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

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.

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Show us your work! – Salech Alhasov May 10 '12 at 13:41
I don't really have any. – user138246 May 10 '12 at 13:42
$$\frac{\sin\,3x}{\sin\,x}=2\cos\,2x+1$$ might be useful... – J. M. May 10 '12 at 13:42
Cancel the sines and $x$'s to get $$ \lim_{x\to 0}\frac{\sin{6x}}{\sin{2x}}=\lim_{x\to 0}\frac{6}{2}=3. $$ – Cam McLeman May 13 '12 at 1:18
@Cam: sounds awfully like a "Lucky Larry"... :) – J. M. May 13 '12 at 4:26
up vote 5 down vote accepted



And you returning to your previous problem.

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Maybe you should divide by $6x,2x$ to help the PO understand this hint – Belgi May 10 '12 at 13:46
I have no idea why but I did not realize that you could do that to the problem. – user138246 May 10 '12 at 13:46
@jordan: The only way to solve such these limits is what Salech pointed above. Use the HINT and note that When $x$ tents to $0$, it will be Ok, if you write your fraction as above. No more tricks needed. – Babak S. May 10 '12 at 13:48
Well, maybe another "trick": ${\sin 6x\over x}= 6\cdot{\sin 6x\over 6x}$ and ${\sin 2x\over x}=2\cdot{\sin 2x\over 2x}$. – David Mitra May 10 '12 at 13:53
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. – TenaliRaman May 10 '12 at 14:11

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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Am I suppose to have that triple angle identity memorized? – user138246 May 10 '12 at 15:22
Nah, obviously not ;) But it's obtained by applying the usual identities a couple of times. I always have to look them up myself. You just need to know that you can reduce the factor somehow. – Simon Markett May 10 '12 at 15:37
Somehow I can't help but feel that using l'Hôpital here is like cutting toothpicks with a chainsaw... – J. M. May 13 '12 at 4:28

$\lim_{x\rightarrow 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\rightarrow 0}\frac{\sin(6x)\cdot(6x/6x)}{\sin(2x)\cdot(2x/2x)}=\lim_{x\rightarrow 0}\frac{6x}{2x}\frac{\frac{\sin(6x)}{6x}}{\frac{\cos(2x)}{2x}}=\lim_{x\rightarrow 0}\frac{6x}{2x}\cdot\lim_{x\rightarrow 0}\frac{\lim_{x\rightarrow 0}\frac{\sin(6x)}{6x}}{\lim_{x\rightarrow 0}\frac{\sin(2x)}{2x}}=\frac{6}{2}\cdot\frac{1}{1}=3.$

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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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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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Even though in this case the limit result is not hard to get by playing with trigonometric identities, exploiting knowledge about $\lim_{u\to 0}\frac{\sin u}{u}$ seems more appropriate, since this is a calculus course. Moreover, the idea deals with equal ease with $\lim_{x\to 0}\frac{\sin(\sqrt{2} x)}{\sin x}$. – André Nicolas May 10 '12 at 14:03
Thank you @SamJones – Likurg May 10 '12 at 14:13
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. might be a good place to find out more. – Sam Jones May 10 '12 at 15:07

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