2
$\begingroup$

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 ?

$\endgroup$
  • 5
    $\begingroup$ Calculating this limit should be able to give the original limit. You can also recall that $\sin(2y)=2\sin(y)\cos(y)$. $\endgroup$ – Keen-ameteur Feb 25 at 16:29
  • 2
    $\begingroup$ Notice that $$\frac{\frac{\sin(4x)}{4x}}{\frac{\sin(2x)}{2x}} = \frac{2x}{4x} \frac{\sin(4x)}{\sin(2x)} = \frac{1}{2}\frac{\sin(4x)}{\sin(2x)}.$$ $\endgroup$ – Viktor Glombik Feb 25 at 16:31
7
$\begingroup$

As Keen-ameteur points out, we can use:

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

You are pretty close,

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.