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Differantiation proof question

  1. Jan 3, 2009 #1
    there is a function f(x) which is differentiable on (a,+infinity)
    suppose lim [f(x)]/x =0 as x->+infinity
    prove that lim inf |f'(x)|=0 as x->+infinity ?

    does this expression lim inf f'(x)=0 has to be true
    if not
    present a disproving example

    ?

    i got this solution :
    First consider [itex]\lim_{m\to\infty}\frac{f(2m)}{m}[/itex], let [itex]2m=x[/itex] and this limit becomes [itex]2\lim_{x\to\infty}\frac{f(x)}{x}=0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(2x)}{x}[/itex] exists and equals [itex]0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(x)}{x}=0\implies \lim_{x\to\infty}\frac{f(2x)-f(x)}{x}=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0[/itex]. So now consider the interval [itex][x,2x][/itex] and apply the MVT letting x approach infinity.

    I tried to follow your logic.
    and i got a few question regarding it??
    http://img82.imageshack.us/my.php?image=14440292jz6.gif
     
  2. jcsd
  3. Jan 3, 2009 #2
    L'Hospital rule?
     
  4. Jan 3, 2009 #3
    what about it?
     
  5. Jan 3, 2009 #4
    I thought you could use L'Hospital here... using that
    you get
    lim .. f'(x)/1 = 0
    but f(x)/x may not be in the form of inf/inf etc.

    I couldn't understand the solution.
     
    Last edited: Jan 3, 2009
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