# Not sure how to evaluate $\lim_{x\to 0}\frac{\sin6x}{\sin2x}$ (without l'Hospital) [duplicate]

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

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

• btw, the answer is wrong. See below. Apr 9, 2015 at 14:54
• Sorry, that was a typo. I corrected the correct solution above. Apr 12, 2015 at 23:08
• why make this so hard? use the fact that $\sin(small) = \small + \cdots$ so that $sin(2x) = 2x + \cdots, \sin(6x) = 6x + \cdots.$ therefore $\frac{\sin 2x}{\sin 6x} = \frac{2x}{6x} + \cdots \to \frac 13$
– abel
Apr 13, 2015 at 1:10
• I'm not following what you did there, Apr 18, 2015 at 19:36

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?

• Technically you don't even need $x$ close to zero to write what you've written. :)
– Ian
Apr 9, 2015 at 14:56
• No, but I wanted to avoid $2x = k\pi$. You know what I mean! ;-) Apr 9, 2015 at 14:56
• @TedShifrin What do you mean? I just mean the answer is a garbage answer (the "explanation" provided to the OP, that is). Apr 9, 2015 at 15:12
• @TedShifrin Which answer do you think is correct? The one by Simon S or the one from the presumed solutions manual? Apr 9, 2015 at 15:14
• 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. Apr 9, 2015 at 15:17

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 $\sin(2x+2x)$.

• Aw come on. This is not an appropriate solution for a beginning calculus question. Apr 9, 2015 at 15:15
• 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 $\lim_{x \to 0}(\sin x)/x = 1$. Apr 9, 2015 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 $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ Apr 10, 2015 at 14:13
• @ParamanandSingh, Nice remark. +1 Jun 14, 2015 at 6:07
• @Idris: see another example of this approach math.stackexchange.com/a/1323410/72031 Jun 14, 2015 at 6:20

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.

• 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: $$n!\sim_\infty\sqrt{2\pi n}\Bigl(\frac ne\Bigr)^n.$$

• Someone here doesn't like asymptotic analysis? Apr 9, 2015 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.math.stackexchange.com/a/1226451/72031 Apr 9, 2015 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 $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ Apr 10, 2015 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. Apr 10, 2015 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. Apr 12, 2015 at 23:12