Question

What are the steps to finding the limit for:
lim tan(6x)/sin(2x)
x==>0

Answer 1

have you tried l'hopital?

Answer 2

No, I have no idea what that is

Answer 3

i am going to guess it is 3

Answer 4

rewrite with only sines and cosines, then ignore the cosine part because as \(x\to 0\) you know \(\cos(x)\to 1\)

Answer 5

that leaves you with
\[\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}\] which is a trivial l'hopital problem, but if you have not got there yet (and i suspect that you have not) then you can rewrite this in the somewhat artificial and cumbersome form as
\[\frac{\sin(6x)}{6x}\times \frac{2x}{\sin(2x)}\times 3\]
then use the fact that
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\] to get and answer of 3