Question
Evaluate $\lim _\limits{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$
$\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$ $=\lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x} \cdot \frac{2 x}{\sin 2 x} \cdot 2\right]$
$=2 \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div\left[\frac{\sin 2 x}{2 x}\right]$
$=2 \cdot \lim _{4 x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div \lim _{2 x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]$
$=2.1 ÷1=2(\text { as } x \rightarrow 0,4 x \rightarrow 0 \text { and } 2 x \rightarrow 0)$
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