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L'Hopital's Rule Help

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data
    lim as x->0 (tan(x)-x)/(sin2x-2x)

    2. Relevant equations
    L'Hopitals rule states that if the limit reaches 0/0, you can take the derivative of the top and the bottom until you get the real limit.

    3. The attempt at a solution

    (sec^2(x)-1)/(2cos2x-2) still 0/0
    2sec^2(x)tan(x)/(-4sin(2x)) still 0/0

    I have pain stakingly taken the derivative twice more and it simply does not seem to reach any end. All help is very much appreciated!
  2. jcsd
  3. Apr 4, 2007 #2


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    You are going to get a nonzero denominator at the next derivative after you've shown. It's a cosine.
  4. Apr 4, 2007 #3
    You are correct and this does give me a non 0/0 answer but the answer is wrong according to the book. It should be -1/4.

    My next derivative is:


    In simpler terms


    The numerator becomes 0 with the limit and the demoninator becomes -8, making the limit 0.

  5. Apr 4, 2007 #4


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    Why aren't you using the product rule on the numerator? What ARE you doing? The derivative of the tan(x) term will be nonzero.
  6. Apr 4, 2007 #5
    I apologize, that was a careless mistake.

    I'll post my solution if it's of any help to anyone...

    the numerator becomes 2(sec^2(x)tan^2(x) + sec^2(x)),
    which of course goes to 2 when pushed to 0
    leaving 2/-8, or -1/4

    Thank you so much!
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