Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$ Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$ 
I'm a little confused on limit trig.  Am i suppose to simplify tan or do I use the derivative quotient rule?
Please Help!!!
 A: Recall the following limits:


*

*$\lim_{x \to 0}\dfrac{\sin(ax)}{x} = a$

*$\lim_{x \to 0}\dfrac{\tan(bx)}{x} = b$


Note that
$$\dfrac{\tan(16x)}{\sin(2x)} = \dfrac{\dfrac{\tan(16x)}{x}}{\dfrac{\sin(2x)}x}$$
Can you finish it off now?
A: $$
\tan16x=\frac{\sin(16x)}{\cos(16x)}
$$
$$
\sin(16x)=2\cdot \sin(8x)\cos(8x)
$$
$$
\sin(16x)=2\cdot 2\cdot\sin(4x)\cdot \cos(4x)\cdot \cos(8x)
$$
$$
\sin(16x)=2\cdot2\cdot2\cdot\sin(2x)\cdot\cos(2x)\cdot\cos(4x)\cdot\cos(8x)
$$
$$
\lim_{x\to0}\frac{\tan16x}{\sin2x}
$$
$$
\lim_{x\to0}\frac{2\cdot2\cdot2\cdot\sin(2x)\cdot\cos(2x)\cdot\cos(4x)\cdot\cos(8x)}{\sin2x}
$$
$$
\lim_{x\to0}2\cdot2\cdot2\cdot\cos(2x)\cdot\cos(4x)\cdot\cos(8x)=2\cdot2\cdot2=8
$$
A: Hint: $$\frac{\tan 16x}{\sin 2x} = \frac{\sin 16x}{\cos 16 x}\frac{1}{\sin 2x} = \frac{\sin 16x}{16 x}\frac{1}{\cos 16 x}\frac{2x}{\sin 2x}\frac{16 x}{2x}= \frac{\sin 16x}{16 x}\frac{2x}{\sin 2x}\frac{8}{\cos 16 x}.$$
A: Depends on what you want to use.
Informally, for small $x$, $\tan x \approx x$ and $\sin x \approx x$, so you would expect the limit to be $8$ ($=\frac{16x}{2x}$).
Formally. Presumably you can use $\lim_{x \to 0} \frac{ \sin x} {x} = 1$.
Then $\frac{\tan (16 x ) } { \sin (2 x) } = \frac{\sin (16 x ) } { \cos (16 x ) \sin (2 x) } = \frac{16x}{2x} \frac{\sin (16 x ) } { 16 x } \frac{2 x } {  \sin (2 x) } \frac{1 } { \cos (16 x ) }$. Hence we have
$$\lim_{x \to 0} \frac{\tan (16 x ) } { \sin (2 x) } = \frac{16}{2} ( \lim_{x \to 0} \frac{\sin (16 x ) } { 16 x } ) ( \lim_{x \to 0} \frac{2 x } {  \sin (2 x) } ) ( \lim_{x \to 0} \frac{1 } { \cos (16 x ) }) = \frac{16}{2}$$
A: Rewrite $\tan 16 x = \frac{\sin 16 x}{\cos 16 x}$, then keep expanding the numerator until you get $8 \sin 2 x \cos 2 x \cos 4x \cos 8x$, then something cancels out and all $\cos ax$ tend to 1. What do you have left?
EDIT: another way is (if you are allowed to do it) is to expand both numerator and denominator in Maclaurin series (Taylor series around $x_0=0$). A lot of stuff cancels out, and you get the same solution. 
