Question

Evaluate $\underset{x\to 0}{\lim }$$\frac{sin4x}{sin2x}=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }$$\frac{sin4x}{sin2x}$
$=\underset{x\to 0}{\lim }$$\left[\frac{sin4x}{4x}\cdot \frac{2x}{sin2x}\cdot 2\right]$
$=2\cdot \underset{x\to 0}{\lim }$$\left\{\left[\frac{sin4x}{4x}\right]\cdot \left[\frac{sin2x}{2x}\right]\right\}$
$=2\cdot \underset{4x\to 0}{\lim }$$\left[\frac{sin4x}{4x}\right]\cdot \underset{2x\to 0}{\lim }$$\left[\frac{sin2x}{2x}\right]$
$=2\cdot 1\cdot 1=2$ (as $x\to 0$, $4x\to 0$ and $2x\to 0$)