L'hopitals rule to solve this limit

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Homework Help Overview

The discussion revolves around applying L'Hôpital's rule to evaluate the limit of the expression \(\frac{\ln(\cos(2x))}{(\tan(x))^2}\) as \(x\) approaches 0. Participants are exploring the behavior of the function near this limit point.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply L'Hôpital's rule but expresses difficulty in deriving the numerator and denominator again after obtaining a "0/0" form. Other participants discuss the derivatives of the numerator and denominator, questioning the results and exploring identities related to tangent.

Discussion Status

Participants are actively engaging with the problem, sharing their derivatives and identities. There is a mix of interpretations regarding the limit's form, and some guidance has been offered on derivatives and identities, but no consensus has been reached on the next steps.

Contextual Notes

There is a mention of the original poster being new to the forum and a concern about forum etiquette regarding thread "bumping." The discussion also highlights the challenge of deriving both the numerator and denominator again after reaching an indeterminate form.

Lodve
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I know I have to apply l'hospital's 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


[tex]\frac{ln(cos2x)}{(tanx)^2}[/tex] The variable x is closely 0.
 
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Bump!
 


"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
[tex]\frac{-2sin(2x)}{cos(2x)}= -2 tan(2x)[/tex]

The derivative of [itex](tan(x))^2[/itex] is
[tex]2 tan(x)sec^2(x)= 2\frac{sin(x)}{cos^3(x)}[/tex]

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


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.
 


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


Thank you for helping me :D Appreciate it ;)
 

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