Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?
 A: You don't need to know any special limits or derivatives, you can do it with trig identities:
From
$$\sin3\theta=\sin(\theta+2\theta)=\sin\theta\cos2\theta+\cos\theta\sin2\theta=\sin\theta\cos2\theta+2\sin\theta\cos^2\theta$$
we have, letting $\theta=2x$,
$${\sin6x\over\sin2x}={\sin2x(\cos4x+2\cos^22x)\over\sin2x}=\cos4x+2\cos^22x$$
and thus
$$\lim_{x\to0}{\sin6x\over\sin2x}=\lim_{x\to0}(\cos4x+2\cos^22x)=\cos0+2\cos^20=1+2=3$$
(The final evaluation relies, of course, on knowing that the cosine function is continuous.)
A: As user Daniel has written you can express the ratio $\frac{\sin 6x}{\sin 2x}$ in a way which makes it amenable to the use of standard limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1\ .$$ Notice that $$\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin 2x}{2x} = \lim_{x\to 0} \frac{\sin 6x}{6x}\ .$$ As long as the argument $x$ is not equal to zero (and $\sin 2x \neq 0$) you can prolong the ratio $\frac{\sin 6x}{\sin 2x}$ as follows.
$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{2x} \cdot \frac{2x}{\sin 2x} = \underbrace{\frac{\sin 6x}{6x}}_{\to 1} \cdot \frac{6}{2} \cdot \underbrace{\frac{1}{\frac{\sin 2x}{2x}}}_{\to\frac{1}{1}} \to 1 \cdot 3 \cdot 1 = 3 \ .$$ Hence you obtain the result $$\lim_{x\to 0}\frac{\sin 6x}{\sin 2x} = 3 \ .$$
A: $$\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)}$$ $$= \lim_\limits{x\to 0}\frac{6}{2} \cdot \frac{\frac{\sin(6x)}{6x}}{\frac{\sin(2x)}{2x}}$$ $$=\frac{6}{2}$$ $$=3$$
A: HINT: rewrite the quotient in the form $$\frac{\sin(6x)}{6x}\frac{1}{\frac{\sin(2x)}{2x}}\cdot 3$$
A: $$\frac{\sin(6x)}{{\sin(2x)}} =\frac{\sin(6x)}{6x} \cdot \frac{2x}{{\sin(2x)} }\cdot \frac{6}{2} $$
A: Since the limit of both the numerator and denominator are 0 use L'Hopitals rule. You then get $\lim_{x \to 0} \frac{6cos(6x)}{2cos(2x)}=\frac{6}{2}=3$
A: The key is to use Hopital rule:
$$\begin{align}
\lim_\limits{x\to 0}\frac{\sin(x)}{x} &= \lim_\limits{x\to 0}\frac{\sin'(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{\cos(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{1}{1}\\
&=1
\end{align}
$$
hence
$$
\begin{align}
\lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} &= \lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} \cdot \frac{2x}{6x} \cdot \frac{6}{2}\\
&= \lim_{x\to 0} \frac{\sin(6x)}{6x} \cdot \frac{\sin(2x)}{6x} \cdot 3\\
&= 1 \cdot 1 \cdot 3\\
&=3
\end{align}$$
A: $$\lim_{x\to0}\frac{2x}{\sin2x}\cdot\frac{\sin6x}{6x}=1$$
A: In asymptotic terms, $\displaystyle \lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$ means that "$\sin(x)$ behaves like $x$ around $0$". 
To this understanding, $\sin(6x)$ behaves like $6x$ and $\sin(2x)$ behaves like $2x$ around $0$. 
Therefore, $\displaystyle \frac{\sin(6x)}{\sin(2x)}$ behaves like $\displaystyle \frac{6x}{2x}=3$ around $0$. 
In other words,  $\displaystyle \lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = 3.$
A: Taking a look at this page we find
$$1+2\cos u + 2 \cos 2 u + 2 \cos 3 u + ... + 2 \cos n u=\frac{\sin \frac{2n+1}{2}u}{\sin\frac u2}$$
In the left hand side set $u=4x$ and $n=1$ so that the left hand side becomes $$\frac{\sin 6x}{\sin 2x},$$ therefore we have $$1+2\cos 4x = \frac{\sin 6x}{\sin 2x}$$
