Question

find the limit when x --->0
(cos3x-1)/sin2x

Answer 1

\[\lim_{u \rightarrow 0}\frac{\cos(u)-1}{u}=? \\ \lim_{u \rightarrow 0}\frac{u}{\sin(u)}=?\]

Answer 2

\[\lim_{x \rightarrow 0}\frac{ \cos3x-1 }{ \sin2x }=\lim_{x \rightarrow 0}\frac{ 2x }{ 3x }\frac{ 3x }{2x}\frac{ \cos3x-1}{\sin2x}\]

Answer 3

\[=\lim_{x \rightarrow 0}\frac{ 3 }{ 2}\frac{ \cos3x-1 }{ 3x }\frac{ 2x }{ \sin2x }\]
then use the limit of product is the product of limits
take the limit of each fraction
and you are done!

Answer 4

all the depends on what you know about those two identical limits
@freckles has also gave you the hint

Answer 5

I don't understand the first step

Answer 6

the fist step:
i multiplied top and bottom by 2x and also by 3x
so i can have something that i know its limit

Answer 7

realize that 6x^2/6x^2 is just 1

Answer 8

if for instance i canceled the 2, 3 and x
i would get the initial expression cos3x-1/sin2x, yes?

Answer 9

Do you know the limits @freckles mentioned about or not?

Answer 10

above*