L'hopitals rule to solve this limit

[Lodve]

I know I have to apply L'hopitals rule to solve this equation, but it seems unsolvable in my eyes. I would be thankful if someone here could spend their time solving this one for me :D


\frac{ln(cos2x)}{(tanx)^2} The variable x is closely 0.

 

[Lodve]

Bump!

 

[HallsofIvy]

"Bumping" a thread just to get it jumped to the top of the list, especially after less than an hour is a pretty sure way to get banned from this forum!

The derivative of ln(cos(2x)) is
\frac{-2sin(2x)}{cos(2x)}= -2 tan(2x)

The derivative of (tan(x))^2 is
2 tan(x)sec^2(x)= 2\frac{sin(x)}{cos^3(x)}

It might also help to know that
tan(2x)= \frac{2tan(x)}{1- tan^2(x)}

 

[Lodve]

Sorry about that :P It won't happen again. I'm quite new here as you can see :P Yes I have derived both nominator and denominatorm and I got the same result as you got. However I still get "0/0", and it's much more difficult to derive both nominator and denominator again.

 

[HallsofIvy]

I don't get "0/0". You should have something like
\frac{-2tan(2x)}{2\frac{sin(x)}{cos^3(x)}}
and using the identity for tan(2x) that I mentioned
-\frac{4tan(x)}{1- tan^2(x)}\frac{cos^3(x)}{2sin(x)}[/itex]<br /> = -2\frac{2\frac{sin(x)}{cos(x}}{1- \frac{sin^2(x)}{cos^2(x}}\frac{cos^3(x)}{sin(x)}<br /> = -2\frac{cos^4(x)}{cos^2(x)}= -2cos^2(x)

 

[Lodve]

Thank you for helping me :D Appreciate it ;)