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Limit is equivalent

  1. Dec 22, 2004 #1


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    I don't understand how to show that

    [tex]\lim_{n \rightarrow \infty} \left(1-\frac{a}{n} \left)^{n} = e^{-a} \ \ \forall a \in \mathbb{R}[/tex]

    For exemple, if I say "Let x be the real number such that [itex]n=-ax \Leftrightarrow x=-n/a[/itex]. Then the limit is equivalent to

    [tex]\lim_{-ax \rightarrow \infty} \left(1+\frac{1}{x} \right)^{-ax} = \left(\lim_{-ax \rightarrow \infty} \left(1+\frac{1}{x} \right)^{x} \right)^{-a}[/tex]

    "but [itex]-ax \rightarrow \infty[/itex] is not equivalent to [itex]x \rightarrow \infty[/itex], so I can't conclude that the limit is [itex]e^{-a}[/itex].

    What am I missing here ? :confused:
  2. jcsd
  3. Dec 22, 2004 #2


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    a is constant. true, x will either be going to negative or positive infinity depending on the sign of a, but the definition of e works for either:

    let: [tex]u = -x[/tex]

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

    [tex]= \lim_{-u \rightarrow \infty} (1+\frac{1}{-u} )^{-u} [/tex]

    [tex]= \lim_{u \rightarrow -\infty} (\frac{1}{(1-\frac{1}{u})})^{u} [/tex]

    [tex]= \lim_{u \rightarrow -\infty} (\frac{1+\frac{1}{u}}{(1-\frac{1}{u^2})})^{u} [/tex]

    and the 1/u2 term becomes negligible, giving the result:

    [tex]e = \lim_{u \rightarrow -\infty} (1+\frac{1}{u} )^{u} [/tex]

    edit: that may not be rigorous enough. you can show the bottom of the fraction above goes to 1 by taking the ln and using l'hopitals. in fact, you might want to just do that from the start.
    Last edited: Dec 22, 2004
  4. Dec 22, 2004 #3


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    [tex]\lim_{n \to -\infty}\left(1+\frac{1}{n}\right)^n=e[/tex] is true too.
  5. Dec 22, 2004 #4
    Shouldn't -ax be an integer anyway?

    Ok, the limit of (1+1/n)^n is e when n ->oo and n is an integer

    but what happens when we get the value
    where x is a very large real but not an integer ??

    Shouln't we prove this cases ?
  6. Dec 22, 2004 #5

    matt grime

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    The identity e = lim(1+1/x)^x is true whether x is an integer or not (as you implicitly use yourself).
  7. Dec 23, 2004 #6


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    Elegance is a quality of mathematics:
    [tex]\lim_{n\rightarrow +\infty}(1-\frac{a}{n})^{n}=[\lim_{n\rightarrow +\infty}(1+\frac{1}{\frac{n}{-a}})^{\frac{n}{-a}}]^{-a}=[\lim_{\frac{n}{-a}\rightarrow\pm\infty}(1+\frac{1}{\frac{n}{-a}})^{\frac{n}{-a}}]^{-a}=e^{-a} [/tex]
    ,where i made use of:
    [tex] \lim_{n\rightarrow\pm\infty}(1+\frac{1}{n})^{n}=e [/tex]


    PS.The sign of "a" is irrelevant.It's important for it not to be "0".
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