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Lim((cosx/x^2)-(sinx)/x^3)) x->0 L'hopital

  1. Sep 21, 2004 #1
    Hey Everybody.

    I was just wondering about this query:

    lim((cosx/x^2)-(sinx)/x^3))
    x->0

    My teacher is telling me to use L'hopital, but the problem is that

    the first part (cosx)/(x^2) isn't a "0/0", cosx-> 1 for x->0

    So what should I do, i know the right answer is -1/3, but i need to prove it.

    Plz. Help Me

    Thanks
     
  2. jcsd
  3. Sep 21, 2004 #2

    Zurtex

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    I've never used that method in calculus before, but am I right in thinking that you need to get to a 0/0 situation?

    In which it seems fairly obvious to me that you should rewrite:

    [tex]\frac{\cos x}{x^2} - \frac{\sin x}{x^3}[/tex]

    As:

    [tex]\frac{x \cos x - \sin x}{x^3}[/tex]

    Does that help?
     
  4. Sep 21, 2004 #3
    i wasn't sure which forum, i should post my query in.

    How did you rearrange it??
     
  5. Sep 21, 2004 #4
    a/(c^2) - b/(c^3)
    = ac/(c^3) - b/(c^3)
    = (ac-b)/(c^3)

    -- AI
     
  6. Sep 21, 2004 #5
    Thank you guys, you just made my day beautiful!!
     
  7. Sep 21, 2004 #6

    Hurkyl

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    Homework help.
     
  8. Sep 21, 2004 #7
  9. Sep 21, 2004 #8

    Hurkyl

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    Actually, the modern spelling is l'Hôpital; the 's' is swallowed into the circumflex.
     
  10. Sep 21, 2004 #9
    Right !
    I love the flavor of the past.
     
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