Question

find indicated limit:
lim ln cos 4x / 3x^2
x->0

I know ln(u) = u' / u

u= cos 4x
u' = -4 sinx

-4 sinx / cos 4x ?

Answer 1

Write it using equation tab which is at the bottom left of commenting box?

Answer 2

\[\lim_{x\to 0}\frac{\ln(\cos(4x))}{3x^2}\]

Answer 3

probably needs l'hopital a couple times

Answer 4

first time you get
\[\frac{-4\sin(4x)}{6x\cos(4x)}\]

Answer 5

yeah it needs l hospital couple of times

Answer 6

next one will do it. use the product rule for the denominator

Answer 7

next one would be
\[\frac{ -4\sec^2x }{6}\]
now apply limits

Answer 8

not sure that is right

Answer 9

the derivative of the numerator is \(16\cos(4x)\) and the derivative of the denominator is
\[6(\cos(4x)-4x\sin(4x))\]

Answer 10

sorry its\[\frac{ -16\sec^2{4x} }{ 6 }\]

Answer 11

Thnx @satellite73 for pointing it out!

Answer 12

as
\[\frac{ -4\sin(4x) }{ 6xcos(4x) } = \frac{ -4(\sin{4x}/\cos{4x}) }{ 6x }=\frac{ -4\tan4x }{6x}\]

Answer 13

that is much snappier !!

Answer 14

yeah @satellite73 i was about to say this is the easy way to work out!

Answer 15

Thank you guys!