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Proof involving e^2

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that lim x ---> 0 (1 + 2x)^(1/x) = e^2

    3. The attempt at a solution

    lim x --> 0 log (1 + 2x)^(1/x) = 1
    lim x --> 0 log (1 + 2x)/x = 1

    Not really sure if I'm headed in the right direction here. Extremely new to proofs, thanks for any help.
  2. jcsd
  3. Feb 13, 2010 #2


    Staff: Mentor

    Let y = (1 + 2x)1/x

    Then take the natural log of both sides.
    Then take the limit as x --> 0 of both sides.

    The limit you'll be working with is indeterminate, of the form [0/0], so you can use L'Hopital's Rule to evaluate it.

    At this point you have [tex]lim_{x \to 0} ln y = <\text{some number}>[/tex].

    If a function is continuous, lim (ln f(x)) = ln( lim f(x)), meaning that you can interchange the operations of limit and ln.

    You'll have ln(lim f(x)) = k, so lim f(x) = ek.
  4. Feb 14, 2010 #3
    thank you very much, I think I'm on the right track as l'hopital's rule is giving me 2 which is ln e^2....
  5. Feb 14, 2010 #4


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    A simple change of variable turns this problem into a familiar limit....
  6. Feb 14, 2010 #5


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    BTW when you receive too much help -- e.g. how Mark44 pretty much told you every little thing to do -- you should ignore it if possible (and report it), or set it aside and try to reproduce the outline on your own.

    If you do wind up using the solution he gave you, you should spend some time studying the problem, trying to answer the question: "How would I have figured out this approach to the problem?" and/or "What could I have done to finish the problem the way I was thinking about it?"
  7. Feb 14, 2010 #6
    Is the substitution you're talking about something like letting x = 1/y?
  8. Feb 14, 2010 #7


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    I was thinking 1/(2y), but if 1/y is also something familiar, so that works too.

    P.S. I mentioned it not because you should do it my way, but just to show you another way to do it. It's good to know lots of ways to do things! (And a lot of people seem not to think about limits by substitution. :frown:)
  9. Feb 14, 2010 #8
    I know what you mean... Thanks both of you for your suggestions.
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