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Homework Help: Differantiation question

  1. Feb 20, 2009 #1
    f(x) and g(x) are differentiable on 0
    f(0)=g(0)=0

    calculate
    [tex]
    \lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2}
    [/tex]
    ??

    it looks like cauchys mean value theorem
    F(x)=cos(f(x))
    G(x)=cos(g(x))
    what to do next??
     
  2. jcsd
  3. Feb 20, 2009 #2

    tiny-tim

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    Welcome to the second millenium!

    Hi transgalactic! Congrats on your 1001st post! :smile:
    Hint: use L'Hôpital's rule … twice! :wink:
     
  4. Feb 20, 2009 #3

    HallsofIvy

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    tiny-tim, that was my first thought. But we don't know that f and g are twice differentiable.
     
  5. Feb 20, 2009 #4

    tiny-tim

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    always look on the bright side of life … ♩ ♫ ♪ ♫ ♬ ♩

    oh tush, mr glass-half-empty …

    :biggrin: we don't know that they aren't! :biggrin:

    :smile: happy days are here again
    the skies above are clear again
    let us sing a song of cheer again
    happy days are here again! :smile:
     
  6. Feb 20, 2009 #5
    [tex]
    \lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
    [/tex]
    so the solution is 0
    ??
     
  7. Feb 20, 2009 #6

    tiny-tim

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    uhh? :confused:

    however did you get that?

    always show your full calculations! :rolleyes:
     
  8. Feb 20, 2009 #7
    [tex]
    \lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+f'(x)cos(f(x))-f''(x)sing(x)+f'(x)cosg(x)}{2}
    [/tex]
    ??
     
  9. Feb 20, 2009 #8

    tiny-tim

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    That's better! :smile:

    And if f(0)=g(0)=0, then that = … ?
     
  10. Feb 20, 2009 #9

    HallsofIvy

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    Transgalactic, as much as I hate to pour cold water on the party, you started by saying "if f and g are differentiable at 0". Is that all you are given or are you given that the second derivative exists?
     
  11. Feb 20, 2009 #10
    its said that" they are differentiable on 0

    but not necessarily differentiable around 0
     
  12. Feb 20, 2009 #11

    tiny-tim

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    This is a bit of a technicality … we'll discuss why in a moment …

    but first, just answer my previous question: what is that limit equal to at x = 0?

    (oh, btw, just noticed :redface: … you should have had an f'(x)2 somewhere in that :wink:)
     
  13. Feb 20, 2009 #12
    i fixed it to
    [tex]
    \lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} =\lim _{x->0}\frac{-f'(x)sin(f(x))-f'(x)sing(x)}{2x}=\lim _{x->0}\frac{-f''(x)sin(f(x)+(f'(x))^2cos(f(x))-f''(x)sing(x)+(f'(x))^2cosg(x)}{2}
    [/tex]

    after i did lhopital twice i got
    [tex]
    \lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-f''(x)sin0+(f'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(f'(x))^2}{2}
    [/tex]

    ??
     
  14. Feb 20, 2009 #13

    Dick

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    I'm with Halls. You only know the derivatives exist AT zero. You can't even apply l'Hopital the first time, you can't take the limit of something you only know exists at a single point. I'd suggest using the series expansion for cos(x) and difference quotients.
     
  15. Feb 20, 2009 #14
    what is the link between the ability of doing lhopital law
    and differentiability of a function?
     
  16. Feb 20, 2009 #15

    Dick

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    Take f(x)=x^2*sin(1/x^2), f(x)=0. That's differentiable at 0. In fact, it's differentiable everywhere. But it's not continuously differentiable at 0. You can't do anything with l'Hopital at 0. But that doesn't mean your limit doesn't exist. So the problem with l'Hopital is more than 'technical'.
     
  17. Feb 20, 2009 #16

    tiny-tim

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    oh, transgalactic, however do you manage to make these mistakes?

    you started with an expression that was half f and half g,

    but you ended losing all the g :cry:

    try again :smile:

    (and then we'll sort out the technicalitites :wink:)
     
  18. Feb 21, 2009 #17
    fixed it
    [tex]
    \lim _{x->0}\frac{-f''(x)sin(0)+(f'(x))^2cos(0)-g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{(f'(x))^2+(g'(x))^2}{2}
    [/tex]

    ??
     
    Last edited: Feb 21, 2009
  19. Feb 21, 2009 #18

    tiny-tim

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    much better! :smile:

    (though you still need to fiddle around with those pluses and minuses :wink:)

    EDIT: oh, and since f'(0) and g'(0) are both defined as existing, is there any reason why you can't use them at the end? :smile:
     
    Last edited: Feb 21, 2009
  20. Feb 21, 2009 #19
    fixed it
    [tex]
    \lim _{x->0}\frac{-f''(x)sin(0)-(f'(x))^2cos(0)+g''(x)sin0+(g'(x))^2cos0}{2}=\lim _{x->0}\frac{-(f'(x))^2+(g'(x))^2}{2}
    [/tex]

    ??
     
    Last edited: Feb 21, 2009
  21. Feb 21, 2009 #20
    what now??
     
  22. Feb 21, 2009 #21

    D H

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    Transgalactic, could you state the question exactly as written in your text? In particular, the phrase "f(x) and g(x) are differentiable on 0" does not make much sense (to me, at least). Does the text say something slightly different than "on 0"?
     
  23. Feb 21, 2009 #22
    f(x) and g(x) are differentiable on 0.
    f(0)=g(0)=0
    calculate "the presented limit"
    notice:
    f(x) and g(x) are not necessarily differentiable around 0
     
  24. Feb 21, 2009 #23

    tiny-tim

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    yes! :smile:

    and now how about using f'(0) and g'(0)?​
     
  25. Feb 21, 2009 #24
    how??
    its differentiable on
    but i dont know the final value of the derivative around
    ??
     
  26. Feb 21, 2009 #25

    HallsofIvy

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    What you have written assumes the derivative exists around 0. (In fact, as I have pointed out, this whole approach assumes f and g are twice differentiable in some neighborhood of 0)

    If the derivative does exist in some neighborhood of 0, then, although a derivative is not necessarily continuous, it does satisfy the "intermediate value property" so [itex]\lim_{x\rightarrow 0} f'(x)= f'(0)[/itex] and [itex]\lim_{x\rightarrow 0} g'(x)= g'(0)[/itex]. I presume you do not know the actual values of f'(0) and g'(0) but at least
    [tex]\frac{g'(0)^2- f'(0)^2}{2}[/tex]
    is simpler than what you have.
     
    Last edited by a moderator: Feb 21, 2009
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