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Limit problem / l'hopital's rule

  1. Jan 25, 2012 #1
    The limit n->infinity I have to compute is:


    [itex]\frac{n\cdot \log ^{5}(n)}{n^{2}}[/itex]


    Should I use L'hopital's rule? If I do, I have a problem:

    First I simplify and get:

    [itex]\frac{ \log ^{5}(n)}{n}[/itex]

    Taking the derivative of the top, and the bottom leads to:

    [itex]\frac{\frac{ 5\log ^{4}(n)}{n}}{1}[/itex]

    At this point, we can see that the numerator is approaching 0, as n increases.

    There 0/1 = 0

    Is this correct?

    Thanks
     
    Last edited: Jan 25, 2012
  2. jcsd
  3. Jan 25, 2012 #2

    HallsofIvy

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    Why is it "obvious" (since you state it without proof) that [itex]5log^4(n)/n[/itex] goes to 0 when it was not obvious for [itex]log^5(n)/n[/itex]?

    What do you get if you use L'Hopital again? And Again?
     
  4. Jan 25, 2012 #3
    I just plugged in very large numbers and saw that it will approach zero. I guess I could have done that with the initial function as well.
     
  5. Jan 25, 2012 #4

    Dick

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    The conclusion is correct. But I don't see why [itex]\frac{ \log ^{4}(n)}{n} \rightarrow 0[/itex] is any more obvious than [itex]\frac{ \log ^{5}(n)}{n} \rightarrow 0[/itex]. Why don't you keep applying l'Hopital until the log disappears?
     
  6. Jan 25, 2012 #5
    Since the denominator becomes one, should I just continue taking the derivative of the numerator, until log disappears?

    Thanks

     
  7. Jan 25, 2012 #6

    Dick

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    You applied l'Hopital to [itex]\frac{ \log ^{5}(n)}{n}[/itex] correctly. The original denominator of n became 1 but now you get a new denominator of n coming from the derivative of the log.
     
  8. Jan 25, 2012 #7
    I think I see where this is going. I'll keep taking the derivative and eventually end up with a constant over n. Correct?

     
  9. Jan 25, 2012 #8

    Dick

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    Yessss.
     
  10. Jan 25, 2012 #9
    Thank you so much!
     
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