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Limit of differentiable functions question

  1. Mar 13, 2009 #1
    f(x) and g(x) are differentiable on 0
    f(0)=g(0)=0
    [tex]
    \lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2
    [/tex]

    because i can use (sin x)/x=1 here
    is it ok??
     
  2. jcsd
  3. Mar 13, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, you don't have "sin(x)/x", you have
    [tex]\frac{sin(\frac{f(x)+ g(x)}{2})}{x^2}[/itex]

    whether this converges at all depends on how fast f(x) and g(x) converge to 0.
     
  4. Mar 13, 2009 #3
    i can split x^2 into xx
    and i have two sinuses ,one for each x
    so i do have it
    the only thing that bothers me is
    its sin x/x
    the stuff inside sinus doesnt equal x
    ??
    but they both go to zero
    ??
     
  5. Mar 13, 2009 #4
    Well, it looks rather simple to me, but i might be totally wrong as well. Anyhow, i hope u won't kill me for this :yuck:


    I can use the fact that both f(x) and g(x) are differentiable at 0...this means that they both are continuous at 0 as well. THis is good to know.

    NOw i will alsu use the fact that cos(x) is continuous everywhere.this is good to know too.

    Now:


    [tex] \lim_{x\rightarrow 0}\frac{cosf(x)-cosg(x)}{x^2}[/tex]

    Applying what i just said above, we see that we get a limit of the form 0/0.

    So, applying l'hopitals rule twice, will take you home safe.

    Edit: DO you know anything about f'(0) and g'(0)??
     
  6. Mar 13, 2009 #5
    nope

    you are correct
    thanks:)
     
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