How to find the limit $$\lim\limits_{x \to 0}\dfrac{\sin4x}{\sin2x}\,?$$
Should I do $$\lim\limits_{x \to 0}\dfrac{\dfrac{\sin4x}{4x}}{\dfrac{\sin2x}{2x}}\,?$$
This doesn't seem to look right, could you show me the way ?
$\endgroup$ 24 Answers
$\begingroup$As Keen-ameteur points out, we can use:
$\endgroup$ $\begingroup$$$\lim_{x \to 0} \frac {\sin 4x}{\sin 2x}=\lim_{x \to 0} \frac {2\sin 2x \cos 2x}{\sin 2x}=\lim_{x \to 0}2\cos (2x)=2\cos (2\cdot 0)=2\cos 0=2\cdot 1=2$$
You can indeed write :
$$ \frac{\sin(4x)}{\sin(2x)} = 2 \cdot\frac{\sin(4x)}{4x} \cdot \frac{2x}{\sin(2x)} $$
And now, because$$\frac{\sin(x)}{x} \xrightarrow{x \rightarrow 0} 1 $$
you get $$\lim_{x \rightarrow 0} \frac{\sin(4x)}{\sin(2x)} = 2.$$
$\endgroup$ $\begingroup$You are pretty close,
$$\lim\limits_{x \to 0}\dfrac{\sin4x}{\sin2x}=\lim\limits_{x \to 0}\dfrac{\dfrac{\sin4x}{4x}}{\dfrac{\sin2x}{\color{green}{2\cdot}2x}}=2.$$
$\endgroup$ $\begingroup$You can't arbitrarily throw in a $4x$ and $2x$, since that changes the value of the expression. What you can do, however, is multiply by $\frac{4x}{4x}$ and $\frac{2x}{2x}$:
$$\lim\limits_{x \to 0}\frac{\sin4x}{\sin2x} = \lim\limits_{x \to 0}\frac{\frac{4x}{4x}\sin4x}{\frac{2x}{2x}\sin2x}=\cdots$$
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