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

I know I can use L'Hospital's but I want to understand this particular explanation. They seem to skip something, and I'm not seeing the connection:

The limit is $\frac{6}{2}=3$ since $\lim_{x\to 0}\frac{\sin(x)}{x}=1$

JackOfAll
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3 Answers3

7

For $x \neq 0$ and $x$ close to zero, we have

$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{\sin 2x} = 3 \cdot \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}$$

See what to do now?

Simon S
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    Technically you don't even need $x$ close to zero to write what you've written. :) – Ian Apr 09 '15 at 14:56
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    No, but I wanted to avoid $2x = k\pi$. You know what I mean! ;-) – Simon S Apr 09 '15 at 14:56
  • Nice--always amuses me not just how bad solutions are for math solutions manuals (that is where I'm assuming this explanation comes from) but also how they are flat wrong. No chance of a typo for the author to save face here... +1 – Daniel W. Farlow Apr 09 '15 at 14:59
  • Not sure what you're "hinting" at here, @MagicMan. – Ted Shifrin Apr 09 '15 at 15:10
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    @TedShifrin What do you mean? I just mean the answer is a garbage answer (the "explanation" provided to the OP, that is). – Daniel W. Farlow Apr 09 '15 at 15:12
  • @MagicMan: I think it's correct. What is your problem? – Ted Shifrin Apr 09 '15 at 15:12
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    @TedShifrin Which answer do you think is correct? The one by Simon S or the one from the presumed solutions manual? – Daniel W. Farlow Apr 09 '15 at 15:14
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    You're commenting on @SimonS's answer, so naturally I presumed your comments related to his explanation. Very confusing. BTW, I doubt the explanation the OP give came from any solutions manual. It probably came from a tutor. – Ted Shifrin Apr 09 '15 at 15:17
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    @TedShifrin Interesting. I didn't think it would be possible for someone to mistake my comment to be deliberately belittling the answerer...only to give him +1 at the end. I can see the possibility for confusion though. Well, now we're on the same page--Simon S has the right way; the solutions manual with $2$ as the answer is clearly wrong. – Daniel W. Farlow Apr 09 '15 at 15:19
  • I am not following the solution, and am more interested in understanding the explanation given in the OP. It is written as if that limit makes the answer obvious with almost no work. Did they skip a major leap? I have no idea how to connect the original problem to $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ – JackOfAll Apr 10 '15 at 14:13
  • As $\lim_{x\to 0}{\sin x \over x} = 1$, it follows that for any $a \neq 0$ that

    $$\lim_{x\to 0}{\sin ax \over ax} = 1$$ This can be proven formally using an $\epsilon-\delta$ proof if you have seen those. If not, this should hopefully be intuitively reasonable as $x \to 0$ iff $ax \to 0$ iff $x/a \to 0$.

    Given all that, it follows that the expression above that

    $$\lim_{x \to 0} \frac{\sin 6x}{\sin 2x} = 3 \cdot \left( \lim_{x \to 0} \frac{\sin 6x}{6x} \right) \left( \lim_{x \to 0} \frac{2x}{\sin 2x} \right) = 3 \cdot 1 \cdot 1 = 3$$

     – Simon S Apr 10 '15 at 15:34
  • The argument in the original post is almost the same. Corrected and connecting a couple of the dots, it could be conceived of as this:

    $$\lim_{x \to 0} \frac{\sin 6x}{\sin 2x} = \left( \lim_{x \to 0} \frac{\sin 6x}{x} \right) \left( \lim_{x \to 0} \frac{x}{\sin 2x} \right) = \frac{\lim_{x \to 0} \frac{\sin 6x}{x} }{ \lim_{x \to 0} \frac{\sin 2x}{x} } = \frac{6 \cdot \lim_{x \to 0} \frac{\sin 6x}{6x} }{2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} } = {6 \over 2 } = 3$$

     – Simon S Apr 10 '15 at 15:38
  • What enrages me about the original explanation is that is so off-handedly implies the connection is obvious, and requires no actual work. There is no way in hell they intended the reader to be deriving all this stuff in these replies. Sadly, I still have absolutely no idea how lim sin(x)/x has anything to do with the 6x/2x problem. Sigh. Can someone please just spell it out for me, and please don't leave the last step out, all cute so I can finish. This has dragged on way longer than it should. – JackOfAll Apr 12 '15 at 23:10
  • I don't understand what you are now asking. I have been explicit in my previous two comments, giving two solutions. Which part do you not understand? – Simon S Apr 12 '15 at 23:15
  • I don;t understand how that can be the solution, yet, the manual implies it's much more direct and obvious. Single worst "solution" I have ever seen in a book. You're just supposed to know how to split up the fraction like that? – JackOfAll Apr 12 '15 at 23:17
  • How did you introduce that division step? (Step 2 to Step 3) – JackOfAll Apr 12 '15 at 23:20
  • If $\lim_{x \to a} f(x)$ exists and is not equal to zero, then

    $$\lim_{x \to a} {1 \over f(x)} = {1 \over \lim_{x \to a} f(x)}$$

     – Simon S Apr 12 '15 at 23:23
2

Another way is to show that $$ \frac{\sin 6x}{\sin 2x} = \cos 4x + 2\cos^2(2x) $$ Which can be proven by applying $\sin(A+B)=\sin A \cos B \sin B \cos A$ twice. First on $\sin(4x+2x)$ and then on $\sin(2x+2x)$.

N3buchadnezzar
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    Aw come on. This is not an appropriate solution for a beginning calculus question. – Ted Shifrin Apr 09 '15 at 15:15
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    While it is a slightly difficult route to answer I will still give an upvote. So +1. A better approach would be to use $\sin 3a = 3\sin a - 4\sin^{3}a$ and putting $a = 2x$ we get $$\frac{\sin 6x}{\sin 2x} = 3 - 4\sin^{2}(2x)$$ and then limit is $3$ as $x \to 0$. This is based on the the fact that $\lim_{x \to 0}\sin x = 0$ and this is slightly easier to prove compared to the limit $\lim_{x \to 0}(\sin x)/x = 1$. – Paramanand Singh Apr 09 '15 at 22:46
  • I am not following the solution, and am more interested in understanding the explanation given in the OP. It is written as if that limit makes the answer obvious with almost no work. Did they skip a major leap? I have no idea how to connect the original problem to $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ – JackOfAll Apr 10 '15 at 14:13
  • @ParamanandSingh, Nice remark. +1 – Idris Addou Jun 14 '15 at 06:07
  • @Idris: see another example of this approach http://math.stackexchange.com/a/1323410/72031 – Paramanand Singh Jun 14 '15 at 06:20
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It's shorter with equivalents: $$\frac{\sin 6x}{\sin 2x}\sim_0\frac{6x}{2x}=3$$ hence the limit is $3$.

Addendum :For people who haven't studied this notion, here are a few explanations: we say that two functions, defined and $\neq 0$ in a neighbourhood of $a$ (except perhaps at $a$ itself) are equivalent at $a$ if: $$ \lim_{x\to a}\frac{f(x)}{g(x)}=1 $$ This is denoted $f(x)\sim_a g(x)$. It is indeed an equivalence relation between functions which are defined in some neighbourhood of $a$, except perhaps at $a$ itself.

Main facts about equivalence:

  • Let $l$ be a number. $f(x)\sim_a l\iff\lim_{x\to a} f(x)=l $.
  • If $f(x)\sim_a g(x)$ and $f_1(x)\sim_a g_1(x)$, then $$\begin{cases}f(x)f_1(x)\sim_a g(x)g._1(x)\\[1ex] \dfrac{f(x)}{f_1(x)}\sim_a \dfrac{g(x)}{g_1(x)}\end{cases} $$
  • However, equivalence is not compatible with addition or subtraction.
  • The first nonzero term of the Taylor polynomial of a function is the simplest equivalent to that function. Thus: $$\sin ax \sim_0 ax, \quad\tan ax \sim_0 ax\quad \ln(1+x)\sim_0 x$$

We often may use equivalents to find limits. The main advantage of the method is getting rid of irrelevant technical details by replacing more or less complicated expressions by simpler ones.

One of the most famous equivalences is Stirling's formula for approximating big factorials: $$n!\sim_\infty\sqrt{2\pi n}\Bigl(\frac ne\Bigr)^n.$$

Bernard
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  • Someone here doesn't like asymptotic analysis? – Bernard Apr 09 '15 at 16:47
  • This answer is correct and hence should be upvoted. +1. Also this is same as the answer by Simon S but uses a very compact notation of $\sim $ and while this is very useful once you get the hang of it, it should be used with caution. See my answer here regarding the caution needed when applying this technique.http://math.stackexchange.com/a/1226451/72031 – Paramanand Singh Apr 09 '15 at 22:50
  • I am not following the solution, and am more interested in understanding the explanation given in the OP. It is written as if that limit makes the answer obvious with almost no work. Did they skip a major leap? I have no idea how to connect the original problem to $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ – JackOfAll Apr 10 '15 at 14:13
  • The link might be that $\lim_{x\to 0}\frac{\sin x}x=1$ is exactly the definition of $\sin x\sim_0 x$. All you have to know in addition is that it is an equivalence relation (between functions defined in a neighbourhood of $0$), that it is compatible with multiplication and division, and that a function $f$ has limit $l$ at $0$ if and only if $f(x)\sim_0 l$. The advantage of this notion is you don't have to care for irrelevant technical details. – Bernard Apr 10 '15 at 14:48
  • No clue what the tilda deal is, but there is no way that original explanation intended there to be any real work, or bizarre notation. Thanks for trying. – JackOfAll Apr 12 '15 at 23:12