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Math Help - Why does this limit equal e?

  1. #1
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    Why does this limit equal e?

    In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

    (1+1/n)^n as n->00 (infinity)

    and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!
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  2. #2
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    Because that's a definition of e.
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  3. #3
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    Quote Originally Posted by ilovecalculus View Post
    In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

    (1+1/n)^n as n->00 (infinity)

    and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!
    there may be a reason he didn't show you why. hope you can follow this ...

    let y = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

    also note that 1 + \frac{1}{n} = \frac{n+1}{n} ...

    \ln{y} = \lim_{n \to \infty} \ln\left(\frac{n+1}{n}\right)^n

    \ln{y} = \lim_{n \to \infty} n \ln\left(\frac{n+1}{n}\right)

    \ln{y} = \lim_{n \to \infty} \frac{\ln(n+1) - \ln{n}}{\frac{1}{n}}

    use L'Hopital's rule ...

    \ln{y} = \lim_{n \to \infty} \frac{\frac{1}{n+1} - \frac{1}{n}}{-\frac{1}{n^2}}

    \ln{y} = \lim_{n \to \infty} \frac{-\frac{1}{n(n+1)}}{-\frac{1}{n^2}}

    \ln{y} = \lim_{n \to \infty} \frac{n}{n+1}

    \ln{y} = 1

    y = e

    to check it yourself, graph the function y = \left(1 + \frac{1}{x}\right)^x in your calculator and see what value the function approaches as x gets large.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ilovecalculus View Post
    In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

    (1+1/n)^n as n->00 (infinity)

    and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!
    It can be proved using L'Hopital's rule. If you set L=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n, apply natural logs to both sides to get

    \ln L=\lim_{n\to\infty}\frac{\ln\!\left(1+\frac{1}{n}\  right)}{\frac{1}{n}}.

    It turns out that \ln L=1\implies L=e^1=e.

    So we can say that \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e.

    Does this clarify things?
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  5. #5
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    Thanks to everyone! It makes a lot more sense now, and I guess it's just something I'll have to remember for future reference!
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