/sci/ - Science & Math

[derp No.4310266]

Hey guys!
I'm trying to calculate:
Lim x->0
(Ln(cos 2x)/Ln(cos3x)) it's 0/0 so I'm using l'hopital's rule and calculate the derivative
((Ln(cos2x))' = (1/cos2x)*(-sin2x)*x
((Ln(cos3x))' = (1/cos3x)*(-sin3x)*x
the - and x go away. then I'm getting tg2x/tg3x
Then I'm getting 0/0 again so I'm using l'hopital's rule again and I get cos^2 3x/cos^2 2x and that's 1/1=1

However the answer is given as 4/9, so, what am I doing wrong?

[derp No.4310290]

bump

[Anonymous No.4310314]

>>4310290
>((Ln(cos2x))' = (1/cos2x)*(-sin2x)*x

no I dont think so, try that again.

[Anonymous No.4310323]

keep using l'hopitals rule faggot
Log(cos) -> -tan -> -sec^2
Sec(0)=1
There you go, you've gotten rid of your 0/0 problem
you're not taking derivatives correctly

[derp No.4310339]

>>4310314
DUH!
It's (1/cos2x)(-sin2x)*2

IMANIDIOT.JPG

[Anonymous No.4310351]

>>4310339
also known as -2tan(2x)