Question

$\displaystyle \lim_{x \to 0}$$\frac{sin\,3x}{sin\,2x}=$

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

Answer 1

Answer: (C)

$\displaystyle \lim_{x \to 0}$ $\frac{sin\,3x}{sin\,2x}$
$=\displaystyle \lim_{x \to 0}$$\left(\frac{sin\,3x}{3x}\times\frac{2x}{sin\,2x} \times \frac{3}{2}\right)$
$=\frac{3}{2} \cdot$$\displaystyle \lim_{3x \to 0}$$\frac{sin\,3x}{3x} \times$$\displaystyle \lim_{2x \to 0}$$\frac{2x}{sin\,2x}$
$(\because$ as $x \to 0$, $3x \to 0$ and $2x \to 0)$
$=\frac{3}{2} \times 1\times1=\frac{3}{2}$.