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Rudin 8.5a

  1. May 7, 2008 #1
    [SOLVED] rudin 8.5a

    1. The problem statement, all variables and given/known data
    Find the following limit:

    [tex]\lim_{x \to 0} \frac{e-(1+x)^{1/x}}{x}[/tex]

    2. Relevant equations
    [tex]e = \lim_{x\to 0} (1+x)^{1/x}[/tex]

    [tex]a^b = e^{b \log a}[/tex]

    3. The attempt at a solution
    Is this the derivative of something? I doubt it is but that is the only way I would know how to do this.
  2. jcsd
  3. May 8, 2008 #2
  4. May 8, 2008 #3
    Well, you could use l'hopital rule here, but I guess you havent learnt it yet.
    besides this I don't see a handy trick here.
  5. May 8, 2008 #4
    Of course I know l'Hopital. Using it here requires me to calculate:

    [tex]\lim_{x \to 0} \left(\frac{1}{x+x^2}-\frac{\log (1+x)}{x^2} \right) \exp \frac{\log (1+x)}{x}[/tex]

    which I have no idea how to do. Of course I could probably use l'Hopital again but then it would just get messier. Of course I only need to find the limit of the expression in parenthesis but I am not really even sure if that exists.
    Last edited: May 8, 2008
  6. May 8, 2008 #5


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    Homework Helper

    first show
    then squeeze
  7. May 9, 2008 #6
    the derivative of the numerator is:
    now you should be taking l'hopital on :((x/(1+x))-log(x+1))/x^2
    you can check if you got the right answer through mathematica.

    you shouldn't pay much attention to those cacluations' problems, pay more attention to theoretical questions in rudin's book.
  8. May 9, 2008 #7
    Yay, I got it:

    [tex]\lim_{x \to 0} \frac{(x/(1+x))-\log(x+1)}{x^2} = \lim_{x \to 0} \frac{1/(1+x)^2-1/(1+x)}{2x} = \lim_{x \to 0} \frac{-2/(1+x)^3+1/(1+x)^2}{2} = -1/2[/tex]

    So the answer is e/2. Apparently lurflurf was right but I have no idea how he got those inequalities.

    I think these computational problems are good preparation for the Putnam.
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