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Trig limit

  1. May 26, 2008 #1
    lim (2/x^2)-(1/(1-cos(x)))
    x-->0

    i have tried to use l'hopital's rule, but i keep on getting -1/6 however from graphing the function on my graphics calculator i know that it is equal to zero

    any help is appreciated
     
  2. jcsd
  3. May 26, 2008 #2

    Defennder

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    I graphed it on graphmatica, and it looks like -1/6 is correct.

    EDIT: Looks like you have to apply L-Hospital rule a hell lot of times before you can get the limit.
     
    Last edited: May 26, 2008
  4. May 26, 2008 #3

    malawi_glenn

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    I get -1/6 too, by graphing with calculator and evaluting it.
     
  5. May 26, 2008 #4

    Gib Z

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    It's quite simple when we take [tex]\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)[/tex].
     
  6. May 26, 2008 #5
    yeh after you do algebraic manipulation to get it in the indeterminate form of

    2-2cos(x)-x^2
    -----------------
    (x^2)-(x^2)cos(x)

    you have to apply l'hospital's rule 4 times to get a non indeterminate form which is -1/6
    the only problem is when i substitute in say 0.00001 or -0.00001 into the original equation, i get zero, i don't understand why
     
  7. May 26, 2008 #6

    malawi_glenn

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    crappy calculator, cancellation etc.
     
  8. May 27, 2008 #7

    Gib Z

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    Try using the series truncation I gave, after a little algebraic manipulation the answer comes almost immediately, its quite easy =]
     
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