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$\begingroup$

$\lim_{x\to 0}\frac{\sin{6x}}{\sin{2x}}$

I have no idea at all on how to proceed. I am guessing there is some trig rule about manipulating these terms in some way but I can not find it in my notes.

I tried to make $\tan$ into $\dfrac\sin\cos$

$\frac{\sin6x}{\cos6x} \times \frac{1}{\sin2x}$

But this doesn't get me anywhere as far as I can tell.

  • 2
    @Cam: sounds awfully like a "Lucky Larry"... :)2012-05-13

5 Answers 5

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I also feel compelled to mention that, in general,

$\lim_{x\to 0}\frac{\sin(Ax)}{\sin(Bx)}=\frac{A}{B}$

which was proved for your case by others above.

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Three different proofs:

By L'Hôpital: $\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\to 0}\frac{6\cos(6x)}{2\cos(2x)}=\frac 62=3$

By trigonometric identity: We have $\sin(3x)=\sin(x)(4\cos^2(x)-1)$ and therefore $\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\to 0}\frac{\sin(2x)(4\cos^2(2x)-1)}{\sin(2x)}=\lim_{x\to 0}(4\cos^2(2x)-1)=4-1=3$

If you now $\lim_{x\to 0}\frac{\sin(x)}{x}=1$: $\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\to 0}3\frac{\frac{\sin(6x)}{6x}}{\frac{\sin(2x)}{2x}}=3\frac11=3$

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    Somehow I can't help but feel that using l'Hôpital here is like cutting toothpicks with a chainsaw...2012-05-13
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Hint:

$\frac{\sin{6x}}{\sin{2x}}=\frac{\frac{\sin{6x}}{x}}{\frac{\sin{2x}}{x}}$

And you returning to your previous problem.

  • 2
    I prefer this method (+1) over the others because it uses the more primary and useful limit $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$. This is a very useful thing to learn and remember. It tells you that for small enough x, sin(x) can be simply be approximated by x.2012-05-10
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$\lim_{x\rightarrow 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\rightarrow 0}\frac{\sin(6x)\cdot(6x/6x)}{\sin(2x)\cdot(2x/2x)}=\lim_{x\rightarrow 0}\frac{6x}{2x}\frac{\frac{\sin(6x)}{6x}}{\frac{\cos(2x)}{2x}}=\lim_{x\rightarrow 0}\frac{6x}{2x}\cdot\lim_{x\rightarrow 0}\frac{\lim_{x\rightarrow 0}\frac{\sin(6x)}{6x}}{\lim_{x\rightarrow 0}\frac{\sin(2x)}{2x}}=\frac{6}{2}\cdot\frac{1}{1}=3.$

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Try something like this

$\sin6x=2(\sin 3x \cos3x)$

$\sin2x=2 \sin x \cos x $

$\lim_{x\to 0}(\cos x) = 1$

You will have $\lim_{x\to 0}\frac{\sin3x}{\sin x}$ so using @J.M. $\lim_{x\to 0}(2\cos2x + 1) = 3$.

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    No problem, this site uses some form of latex for displaying equations, you have to surround the mathematics with \$ for an inline piece of mathematics and \$\$ for a "displayed" piece. http://en.wikipedia.org/wiki/LaTeX might be a good place to find out more.2012-05-10