Question

$\underset{x\to 0}{\lim }$$\frac{sin3x}{sin2x}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $0$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }$ $\frac{sin3x}{sin2x}$
$=\underset{x\to 0}{\lim }$$\left(\frac{sin3x}{3x}\times \frac{2x}{sin2x}\times \frac{3}{2}\right)$
$=\frac{3}{2}\cdot$$\underset{3x\to 0}{\lim }$$\frac{sin3x}{3x}\times$$\underset{2x\to 0}{\lim }$$\frac{2x}{sin2x}$
$(\because$ as $x\to 0$, $3x\to 0$ and $2x\to 0)$
$=\frac{3}{2}\times 1\times 1=\frac{3}{2}$.