limit involving trig functions I'm not sure how to solve this limit.
$$
\lim_{x\to 0} \frac{\tan 6x}{\sin 2x}
$$
After some rearranging I get this.
$$
\lim_{x\to 0} \frac{\sin 6x}{\cos 6x} \cdot \lim_{x\to0} \frac{1}{\sin 2x}
$$
I know the limit is 3, but how do you get there using algebra? What am I missing?
 A: $$\frac{\tan{6x}}{\sin{2x}} = \frac{\tan{6x}}{6x}\cdot\frac{2x}{\sin{2x}}\cdot \frac{6}{2}$$
And 
$$\lim_{x\rightarrow 0} \frac{\tan{6x}}{6x} = 1$$
$$\lim_{x\rightarrow 0} \frac{2x}{\sin{2x}} = 1$$
I'll leave the rest to you.
A: Notice, $$\lim_{x\to 0}\frac{\tan 6x}{\sin 2x}=\lim_{x\to 0}\frac{6x\left(\frac{\tan 6x}{6x}\right)}{2x\left(\frac{\sin 2x}{2x}\right)}$$
$$=3\lim_{x\to 0}\frac{\left(\frac{\tan 6x}{6x}\right)}{\left(\frac{\sin 2x}{2x}\right)}$$
$$=3\cdot \frac{\lim_{x\to 0}\frac{\tan 6x}{6x}}{\lim_{x\to 0}\frac{\sin 2x}{2x}}$$
$$=3\cdot \frac{1}{1}=\color{red}{3}$$
A: $$\lim_{x\to0}\space\frac{\tan(6x)}{\sin(2x)}=$$

Use $\tan(6x)=\frac{\sin(6x)}{\cos(6x)}$:

$$\lim_{x\to0}\space\frac{\frac{\sin(6x)}{\cos(6x)}}{\sin(2x)}=$$
$$\lim_{x\to0}\space\frac{\sin(6x)}{\cos(6x)\sin(2x)}=$$
$$\left[\lim_{x\to0}\space\frac{1}{\cos(6x)}\right]\left[\lim_{x\to0}\space\frac{\sin(6x)}{\sin(2x)}\right]=$$
$$\left[\space\frac{1}{\cos(6\cdot0)}\right]\left[\lim_{x\to0}\space\frac{\sin(6x)}{\sin(2x)}\right]=$$
$$\left[\space\frac{1}{1}\right]\left[\lim_{x\to0}\space\frac{\sin(6x)}{\sin(2x)}\right]=$$
$$\left[1\right]\left[\lim_{x\to0}\space\frac{\sin(6x)}{\sin(2x)}\right]=$$
$$\lim_{x\to0}\space\frac{\sin(6x)}{\sin(2x)}=$$
$$\lim_{x\to0}\space\frac{\frac{\text{d}}{\text{d}x}\left(\sin(6x)\right)}{\frac{\text{d}}{\text{d}x}\left(\sin(2x)\right)}=$$
$$\lim_{x\to0}\space\frac{6\cos(6x)}{2\cos(2x)}=$$
$$\lim_{x\to0}\space\frac{3\cos(6x)}{\cos(2x)}=$$
$$\frac{3\cos(6\cdot0)}{\cos(2\cdot0)}=\frac{3\cos(0)}{\cos(0)}=\frac{3}{1}=3$$
A: With $t=2x$,
$$
\lim_{x\to 0} \frac{\tan(6x)}{\sin(2x)}=\lim_{t\to 0} \frac{\sin(3t)}{\sin(t)}\lim_{t\to 0} \frac1{\cos(3t)}=
\lim_{t\to 0} \frac{3\sin(t)-4\sin^3(t)}{\sin(t)}\cdot1=3-4\cdot0.$$
No need for extra limit formulas.
