Question

lim x->0 tan2x/sin4x

Answer 1

tan of zero is zero, sin of zero is zero. think you need l'hopitals on this. have you done derivatives or is this precalc?

Answer 2

answer is 1/2
you do not need l'hopital rule .... of course it will work either way

im guessing @hewsmike will provide explanation

Answer 3

From\[sin(A+B) = sinAcosB + cosAsinB\]\[sin(4x) = 2sin(2x)cox(2x)\]then\[\frac{tan(2x)}{sin(4x)}=\frac{sin(2x)}{cos(2x)}\frac{1}{2sin(2x)cos(2x)}\]\[=\frac{1}{2}\frac{1}{cos^{2}(2x)}\]so as x -> 0, 2x -> 0, cos(2x) -> 1 and the result follows.

Answer 4

thanks

Answer 5

$$\lim \limits_{x\to 0} \frac{\sin x}{x}=1$$ and
$$\lim \limits_{x\to 0} \frac{\tan x}{x}=\lim \limits_{x\to 0} \frac{\sin x}{x}\frac{1}{\cos x}=\lim \limits_{x\to 0} \frac{\sin x}{x}\lim \limits_{x\to 0} \frac{1}{\cos x}=1$$
from this,
$$\lim \limits_{x\to 0} \frac{\tan 2x}{\sin 4x}=\lim \limits_{x\to 0} \frac{1}{2}\frac{\tan 2x}{2x}\frac{4x}{\sin 4x}=\frac{1}{2}\left[\lim \limits_{x\to 0} \frac{\tan 2x}{2x}\right]\frac{1}{\lim \limits_{x\to 0} \frac{\sin 4x}{4x}}=\frac{1}{2}\times1\times1=\frac{1}{2}$$