Question

the limit as x-> 0 for
cos(2x)-1 / cos(3x)-1

Answer 1

ok if you plug in 0 you get 0/0
so you can use l'hospital's rule
-2sin(2x)/-3sin(3x) give me just a sec i'm going to use the equation thingy

Answer 2

\[\lim_{x \rightarrow 0}\frac{2}{3} \frac{\sin(2x)}{2x}\frac{\sin(3x)}{3x}6x^2\]

Answer 3

2/3(1)(1)(6)(0)=0

Answer 4

it says the answer is 4/9

Answer 5

\[\lim_{x \rightarrow 0} \frac{\cos(2x)-1}{\cos(3x)-1}\]

Answer 6

is that right?

Answer 7

yep

Answer 8

ok let me see

Answer 9

oops i see a mistake

Answer 10

i put sin(3x) in the numerator do you see that?

Answer 11

\[\lim_{x \rightarrow 0}\frac{2}{3}\frac{\sin2x}{2x}\frac{3x}{\sin(3x)}\frac{2x}{3x}\]

Answer 12

2/3*1*1*2/3=4/9

Answer 13

got it? do you remember what sinx/x goes to as x goes to 0?