L'hopitals rule to solve this limit

  • Thread starter Lodve
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  • #1
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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


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

  • #2
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Bump!
 
  • #3
HallsofIvy
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"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]
 
  • #4
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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.
 
  • #5
HallsofIvy
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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]
[tex]= -2\frac{2\frac{sin(x)}{cos(x}}{1- \frac{sin^2(x)}{cos^2(x}}\frac{cos^3(x)}{sin(x)}[/tex]
[tex]= -2\frac{cos^4(x)}{cos^2(x)}= -2cos^2(x)[/tex]
 
  • #6
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Thank you for helping me :D Appreciate it ;)
 

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