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L'hopital's rule, indeterminate forms

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]lim_{x -> \infty} \left( \frac{x}{x+1} \right) ^ {x}[/itex]



    3. The attempt at a solution

    So I did e^whole statement with ln(x/(x+1))*x, after that I multiplied that expression by 1/x/1/x, then I go ln(x/(x+1)/1/x, I tried taking derivative of top and bottom but it doesn't help with finding an answer.
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2
    It did for me. Try it again, maybe.
     
  4. Feb 27, 2014 #3
    I wouldn't e^ln(statement).I would write it as

    ##y= \displaystyle \lim_{x \to \infty} \left( \frac{x}{x+1} \right)^x##
    ##\ln{y}= \displaystyle \lim_{x \to \infty} \ln{\left[\left( \frac{x}{x+1} \right)^x\right]}##

    Then you eventually solve for y. What do you get when you perform l'Hospital's rule? This is the correct procedure so far.
     
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