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Math Help - Help with limit

  1. #1
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    Help with limit

    I need help finding the limit of

    e^(-n) * (1 + 1/n)^(n^2)
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  2. #2
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    n is natural, and n->+inf.
    The problem is you can't separate the expression and calculate the product of both limits...
    Last edited by gunt; October 28th 2006 at 10:19 AM. Reason: Was a reply to user Aglaia
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  3. #3
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    Hello, gunt!

    I assume that n \to \infty


    \lim_{n\to\infty}\left[e^{-n}\cdot\left(1 + \frac{1}{n}\right)^{n^2}\right]

    We have: . \frac{\left(1 + \frac{1}{n}\right)^{n^2}}{e^n} \;= \;\frac{\left(1 + \frac{1}{n}\right)^{n\cdot n}}{e^n} \;=\;\frac{\left[\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n}

    Then: . \lim_{n\to\infty} \frac{\left[\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;= \;\frac{\left[\lim_{_{n\to\infty}}\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;=\;\frac{e^n}{e^n} \;=\;1

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  4. #4
    TD!
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    Quote Originally Posted by Soroban View Post
    Then: . \lim_{n\to\infty} \frac{\left[\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;= \;\frac{\left[\lim_{_{n\to\infty}}\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;=\;\frac{e^n}{e^n} \;=\;1
    This can't be correct:
    - the denominator depends on 'n' as well so you can't just shift the limit to the numerator only.
    - in the numerator, you ignored the top exponent n, applied the standard limit and then just put the exponent back.

    \begin{array}{*{20}l}<br />
   {\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)^{n^2 }  = \mathop {\lim }\limits_{n \to \infty } \left( {\left( {1 + \frac{1}{n}} \right)^n } \right)^n } & { \ne \left( {\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n } \right)^n  = e^n }  \\<br />
   {} & { = \mathop {\lim }\limits_{n \to \infty } \left( e \right)^n  =  + \infty }  \\<br />
\end{array}<br />
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  5. #5
    Forum Admin topsquark's Avatar
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    Hmmmmm....

    That's the same "error" that TD pointed out (if error it is). And it appears I lied... The expression with n = 1000 (as far as my calculator will go) appears to be decreasing but fairly flat (first derivative at n = 1000 is -2 x 10^(-7)) and only at 0.6 or so.

    I can't say why TPH and Soroban's methods would be wrong, but I don't think the answer is 1.

    -Dan
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  6. #6
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    According to Maple,

    lim(as x approaches infinity) [((1+(1/x))^x)^x]/(e^x) = 1/sqrt(e)

    Not 1.
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  7. #7
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    Quote Originally Posted by AfterShock View Post
    According to Maple,

    lim(as x approaches infinity) [((1+(1/x))^x)^x]/(e^x) = 1/sqrt(e)

    Not 1.
    You seem to be right, I graphed it too.

    However, I am confused why my results produced the incorrect answer. It is certainly that the limit composition rule failed, for some reason. Perhaps, when I said it works for \pm \infty was wrong, I know that it works for numbers, I just always assumed it works for infinite limits also. Apparently it seems by assumption was wrong.

    Though I can prepared to be wrong, and think I am. What you shown was not a proof . It needs to be formal.

    EDIT. I realized my mistake.
    What I said about composition functions holds true. But I have not properly expressed the outer function, I assumed it was raised to the x, but that does not succesfully express the outer function.
    Last edited by ThePerfectHacker; October 28th 2006 at 05:03 PM.
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  8. #8
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    I tried a few algebric manipulations and got stuck on each one of them.

    It's giving me more trouble than I expected. It looked so innocent...
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  9. #9
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    Quote Originally Posted by AfterShock View Post
    According to Maple, lim(as x approaches infinity) [((1+(1/x))^x)^x]/(e^x) = 1/sqrt(e) Not 1.
    MathCad also gave that as the answer.
    But try as I have, I cannot justify it.
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  10. #10
    TD!
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    Quote Originally Posted by topsquark View Post
    That's the same "error" that TD pointed out (if error it is). And it appears I lied... The expression with n = 1000 (as far as my calculator will go) appears to be decreasing but fairly flat (first derivative at n = 1000 is -2 x 10^(-7)) and only at 0.6 or so.

    I can't say why TPH and Soroban's methods would be wrong, but I don't think the answer is 1.
    I understand that this problem was posted and answered somewhere else too? I don't know TPH's solution, but I think I've explained why Soroban's isn't correct. The answer definitely isn't 1, but exp(-1/2).
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD! View Post
    I understand that this problem was posted and answered somewhere else too? I don't know TPH's solution, but I think I've explained why Soroban's isn't correct. The answer definitely isn't 1, but exp(-1/2).
    Did I do this right?

    \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) }

    By a Taylor expansion (since 1/n is so small when we take the limit) we get that:
    ln \left ( 1 + \frac{1}{n} \right ) = \frac{1}{n} - \frac{1}{2n^2} +...

    So:
    \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) } = e^{n^2 \cdot \left ( \frac{1}{n} - \frac{1}{2n^2} +... \right ) } = e^{ \left ( n - \frac{1}{2} +... \right ) }

    So for the problem:
    \lim_{n \to \infty}e^{-n} \left ( 1 + \frac{1}{n} \right )^{n^2} = \lim_{n \to \infty}e^{-n} e^{ \left ( n - \frac{1}{2} +... \right ) } = \lim_{n \to \infty}1 \cdot e^{-1/2} = e^{-1/2}

    -Dan
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  12. #12
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    Great! Thanks for your help.
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  13. #13
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    Quote Originally Posted by topsquark View Post
    Did I do this right?

    \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) }

    By a Taylor expansion (since 1/n is so small when we take the limit) we get that:
    ln \left ( 1 + \frac{1}{n} \right ) = \frac{1}{n} - \frac{1}{2n^2} +...

    So:
    \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) } = e^{n^2 \cdot \left ( \frac{1}{n} - \frac{1}{2n^2} +... \right ) } = e^{ \left ( n - \frac{1}{2} +... \right ) }

    So for the problem:
    \lim_{n \to \infty}e^{-n} \left ( 1 + \frac{1}{n} \right )^{n^2} = \lim_{n \to \infty}e^{-n} e^{ \left ( n - \frac{1}{2} +... \right ) } = \lim_{n \to \infty}1 \cdot e^{-1/2} = e^{-1/2}

    -Dan
    I was also thinking about using a series expansion on this one, but I think you made it as simple as possible. Good job.

    It looks correct to me. Again as I said the limit composition rule is used here.

    The only think I am concerned about is whether a distribution of the limit to each individual term is allowed in an infinite series. I think yes, I believe you can always do it. Or whether the series need to be absolutely convergent, but I think you are correct about this one.
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  14. #14
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    Quote Originally Posted by ThePerfectHacker View Post
    I was also thinking about using a series expansion on this one, but I think you made it as simple as possible. Good job.

    It looks correct to me. Again as I said the limit composition rule is used here.

    The only think I am concerned about is whether a distribution of the limit to each individual term is allowed in an infinite series. I think yes, I believe you can always do it. Or whether the series need to be absolutely convergent, but I think you are correct about this one.
    I was wondering about the same point myself. Unfortunately (and you can blame this on my being a Physicist) I have a fairly poor knowledge of how to work with infinite series, so I can't even begin to figure out how to approach that question.

    -Dan
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  15. #15
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    Quote Originally Posted by topsquark View Post
    I was wondering about the same point myself. Unfortunately (and you can blame this on my being a Physicist) I have a fairly poor knowledge of how to work with infinite series, so I can't even begin to figure out how to approach that question.
    Not to offend you but I realized that in my engineering class. The professor just manipulates that stuff and differencials in an way he wants to.

    It some ways I agree with that approach. I think I am the only mathemation in my differencial equations class. I have made some remarks about the professors' solution after class, he told me he would be concered with it if he was in a roomful of mathemations but engineers do not need to know that stuff. In fact, If I want to for example, differencial equations on a serious level I would first learn how to solve them informally and incorrectly. Then I would learn the theory, so that I not have to memorize techniques to how to solve them, that stuff I would already know and only be concerned with about the theory.
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