Not sure how to evaluate $\lim_{x\to 0}\frac{\sin6x}{\sin2x}$ (without l'Hospital) $$\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}$$
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 $\lim_{x\to 0}\frac{\sin(x)}{x}=1$
 A: 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?
A: 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 $\sin(2x+2x)$.
A: 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:
$$n!\sim_\infty\sqrt{2\pi n}\Bigl(\frac ne\Bigr)^n.$$
