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The number e as a limit

  1. Nov 29, 2015 #1
    1. The problem statement, all variables and given/known data
    I have lim of n > infinity (1+1/n)^n

    2. Relevant equations

    3. The attempt at a solution
    I know that I must use l'hospital rule and setting ln y = n ln (1+1/n)

    And after lim n ln (1+1/n) as n approaches infinity.

    After what do I do ?
  2. jcsd
  3. Nov 29, 2015 #2


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    Try writing it as$$\frac{\ln(1 + \frac 1 n)}{\frac 1 n}$$before using L'Hospital's rule and taking the limit.
  4. Nov 29, 2015 #3
    OK but I'll have to take the natural logarithm right ?
  5. Nov 29, 2015 #4


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    Didn't you already take the logarithm to get that expression?
  6. Nov 29, 2015 #5


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    You've either applied LH incorrectly, or I'm misreading you. How did you apply LH (how did you set it up) when you're at the step: $$exp[\lim_{n\to\infty}(n\log(1+\frac{1}{n})] $$

    Never mind, didn't see the new post update. :) Ignore me LC's already said the same thing.
  7. Nov 29, 2015 #6
    Wait, what I just wrote was correct or not ? Am I in the right direction ?
  8. Nov 29, 2015 #7


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    Follow post #2.
  9. Nov 29, 2015 #8


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    For this to work out correctly, you need to keep track of your equations. Each line you write should be an equation.
    Let ##y = (1 + 1/n)^n##
    ##\ln y = n \ln(1 + 1/n) = \frac{\ln(1 + 1/n)}{1/n}##
    Now take limits of both sides, after which you can apply L'Hopital's Rule.

    Again, take care to work with equations at each step.
  10. Nov 29, 2015 #9
    I'm sorry if it takes time to respond, I have like several question to answer at the same and I must give this tomorrow lol Don't worry, I keep track of what I write on paper. I'll respond here if something I'm having a problem with.
  11. Nov 29, 2015 #10
    Ok thanks it worked
  12. Nov 29, 2015 #11
    Yes, but it is extremely rude and time consuming for the people on this site. How are we to know the work you present means.
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