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Limit (L'hopitals) trig functions

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data
    use l'hopital's to evaluate the limit.


    2. Relevant equations
    limit (∅->0) ∅-3sin∅cos ∅
    --------------------
    tan∅- ∅



    3. The attempt at a solution
    i take the derivatives of the top and bottom, and use trig identity for sec^2x - 1 to tan^x.
    i get 6(cosx)^2 annd get the value of 6. but the answer is -∞.


    6(sinx)^2
    ----------
    (tanx)^2


    6(cosx)^2
     
  2. jcsd
  3. Mar 3, 2013 #2

    SammyS

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    What is the derivative of -3sin(θ)cos(θ) ? (It's not 6 sin2(θ) . )
     
  4. Mar 3, 2013 #3

    mfb

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    The same fraction, but easier to read with LaTeX:
    $$\lim_{x \to 0} \frac{x-3\sin(x)\cos(x)}{\tan(x)-x}$$
     
  5. Mar 4, 2013 #4
    those values were actually taking out of a solution manual, but i can check again.
     
  6. Mar 4, 2013 #5
    using the product rule:

    -3( (cosx)^2 - (sinx)^2 ) == cos2x

    which means:

    1-3(cos2x)
    ------------
    (tanx)^2


    further:


    1-(3/2)sin(2x)
    -----------------
    2tanx * (secx)^2

    i'm still not getting to -infinity, and it seems to be getting complicated in the denominator.
     
    Last edited: Mar 4, 2013
  7. Mar 4, 2013 #6

    SammyS

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    Yes, using the product rule, [itex]\displaystyle \ \ \frac{d}{dx}\left(\sin(xcos(x)\right)=\cos^2(x)-\sin^2(x)\ \ [/itex] and that's equal to [itex]\ \cos(2x)\ .\ [/itex]

    So, you have [itex]\displaystyle \ \lim_{x \to 0} \frac{x-3\sin(x)\cos(x)}{\tan(x)-x}= \lim_{x \to 0}\,\frac{1-3\cos(2x)}{\tan^2(x)}\,,\ [/itex] which is not of indeterminate form, and gives the result you're looking for.
     
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