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L'Hospitals Rule

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Given f is differentiable on (0,[tex]\infty[/tex])
    Given [tex]lim_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
    S.T lim f(x)=L and lim f'(x)=0
    Hint f(x)=e[tex]^{x}[/tex]f(x)/e[tex]^{x}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    A Lim [tex]_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
    Then for some [tex]\epsilon[/tex]>0
    |f(x)+f'(x)-L|<[tex]\epsilon[/tex]


    Tried different approaches by substituting for f(x) and f'(x) based on the hint. But did not help. I tried to get it to a L/infinity form so f'(x)=0 but could not.
     
  2. jcsd
  3. Apr 30, 2009 #2
    Use l'hopital rule on f(x) = [e^xf(x)] / e^x
     
  4. Apr 30, 2009 #3
    its unclear what you're actually asking. f(x)=exp(x)*f(x)*exp(-x) makes no sense to me.

    i'd try proof by contradiction and show all such conditions cannot hold.

    lets say f(x) is merely an increasing function,
    (1) is it possible that a bounded f(x) has nonzero slope everywhere?
    (2) is it possible that an unbounded f(x) eventually has zero slope?

    [edit: oh okay yeah, use the hint.]
     
    Last edited: Apr 30, 2009
  5. Apr 30, 2009 #4

    djeitnstine

    User Avatar
    Gold Member

    Just a side note, you do not have to use separate wraps for Latex, simply use one so that we can understand it better =]
     
  6. May 1, 2009 #5
    How?

    I can see that x->infinity, e^xf(x)/e^x is of the infinity . limx->inf f(x)/infinity. Since we don't know anything about f(x) except it s continuous and differentiable on (0,infnty), can i make the conclusion that it is not= 0 hence is of the infnty/infnty form.
    Proceeding with that thought:
    lim x->infnty f'(x)= limx->infnty [(e^xf(x) +e^x)f'(x))/e^x]. Now this is infnty . L/infnty form. What do I do after.I can see that lim x->infnty f'(x) Not=L but how do I show it is 0.
     
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