Differantiation question ..

  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. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Welcome to the second millenium!

    Hi transgalactic! Congrats on your 1001st post! :smile:
    Hint: use L'Hôpital's rule … twice! :wink:
     
  4. HallsofIvy

    HallsofIvy 40,395
    Staff Emeritus
    Science Advisor

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

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    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. [tex]
    \lim _{x->0}\frac{-cosf(x)+cosg(x)}{2}
    [/tex]
    so the solution is 0
    ??
     
  7. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    uhh? :confused:

    however did you get that?

    always show your full calculations! :rolleyes:
     
  8. [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. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    That's better! :smile:

    And if f(0)=g(0)=0, then that = … ?
     
  10. HallsofIvy

    HallsofIvy 40,395
    Staff Emeritus
    Science Advisor

    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. its said that" they are differentiable on 0

    but not necessarily differentiable around 0
     
  12. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    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. 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. Dick

    Dick 25,824
    Science Advisor
    Homework Helper

    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. what is the link between the ability of doing lhopital law
    and differentiability of a function?
     
  16. Dick

    Dick 25,824
    Science Advisor
    Homework Helper

    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. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    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. 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. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    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. 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. what now??
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?