Trig limit of $\lim\limits_{x\to 0}\frac{\sin{6x}}{\sin{2x}}$ $$\lim_{x\to 0}\frac{\sin{6x}}{\sin{2x}}$$
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.
 A: 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 $\lim_{x\to 0}\frac{\sin(x)}{x}=1$:
$$\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\to 0}3\frac{\frac{\sin(6x)}{6x}}{\frac{\sin(2x)}{2x}}=3\frac11=3$$
A: Hint:
$$\frac{\sin{6x}}{\sin{2x}}=\frac{\frac{\sin{6x}}{x}}{\frac{\sin{2x}}{x}}$$
And you returning to your previous problem.
A: $\lim_{x\rightarrow 0}\frac{\sin(6x)}{\sin(2x)}=\lim_{x\rightarrow 0}\frac{\sin(6x)\cdot(6x/6x)}{\sin(2x)\cdot(2x/2x)}=\lim_{x\rightarrow 0}\frac{6x}{2x}\frac{\frac{\sin(6x)}{6x}}{\frac{\cos(2x)}{2x}}=\lim_{x\rightarrow 0}\frac{6x}{2x}\cdot\lim_{x\rightarrow 0}\frac{\lim_{x\rightarrow 0}\frac{\sin(6x)}{6x}}{\lim_{x\rightarrow 0}\frac{\sin(2x)}{2x}}=\frac{6}{2}\cdot\frac{1}{1}=3.$
A: 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. $\lim_{x\to 0}(2\cos2x + 1) = 3$.
A: 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.
A: There are way too many ways to do this , this is just meant to be an addition to the many approaches that can be used;
If $\lambda(x)$ is an infinitesimal as $x \to 0$ we've
$$\sin(\lambda(x)) \sim \lambda(x)$$
Therefore for the present case the limit is just $\frac{6}{2}=3$
