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E as a limit

  1. Feb 27, 2008 #1
    Show that:

    lim (1-1/n)^n=1/e
    n->infinity

    I don't really know where to begin...
     
  2. jcsd
  3. Feb 27, 2008 #2
    Well, I guess it depends what you are allowed to start with. You could try to use the limit deffinition which is used to define interest which is compounded continuously. Another thought would be to do a series expansion. Does your book give you a deffinition of e?
     
  4. Feb 27, 2008 #3

    lurflurf

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    first what can you start with for e?
    What is given as the definition of e?
    by elementaty algebra
    (1-1/n)^n=1/(1+1/(n-1))^n
    lim (1+1/(n-1))^n=e
    n->infinity
     
  5. Feb 28, 2008 #4

    nicksauce

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    Try taking the logarithm of both sides, then applying L'hopital's rule.
     
  6. Feb 28, 2008 #5

    lurflurf

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    That won't work because we are trying to establish the equality, thus we can not assume it to be true. Something similar works however.
    let lim denote the limit n->infinity n a natural number

    we wish to show
    lim (1-1/n)^n=1/e
    consider
    log(lim (1-1/n)^n)=lim log((1-1/n)^n)
    supose we know log is continuous at 1, then
    log(lim (1-1/n)^n)=lim log((1-1/n)^n
    perhaps we know log(x^y)=y*log(x), thus
    log(lim (1-1/n)^n)=lim n*log((1-1/n)^n)
    log(lim (1-1/n)^n)=lim -log(1-1/n)/(-1/n)
    log(1)=0 so
    log(lim (1-1/n)^n)=lim -[log(1-1/n)-log(1)]/(-1/n)
    log(lim (1-1/n)^n)=-log'(1)
    log(lim (1-1/n)^n)=-1
    taking antilogs of both sides
    lim (1-1/n)^n=exp(-1)=1/e
    as desired
    we do not need L'hopitals rule since we have the definition of the derivative
     
  7. Jan 27, 2011 #6
    Hi, in your proof above, where do we use the fact that "log is continuous at 1"? For derivative definition at the end, or, for interchanging log and limit?

    And one more question if you don't mind.

    " Given f(n) a monotonically increasing sequence going to infinity as n goes to infinity.
    Is it always true that lim(1 - 1/f(n))^f(n) = 1/e? "

    Thanks a lot for your answer above, and any answer you may have for my two questions.
     
  8. Feb 13, 2011 #7

    lurflurf

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    " Given f(n) a monotonically increasing sequence going to infinity as n goes to infinity.
    Is it always true that lim(1 - 1/f(n))^f(n) = 1/e? "
    Yes. In this is the substitution principle
    if lim_{x->a} f(x)=L
    and lim_{x->a}g(x)=a
    then lim_{x->a}f(g(x))=L
    When the variable is n often the limit is taken over integers and in general the domain of the limit variable matters.

    Hi, in your proof above, where do we use the fact that "log is continuous at 1"? For derivative definition at the end, or, for interchanging log and limit?
    For both. "log is continuous at 1" is one of several ways of making sure log is not a messed up function, that is nowhere continuous.

    To define log (for real numbers) it is sufficient to require
    1) for all x,y real log(xy)=log(x)+log(y)
    2) log'(1)=1 (and thus log'(1) exists)
    or
    2*)log(e)=1 and there is at least one point where log is continuous
    Condition 2) serves two purposes
    We assure a nice function and specify the natural logarithm.
     
  9. Feb 13, 2011 #8

    HallsofIvy

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    One serious problem you have with this is that it isn't true!
    What is true is that
    [tex]\lim_{n\to\infty}\left(1- \frac{1}{n}\right)^n= e[/tex]
    not "1/e".

    How you show that depends upon exactly how you have defined "e". In fact, in a recent thread, the question was asked how to show that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] where the formula you have was given as a definition for e. What we can do here is the reverse of what I did in that thread.

    Now, what definition of "e" are you using? Simplest, if you are not going to use that formula itself, is that ln(e)= 1 where ln(x) is defined by
    [tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex].

    You can show from that that "ln(x)" and "[itex]e^x[/itex]" are inverse functions. And, since the derivative of ln(x) is 1/x, it follows that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] itself. Or, if you prefer to define [itex]e^x[/itex] by the fact that its derivative is itself, you can start from that fact.

    Now, of course, the derivative of [itex]e^x[/itex] is given by
    [tex]\lim_{h\to 0}\frac{e^{x+h}- e^x}{h}= \lim_{h\to 0}\frac{e^xe^h- e^x}{h}= e^x\lim_{h\to 0}\frac{e^h- 1}{h}[/tex]

    And, since the derivative is [itex]e^x[/itex] that means we must have
    [tex]\lim_{h\to 0}\frac{e^h- 1}{h}= 1[/tex].

    That, now, means that for h very close to 0,
    [tex]\frac{e^h- 1}{h}\approx 1[/tex]
    so that
    [tex]e^h- 1\approx h[/tex]
    For any h> 0, there exist an integer n such that n> 1/h. With such an n, we have
    [tex]e^{1/n}- 1\approx \frac{1}{n}[/tex]
    so that
    [tex]e^{1/n}\approx 1+ \frac{1}{n}[/tex]
    [tex]e\approx \left(1+ \frac{1}{n}\right)^n[/tex]
    Now taking the limit as n goes to infinity is the same as taking the limit as h goes to 0 so the "approximation" becomes exact:
    [tex]e= \lim_{n\to\infty} \left(1+ \frac{1}{n}\right)^n[/tex]
     
  10. Feb 13, 2011 #9

    arildno

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    Halls:
    You are wrong.
    There is a - in the parenthesis, not a +.
     
  11. Feb 13, 2011 #10
    Unless i am serioysly deceived the thread is 2 years old btw ... And this seems to be an excersice for first year or indroductory course to analysis, so couldnt possible have to do with continuity, derivatives and Lhopital rule. Post #3 also posted 2 years ago seem to do the proper job in answering.
     
  12. Feb 13, 2011 #11

    dextercioby

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    Details, details :))

    Anyways, Halls' post is very good.
     
  13. Feb 14, 2011 #12

    HallsofIvy

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    Except for that minor detail with the negative!

    Ah, well, the operation was succesful even though the patient died.
     
  14. Feb 14, 2011 #13

    arildno

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    :rofl:
     
  15. Feb 14, 2011 #14
    and now that i see again i am seriously deceived cause the thread is almost 3 years old...
     
  16. Feb 14, 2011 #15

    Char. Limit

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    Hmm... it gave me a thought. If

    [tex]\left(1+\frac{1}{n}\right)^n = e[/tex]

    then what is

    [tex]\left(1+\frac{1}{-n}\right)^{-n}[/tex]

    ?

    As n approaches positive infinity, of course.
     
  17. Feb 15, 2011 #16
    Simplify and it becomes clear: [tex]lim_{n \to \infty} \frac{1}{(1 - \frac{1}{n})^n} = \frac{1}{\frac{1}{e}} = e[/tex]. You can do this because the negative n in the fraction turns it into a negative, the negative n in the exponent turns it into division, and the limit of a fraction is equal to the fraction of the numerator's and denominator's limit, when the denominator isn't 0.
     
  18. Feb 15, 2011 #17

    Char. Limit

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    But wolfram alpha tells me the limit should be e.
     
  19. Feb 15, 2011 #18
    See it again. I made a dumb mistake, then edited.
     
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