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Homework Help: Very lost with l'Hospital's rule

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate lim x[tex]\rightarrow[/tex] infinity of ([tex]\frac{x}{x+1}[/tex])[tex]^{}x[/tex], state explicitly the type of the indeterminate form.



    2. Relevant equations



    3. The attempt at a solution
    I somewhat understand how to use l'Hospital's rule when the form is 0/0, but the inf/inf throws me off completely.
    f'(x) = (x/(x+1))^x *ln(x/(x+1))
    but from there or maybe the start I'm lost.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 6, 2007 #2
    whether its 0/0 or inf/inf you proceed the same way...take the derivative of the top function and the derivative of the bottom function and evaluate the limit again
     
  4. Nov 6, 2007 #3

    Dick

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    First you need to change the expression to a 0/0 form. Hint: Take the log first, then try to rearrange that into a 0/0 form.
     
  5. Nov 6, 2007 #4
    I should have posted this sooner, but isn't it in the form 1^inf ?
     
  6. Nov 6, 2007 #5

    Avodyne

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    Yes, that's why you need to take the log first. Find the limit of the logarithm, then exponentiate to get the limit of the original expression.
     
  7. Nov 6, 2007 #6
    so you would get e^limx->inf of xln(x/(x+1)) and then what?
     
  8. Nov 6, 2007 #7

    Dick

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    Uh, rearrange xln(x/(x+1)) into a 0/0 form and apply l'Hopital to find the limit.
     
  9. Nov 6, 2007 #8
    ok I got that much, and I got x/(1/ln(x/(x+1))), but I'm lost as to evaluate that lim with l'Hospitals, would it be (1) / (1/ln(x/(x+1)))', if so I haven't the slightest as to how to get that derivative.
     
  10. Nov 6, 2007 #9

    Dick

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    It's a lot easier to write it as ln(x/(x+1))/(1/x). Use the chain rule to do the ln part.
     
  11. Nov 6, 2007 #10
    hey thanks for all your help guys, i got the answer (e^-1) finally and it was for a quiz so i appreciate it.
     
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