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Math Help - Limits that equal e

  1. #1
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    Limits that equal e

    Ah, yes, this seems like the most famous mathematical term, I have heard before. I was looking at a function whose limit was e as it went to infinity, so I decided to read up about this sort of thing, and found out how amazing e really is, went on a few hour rant reading various things about it. never really clicked before . Now my question is, if you given a limit say \lim_{n\to\infinty}f(n) and it contains some sort of ratio/fraction where n is in the denominator and involves something to the nth power, this is an observation I made, which, of course in math, can get you somewhere very wrong, the observation that if the function contains something along the lines of what I just suggested it will go to a limit equivalent to e?
    as I was suggested this example earlier by redsox apart from the traditional "definition" of e

    \lim_{n\to\infty}f(n)=n^{\frac{1}{ln(n)}}
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  2. #2
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    Think of it this way.

    L=\lim_{n\to \infty}n^{\frac{1}{ln(n)}}

    ln of both sides:

    ln(L)=\lim_{n\to \infty}\frac{1}{ln(n)}\cdot ln(n)

    ln(L)=1

    Therefore, L=e
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  3. #3
    MHF Contributor chisigma's Avatar
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    According to the basic property of exponential...

    a^{b} = e^{b\cdot \ln a} (1)

    ... is...

    n^{\frac{1}{\ln n}} = e^{\frac{\ln n}{\ln n}} = e ,  \forall n (2)

    ... so that it is not necessary that n \rightarrow \infty ...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    According to the basic property of exponential...

    a^{b} = e^{b\cdot \ln a} (1)

    ... is...

    n^{\frac{1}{\ln n}} = e^{\frac{\ln n}{\ln n}} = e ,  \forall n (2)

    ... so that it is not necessary that n \rightarrow \infty ...

    Kind regards

    \chi \sigma
    So this would work for this limit as well.

    \lim_{n\to\infty}n^{\frac{1}{\sqrt{n}}}

    by the property of the exponentional you stated

    n^{\frac{1}{\sqrt{n}}} = e^{\frac{ln\frac{1}{\sqrt{n}}}{ln\frac{1}{\sqrt{n}  }}}
    would that still hold true? forgot the power of n, but you catch my drift
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  5. #5
    MHF Contributor Drexel28's Avatar
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    n^{\frac{1}{\sqrt{n}}}\to1. Informally \sqrt{n} is much too "strong". It will completely overpower the base. I mean there are many more limits that give e. One that is not particularly interesting but comes to mind is \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    n^{\frac{1}{\sqrt{n}}}\to1. Informally \sqrt{n} is much too "strong". It will completely overpower the base. I mean there are many more limits that give e. One that is not particularly interesting but comes to mind is \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e.
    Yes, would the function I just suggested go to infinity? after I had time to work it out I got e^{\frac{ln(n)}{\sqrt{n}}}
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