$

• A. $-\frac{1}{2}$
• B. $\frac{23}{6}$
• C. $-\frac{5}{2}$
• D. $-\frac{7}{2}$

Answer 1

The correct option is C $-\frac{5}{2}$

$\underset{x\to 0}{lim}\frac{\sin 3x+a\sin 2x}{x^3}$
Using expansions, we have:
$\underset{x\to 0}{lim}\frac{(3x-\frac{(3x{)}^3}{3!}+\frac{(3x{)}^5}{5!}-......)+a(2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-......)}{x^3}$
For limit to exist, coefficient of $x=3+2a$ has to be $0$
$\therefore a=-\frac{3}{2}$
So, the value of the limit $\lambda =-\frac{27}{6}-\frac{a\cdot 8}{6}$
$\Rightarrow \lambda =\frac{-5}{2}$