2
$\begingroup$

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

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.

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

Hint:

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

And you returning to your previous problem.

$\endgroup$
  • $\begingroup$ Maybe you should divide by $6x,2x$ to help the PO understand this hint $\endgroup$ – Belgi May 10 '12 at 13:46
  • $\begingroup$ I have no idea why but I did not realize that you could do that to the problem. $\endgroup$ – toby yeats May 10 '12 at 13:46
  • 1
    $\begingroup$ @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. $\endgroup$ – mrs May 10 '12 at 13:48
  • 6
    $\begingroup$ Well, maybe another "trick": ${\sin 6x\over x}= 6\cdot{\sin 6x\over 6x}$ and ${\sin 2x\over x}=2\cdot{\sin 2x\over 2x}$. $\endgroup$ – David Mitra May 10 '12 at 13:53
  • 2
    $\begingroup$ 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. $\endgroup$ – TenaliRaman May 10 '12 at 14:11
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Am I suppose to have that triple angle identity memorized? $\endgroup$ – toby yeats May 10 '12 at 15:22
  • 1
    $\begingroup$ 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. $\endgroup$ – Simon Markett May 10 '12 at 15:37
  • $\begingroup$ Somehow I can't help but feel that using l'Hôpital here is like cutting toothpicks with a chainsaw... $\endgroup$ – J. M. is a poor mathematician May 13 '12 at 4:28
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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. $\lim_{x\to 0}(2\cos2x + 1) = 3$.

$\endgroup$
  • $\begingroup$ 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 $\lim_{x\to 0}\frac{\sin(\sqrt{2} x)}{\sin x}$. $\endgroup$ – André Nicolas May 10 '12 at 14:03
  • $\begingroup$ 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. en.wikipedia.org/wiki/LaTeX might be a good place to find out more. $\endgroup$ – Sam Jones May 10 '12 at 15:07

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.