# How to find $\lim\limits_{x \to 0}\frac{\sin4x}{\sin2x}$?

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 ?

• Calculating this limit should be able to give the original limit. You can also recall that $\sin(2y)=2\sin(y)\cos(y)$. – Keen-ameteur Feb 25 at 16:29
• Notice that $$\frac{\frac{\sin(4x)}{4x}}{\frac{\sin(2x)}{2x}} = \frac{2x}{4x} \frac{\sin(4x)}{\sin(2x)} = \frac{1}{2}\frac{\sin(4x)}{\sin(2x)}.$$ – Viktor Glombik Feb 25 at 16:31

As Keen-ameteur points out, we can use:

$$\lim_{x \to 0} \frac {\sin 4x}{\sin 2x}=\lim_{x \to 0} \frac {2\sin 2x \cos 2x}{\sin 2x}=\lim_{x \to 0}2\cos 2x=2\cos 2\cdot 0=2\cos 0=2\cdot 1=2$$

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 $$\lim_{x \rightarrow 0} \frac{\sin(4x)}{\sin(2x)} = 2.$$

You are pretty close,

$$\lim\limits_{x \to 0}\dfrac{\sin4x}{\sin2x}=\lim\limits_{x \to 0}\dfrac{\dfrac{\sin4x}{4x}}{\dfrac{\sin2x}{\color{green}{2\cdot}2x}}=2.$$

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 $$\frac{2x}{2x}$$:

$$\lim\limits_{x \to 0}\frac{\sin4x}{\sin2x} = \lim\limits_{x \to 0}\frac{\frac{4x}{4x}\sin4x}{\frac{2x}{2x}\sin2x}=\cdots$$