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Thread: limit

  1. #1
    Super Member dhiab's Avatar
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    limit

    limit-43190290_552910198495284_2647754021229559808_n.jpg
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    Re: limit

    Quote Originally Posted by dhiab View Post
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    See here.
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    Re: limit

    Quote Originally Posted by dhiab View Post
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    Also, dhiab, I see that you're a Super Member. Maybe this is at least partially a challenge. Can you address that?


    Let $\displaystyle \ \ ln(X) \ $ mean $\displaystyle \ \ log_e(X)$.

    Let y = $\displaystyle \displaystyle\lim_{n \to \infty}(5n)^{1/n}$

    ln(y) = ln[$\displaystyle \displaystyle\lim_{n \to \infty}(5n)^{1/n}]$

    ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}ln[(5n)^{1/n}]$

    ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\bigg(\dfrac{1}{n}\bigg)ln(5n)$

    ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{ln(5n)}{n}$


    I'll use L'Hopital's Rule:


    ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{\bigg(\dfrac{1}{n}\bigg)}{1}$

    ln(y) = 0

    $\displaystyle e^{ln(y)} = e^0$

    y = 1


    The limit is 1.
    Last edited by greg1313; Oct 8th 2018 at 06:59 AM.
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  4. #4
    Super Member dhiab's Avatar
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    Re: limit

    Hello i'have this solution :


    lim(n--> + infini) (1/n)Ln(5n) = 0 .

    lim(n--> + infini) racine_nième(5n)

    = lim(n--> + infini) (5n)^(1/n)

    = lim(n--> + infini) exp(Ln((5n)^(1/n)))

    = lim(n--> + infini) exp((1/n)Ln(5n)) = 1 .
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    Re: limit

    Quote Originally Posted by dhiab View Post
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    Define $x_n=\sqrt[n]n-1$
    If $n\ge 2$ then
    $ \begin{align*}n&=(1+x_n)^n \\n&\ge\dfrac{n(n+1)}{2}(x_n)^2\\\text{or }(x_n)^2&\le\dfrac{2}{n+1}\\\text{as }n\to\infty\text{ then }(x_n)&\to 0\\\therefore \sqrt[n]n-1&\to 0\\\therefore \sqrt[n]n&\to 1 \end{align*}$
    Thanks from SlipEternal
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    Re: limit

    Quote Originally Posted by greg1313 View Post
    I'll use L'Hopital's Rule:

    ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{\bigg(\dfrac{1}{n}\bigg)}{1}$
    I don't think much of your differentiation.
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    Re: limit

    Quote Originally Posted by Archie View Post
    I don't think much of your differentiation.
    You don't think much what of my differentiation!? You made
    a negative comment without a specific reason attached
    to it. So, you're trolling, and you wasted a post. Learn to make
    constructive comments.
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    Re: limit

    Keep your hair on. It was a joke - that rather backfired as my differentiation wasn't up to much, now I look at it again.
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    Re: limit

    Quote Originally Posted by Archie View Post
    Keep your hair on. It was a joke - that rather backfired as my differentiation wasn't up to much, now I look at it again.
    You're an expert on all sorts of differentiation and integration techniques on multiple
    message board sites.
    Thank you.
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    Re: limit

    Quote Originally Posted by greg1313 View Post
    You don't think much what of my differentiation!? You made
    a negative comment without a specific reason attached to it. So, you're trolling, and you wasted a post.
    Learn to make constructive comments.
    TO: greg1313, O.K. I can understand your being insulted by that comment. But I will add that depending on the level of the course, I would not accepted a solution using differentiation, l'Hopital's Rule. In my analysis courses I always went for more general principles.
    This was one of my favorite question. SEE my post above: $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n} = 1$
    Having proved that, it an easy step to show that for $m\ge 0$ then $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{m} = 1$
    Of course $\sqrt[n]{5n}=\sqrt[n]{5}\sqrt[n]{n}$. Now that is a real teaching moment, a general concept. Not a one time off problem.
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    Re: limit

    Quote Originally Posted by Plato View Post
    Define $x_n=\sqrt[n]n-1$
    If $n\ge 2$ then
    $ \begin{align*} . . . \therefore \sqrt[n]n&\to 1 \end{align*}$
    This is a start/promising. And we are still left to prove it with the
    original radicand of 5n.
    And suppose the radicand were $\displaystyle n^2, \ n^3, \ \ or \ \ even \ \ n^{ \sqrt{n}}$.
    We would need to show that those limits are also 1, or not.
    Last edited by greg1313; Oct 9th 2018 at 07:13 PM.
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    Re: limit

    Quote Originally Posted by greg1313 View Post
    This is a start/promising. And we are still left to prove it with the
    original radicand of 5n.
    And suppose the radicand were $\displaystyle n^2, n^3, or even n^{ \sqrt{n}}$.
    We would need to show that those limits are also 1, or not.
    Surely this is not too much for you:
    $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m} = {1^m} = 1~?$
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    Re: limit

    Quote Originally Posted by Plato View Post
    Surely this is not too much for you:
    $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m} = {1^m} = 1~?$
    That is a trap. If m = n (for instance), it collapses to the nth root of (n raised
    to the n exponent). Then, it is the limit as n approaches infinity
    of just n, which equals infinity.
    Last edited by greg1313; Oct 9th 2018 at 07:28 PM.
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    Re: limit

    Quote Originally Posted by greg1313 View Post
    That is a trap. If m = n (for instance), it collapses to the nth root of (n raised
    to the n exponent). Then, it is the limit as n approaches infinity
    of just n, which equals infinity.
    You do not understand much about notation do you? In $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m}$ anyone claiming to help at this level surely realizes that $n$ is a variable but $m$ is constant. So given your limitation, please in the further consider if you are up to answering a question.
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    Re: limit

    Quote Originally Posted by Plato View Post
    You do not understand much about notation do you?
    anyone claiming to help at this level surely realizes that $n$ is a variable but $m$ is constant. So given your limitation, please in the further consider if you are up to answering a question.
    You do not understand much about not making presumptions, do you!? So, given your limitation in your lack of ability to make a valid discussion, please refrain from making them,
    because you cross into making uncivil posts. Your history of it is clear. Stop doing it, hurry up and understand that. Your post was reported. And you haven't been learning from
    from your bad attitude notifications.
    Last edited by greg1313; Oct 10th 2018 at 08:23 AM.
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