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Proof of e definition?

  1. Jan 22, 2008 #1
    e can be defined as limit as x approaches infinity of (1 + (1/x)) ^ x or limit as x approaches zero of (1 + x) ^ (1/x)

    From my knowledge of limits, this does not make sense. I must be missing something here. Where can I find a proof of these?
  2. jcsd
  3. Jan 22, 2008 #2
    L'Hôpital's rule is needed to prove

    If the limit exists there is some number, L, which the limit will approach.

    [tex]L = \lim_{x\rightarrow\infty} (1+x)^{1/x}[/tex]

    [tex]ln{L} = \lim_{x\rightarrow\infty} \frac{ln{(1+x)}}{x} = \frac{\inf}{\inf}[/tex]

    [tex]ln{L} = \lim_{x\rightarrow\infty} \frac{\frac{d}{dx} ln{(1+x)}}{\frac{d}{dx}x}[/tex]

    [tex]ln{L} = \lim_{x\rightarrow\infty} \frac{1}{1+x} = 1[/tex]

    Therefore we can conclude the number the limit approaches is [tex]ln{L} = 1[/tex] or [tex]L = e^1[/tex]
    Last edited: Jan 22, 2008
  4. Jan 22, 2008 #3
    I think you have made a slight typo in the second line of your calculations.
  5. Jan 22, 2008 #4
    Oops, fixed
  6. Jan 22, 2008 #5
    Depends a little bit on what you mean by proving. If we assume the basic properties of the e already known, then that is a fine calculation. If we actually want to define the e with this limit, then that is not fine.

    For example, if the logarithm is defined as the inverse of the exponential mapping [itex]x\mapsto e^x[/itex], you already have a circular logic there. If you have instead defined the logarithm with something else, like with the integral of 1/x, then its relationship to the exponential functions becomes nontrivial, and the final step [itex]\textrm{log}(L)=1 \implies L=e^1[/itex] requires a proof of its own.
    Last edited: Jan 22, 2008
  7. Jan 22, 2008 #6
    You did not tell what's your problem. If your problem is simply, that you don't understand how

    \big(1+\frac{1}{x}\big)^x \to \big(1+\frac{1}{\infty}\big)^{\infty}


    (1+x)^{1/x} \to (1+0)^{1/0}

    could become something reasonable, then calculations like the one that Feldoh showed, are a good pedagogical start. But if you want to know how e is really defined rigorously, don't be satisfied too soon!
  8. Jan 22, 2008 #7
    Sure, however I think he was just looking for clarification on why the limit is equivalent to e which I did show.
  9. Jan 23, 2008 #8
    expand binomially and cancel and substitute values
  10. Jan 23, 2008 #9


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    What is e?
    The original question was flawed in that it is unclear what is e is.
    The original question could be interpeted in several ways.
    a:=limit as x approaches infinity of (1 + (1/x)) ^ x
    b:=limit as x approaches zero of (1 + x) ^ (1/x)
    we might then ask
    1)are the numbers a and b well defined
    2)is a=b yes
    3)if e is defined in any common way can we show a,b=e

    Which question was asked? And in 3) which common definition are we using
    e:=limit n (a natural number) approaches infinity of (1+1/n)^n
    e:=limit n (a natural number) approaches infinity of n/(n!)^(1/n)
    and many more.
  11. Jan 24, 2008 #10
    That's not my point I don't think he was asking for some super ultra rigorous proof, yes we know there is more to add. But is it needed to demonstrate what he asked so it's understood? No.
  12. Jan 24, 2008 #11

    Gib Z

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    From how he worded the original post, it seemed he was just wondering how those limits could in fact be anything other than 1, assuming from intuition "well, the second term inside the brackets becomes 0, and 1 to the power of anything is just 1, so the limit should be 1".

    It should be taken note of that that kind of intuition rarely works in analysis!

    The easiest way to show that the limit is e, with rigor, is the define the function, f(x) = exp(x), as the unique function with the properties exp(0) = 1, and f'(x) = f(x). Then we merely propose the series of exp(x), see, or if required, show, that the series obviously fulfill those properties. Then, expand the limit we have in question via the binomial theorem, and theres not too much left after that! =]
  13. Jan 25, 2008 #12


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    That begs the question I have, of what "e" does the poster speak.
    We should be given an e to show that the given limits equal it.
    I might just say let e any number equal to either of the given limits making the question trivial.

    Reminder in mathematics there is not usually "the" definition of some object.
    We may ask is a definition defines one and only one object.
    We may ask is to definition define the same object.
    We could ask is a definition meets some criteria like

    Looking again at the original post and defining
    a:=limit as x approaches infinity of (1 + (1/x)) ^ x
    b:=limit as x approaches zero of (1 + x) ^ (1/x)
    it is unclear what type of number x is
    if we assume x a real number we have the problem of several definitions of raising to powers.
    if we assume x a natural number we have a commmon definition (top 10 at least)
  14. Jan 29, 2008 #13
    eh, was looking around for the answer to give to someone i know
    found this thread and decided to add to it

    another way to prove (from my calc textbook):

    let f(x) = lnx
    then f'(x)=1/x
    so f'(1)=1

    by using the definition of derivative:
    f'(1)= lim h->0 [f(1+h)-f(1)] / h
    = lim x->0 [f(1+x)-f(1)] / x
    = lim x->0 [ln(1+x)-ln(1)] / x
    = lim x->0 [ln(1+x)-0] / x
    = lim x->0 ln(1+x) / x
    = lim x->0 (1/x) ln(1+x)
    = lim x->0 ln(1+x)^(1/x)
    = ln (lim x->0 (1+x)^(1/x)) (since ln is continuous)

    because f'(1) = 1 (and f'(1)=ln (lim x->0 (1+x)^(1/x)))

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

    lim x->0 (1+x)^(1/x) = e
    Last edited: Jan 29, 2008
  15. Jan 29, 2008 #14
    Using that forumla the limit should be as x goes to 0, shouldn't it? Otherwise the 2nd last line doesn't make sense...
  16. Jan 30, 2008 #15

    Gib Z

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    Yes it seems Feldoh got a tiny bit confuddled. Hard to blame him though, since theres a limit very similar to that which also gives e. Yes he meant limit as x approaches 0 in every limit, and the second line he meant 0/0 form, not infinity/infinity. The Rest works out.
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