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L'Hospital's rule question.

  1. Apr 21, 2008 #1
    so i've been working on this for a while now and i just cant get it.

    Evaluate the limit using L'Hospital's rule if necessary

    lim X -> INF
    (x/x+1) ^7x

    i know i need to do...


    exp(lim x-> inf 7x ln(x/x+1))

    from there i'm stuck.

    i appreciate any help
     
  2. jcsd
  3. Apr 21, 2008 #2
    You could break it up into numerator and denominator, f(x)/g(x) lets say, where f(x) = x^7x and g(x) = (x+1)^7x l'hopital's rule says that if both f(x) and g(x) are divergent, then x--> inf f(x)/g(x) = x-->inf f'(x) / g'(x) , so you take the derivative of the top and bottom separately... you're doing to have to do this until its not divergent anymore... or 8 times? i guess?
    i see what you're going for with the exp(7xln...) but i think thats probably harder / more work. Good luck
     
    Last edited: Apr 21, 2008
  4. Apr 21, 2008 #3

    Dick

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    7x*ln(x/(x+1)) is a 0*infinity limit. Use l'Hopital again. Write it as 7*ln(x/(x+1))/(1/x). You can make life a little easier by changing ln(x/(x+1)) to -ln((x+1)/x) and simplifying inside the log.
     
  5. Apr 21, 2008 #4

    Dick

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    The limit isn't 1. And taking the log does make it easier.
     
  6. Apr 21, 2008 #5

    exk

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    As Dick said taking the log makes it a lot easier.

    You may have to apply l'Hopital more than once.
     
  7. Apr 21, 2008 #6
    thanks guys i got it by taking ln of the whole thing and l'hopitaling the thing twice. ended up being e^(-7)
     
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