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Using l'Hospitals rule for sequences of functions

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Let (fn) be the sequence of functions defined on [0,1] by fn(x) = nxnlog(x) if x>0 and fn(0)=0. Each fn is continuous on [0,1].

    Let a be in (0,1) and ha(y) = yay := yey log(a). Using l'Hospitals rule or otherwise, prove that limy->+∞ ha(y) = 0.

    Then considering different cases for x and using the previous part, prove that (fn) converges point wise to the zero function.

    2. Relevant equations



    3. The attempt at a solution

    I'm not too sure how to start this off. I rearranged h to ay/(1/y), differentiated top and bottom so I get ay/(-1/y2), and then taking the limit I get 0/0, and I'm not sure what to make of that. Also trying to rearrange the second part of h I get 0/0 when taking the limit of its derivatives.

    What other path can I take to prove that the limit is 0?
     
  2. jcsd
  3. May 27, 2013 #2

    Dick

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    Take the log of y*a^y. Try to show it approaches -infinity as y->infinity. Use that log(a) is negative and show lim y->infinity log(y)/y=0.
     
  4. May 28, 2013 #3
    I don't understand why I need to take the log. Aren't I changing the function then?
     
  5. May 28, 2013 #4

    Office_Shredder

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    If you really want to use L'Hopital, write it as
    [tex] \frac{y}{a^{-y}} [/tex]
     
  6. May 28, 2013 #5

    Dick

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    That is more economical. But my point for Whistlekins is that if log(r)->-infinity then r goes to 0.
     
  7. May 28, 2013 #6
    Understood.

    Now the second part of the question, can I just use the fact that limits of the product of two functions is equal to the product of the limits of the functions, and the fact that g_n(x) = nx^n is a subsequence of h? Is this all that is required for a proof?
     
  8. May 28, 2013 #7

    Dick

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    Yes, I think so. Now you know nx^n->0 for x in (0,1). So nx^nlog(x) must do the same. x=0 and x=1 are separate cases.
     
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